Ma101 Notes by Thamban Nair

Calculus - I

(MA 101)

M.Thamban Nair

Department of MathematicsIndian Institute of Technology

December 2001

Contents

Preface 6

1 Sequence and Series of Real Numbers 1

1.1 Sequence of Real Numbers . . . . . . . . . . . . . . . . 1

1.1.1 Convergence and Divergence . . . . . . . . . . 1

1.1.2 Monotonic Sequences . . . . . . . . . . . . . . 6

1.1.3 Subsequence . . . . . . . . . . . . . . . . . . . 7

1.1.4 Further Examples . . . . . . . . . . . . . . . . 8

1.1.5 Cauchy sequence . . . . . . . . . . . . . . . . . 12

1.1.6 Additional Exercises . . . . . . . . . . . . . . . 14

1.2 Series of Real Numbers . . . . . . . . . . . . . . . . . . 15

1.2.1 Convergence and Divergence of Series . . . . . 15

1.2.2 Some Tests for Convergence . . . . . . . . . . . 18

1.2.3 Alternating series . . . . . . . . . . . . . . . . . 22

1.2.4 Absolute convergence . . . . . . . . . . . . . . 23

2 Definite Integral 26

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2 Upper and Lower Sums . . . . . . . . . . . . . . . . . 28

2.3 Integrability and integral . . . . . . . . . . . . . . . . 30

2

Contents 3

2.3.1 Riemann sum . . . . . . . . . . . . . . . . . . . 34

2.4 Integral of Continuous Functions . . . . . . . . . . . . 35

2.5 Some properties . . . . . . . . . . . . . . . . . . . . . . 39

2.6 Some results . . . . . . . . . . . . . . . . . . . . . . . . 41

2.6.1 Indefinite Integral . . . . . . . . . . . . . . . . 41

2.6.2 Fundamental Theorem of Integral Calculus . . 43

2.6.3 Applications of Fundamental Theorem . . . . . 44

3 Improper Integrals 47

3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Geometric and Mechanical Applications of Integrals 56

4.1 Computing Area . . . . . . . . . . . . . . . . . . . . . 56

4.1.1 Using Cartesian Coordinates . . . . . . . . . . 56

4.1.2 Using Polar Coordinates . . . . . . . . . . . . . 57

4.1.3 Examples . . . . . . . . . . . . . . . . . . . . . 57

4.2 Computing Arc Length . . . . . . . . . . . . . . . . . 58

4.2.1 Using Cartesian Coordinates . . . . . . . . . . 58

4.2.2 Using Polar Coordinates . . . . . . . . . . . . . 59

4.2.3 Examples . . . . . . . . . . . . . . . . . . . . . 60

4.3 Computing Volume of a Solid . . . . . . . . . . . . . . 62

4.4 Computing Volume of a Solid of Revolution . . . . . . 63

4.5 Computing Area of Surface of Revolution . . . . . . . 64

4.6 Centre of Gravity . . . . . . . . . . . . . . . . . . . . . 64

4.6.1 Centre of gravity of a material line in the plane 65

4 Contents

4.6.2 Centre of gravity of a material planar region . 66

4.7 Moment of Inertia . . . . . . . . . . . . . . . . . . . . 68

4.7.1 Moment of inertia of a material line in the plane 68

4.7.2 Moment of inertia of a circular arc with respectto the centre . . . . . . . . . . . . . . . . . . . 69

4.7.3 Moment of inertia of a material sector in theplane . . . . . . . . . . . . . . . . . . . . . . . 70

4.8 Additional Exercises . . . . . . . . . . . . . . . . . . . 71

5 Sequence and Series of Functions 74

5.1 Sequence of Functions . . . . . . . . . . . . . . . . . . 74

5.1.1 Pointwise Convergence and Uniform Conver-gence . . . . . . . . . . . . . . . . . . . . . . . 74

5.2 Series of Functions . . . . . . . . . . . . . . . . . . . . 80

5.2.1 Examples . . . . . . . . . . . . . . . . . . . . . 82

6 Power Series 85

6.1 Convergence and Absolute convergence . . . . . . . . . 85

6.2 Integration and Differentiation . . . . . . . . . . . . . 88

6.3 Series that can be converted into a power series . . . . 90

7 Fourier Series 91

7.1 Fourier Series of 2π-Periodic functions . . . . . . . . . 91

7.1.1 Fourier Series and Fourier Coefficients . . . . . 91

7.1.2 Even and Odd Expansions . . . . . . . . . . . . 94

7.1.3 Examples . . . . . . . . . . . . . . . . . . . . . 95

7.2 Fourier Series of 2`-Periodic functions . . . . . . . . . 97

7.2.1 Fourier series of Functions on Arbitrary intervals 98

7.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . 99

Contents 5

8 Functions of Several Variables 101

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 101

8.2 Limit and Continuity . . . . . . . . . . . . . . . . . . . 102

8.2.1 Some Topological Notions . . . . . . . . . . . . 105

8.2.2 Two Theorems . . . . . . . . . . . . . . . . . . 106

8.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . 107

8.3.1 Partial Increments and Total Increment . . . . 108

8.3.2 Total Differential, Gradient, Differentiability . 110

8.3.3 Derivatives of Composition of Functions . . . . 111

8.4 Derivatives of Implicitly Defined Functions . . . . . . 113

8.4.1 Level Curve and Level Surface . . . . . . . . . 114

8.5 Directional Derivatives . . . . . . . . . . . . . . . . . . 114

8.6 Taylor’s Formula . . . . . . . . . . . . . . . . . . . . . 116

8.7 Maxima and Minima . . . . . . . . . . . . . . . . . . . 117

8.7.1 Method of Lagrange Multipliers . . . . . . . . . 119

9 Appendix 121

Index 125

Preface

This is based on a course that the author gave to B.Tech. studentsof IIT Madras many times since 1996.

December 2001 M. T. Nair

6

1

Sequence and Series of RealNumbers

1.1 Sequence of Real Numbers

Definition 1.1 A real sequence is a function from the set N ofnatural numbers to the set R of real numbers. If f : N → R is asequence, and if an = f(n) for n ∈ N, then we write the sequence fas (an).

EXAMPLE 1.1 (i) (an) with an = 1 for all n ∈ N – a constantsequence with value 1 throughout.

(ii) (an) with an = n for all n ∈ N.

(iii) (an) with an = (−1)n for all n ∈ N – the sequence takesvalues 1 and −1 alternately.

(iv) (an) with an = 1/n for all n ∈ N.

1.1.1 Convergence and Divergence

A fundamental concept in mathematics is that of convergence. Weconsider convergence of sequences.

Definition 1.2 A real sequence (an) is said to converge to a realnumber a if for every ε > 0, there exists a natural number N (ingeneral depending on ε) such that

|an − a| < ε ∀n ≥ N.

1

2 Sequence and Series of Real Numbers

The number a is called the limit of the sequence (an). If (an)converges to a, then we write

limn→∞

an = a

oran → a as n→∞.

Note that|an − a| < ε ∀n ≥ N

if and only ifa− ε < an < a+ ε ∀n ≥ N.

Thus, limn→∞ an = a if and only if for every ε > 0, there existsN ∈ N such that

an ∈ (a− ε, a+ ε) ∀n ≥ N.

Thus convergence of (an) to x implies that for every ε > 0, an belongsto the open interval (a − ε, a + ε) for all n after some finite stage,and this finite stage may vary according as ε varies.

Exercise 1.1 Show that a sequence (an) converges to a if and onlyif for every open interval I containing x, there exits N ∈ N such thatan ∈ I for all n ≥ N .

EXAMPLE 1.2 (i) The sequences (1/n), ((−1)n/n), (1 − 1n) are

convergent with limit 0, 0, 1 respectively.

(ii) Every constant sequence is convergent.

Theorem 1.1 Limit of a convergent sequence is unique. That, is ifan → a and an → a′ as n→∞, then a = a′.

Proof. Suppose an → a and an → a′ as n → ∞, and supposethat a′ 6= a. Now, for ε > 0, suppose N1, N2 ∈ N be such that

an ∈ (a− ε, a+ ε) ∀n ≥ N1, an ∈ (a′ − ε, a′ + ε) ∀n ≥ N2.

In particular,

an ∈ (a− ε, a+ ε), an ∈ (a′− ε, a′+ ε) ∀n ≥ N := maxN1, N2.

If we take ε < |a − a′|/2, then we see that (a − ε, a + ε) and (a′ −ε, a′ + ε) are disjoint intervals. Thus the above observation leads toa contradiction.

Sequence of Real Numbers 3

Another way of showing a′ = a. : Note that

|a− a′| = |(a− an) + (an − a′)| ≤ |a− an|+ |an − a′|.

Now, for ε > 0, let N1, N2 ∈ N be such that

|a− an| < ε/2 for all n ≥ N1, |a′ − an| < ε/2 ∀n ≥ N2.

Then it follows that

|a− a′| ≤ |a− an|+ |an − a′| < ε ∀n ≥ N := maxN1, N2.

Since this is true for all ε > 0, it follows that a′ = a.

Exercise 1.2 Show that a sequence (an) converges to x if and onlyif for any k ∈ N, (an+k) converges to x.

The following theorem can be easily proved.

Theorem 1.2 Suppose an → a, bn → b as n → ∞. Then we havethe following :

(a) an + bn → a+ b as n→∞,

(b) For every real number c, can → c a as n→∞

(c) If an ≤ bn for all n ∈ N, then a ≤ b.

(d) If an ≤ cn ≤ bn for all n ∈ N, and if a = b, then cn → a asn→∞.

Exercise 1.3 Prove Theorem 1.2.

EXAMPLE 1.3 Consider the sequence (an) with an =(

1 +1n

)1/n

,

n ∈ N. Then limn→∞

an = 1. This is seen as follows: Observe that

1 ≤ an ≤ (1 + 1/n), and use part (d) of the above theorem.

Definition 1.3 A sequence which does not converge is called a di-vergent sequence.

Definition 1.4 If (an) is such that for every M > 0, there existsN ∈ N such that an > M for all n ≥ N , then we say that (an)diverges to +∞.


If (an) is such that for every M > 0, there exists N ∈ N suchthat an < −M for all n ≥ N , then we say that (an) diverges to −∞.

Definition 1.5 If (an) is such that for every anan+1 < 0 for everyn ∈ N, that is an changes sign alternately, then we say that (an) isan alternating sequence .

An alternating sequence converge or diverge. For example, (Ver-ify that) the sequence ((−1)n) diverges, whereas ((−1)n/n) convergesto 0.

Definition 1.6 A sequence (an) is said to be bounded above if thereexits a real number M such that an ≤ M for all n ∈ N; and thesequence (an) is said to be bounded below if there exits a real numberM ′ such that an ≥ M ′ for all n ∈ N. A sequence which is boundabove and bounded below is said to be a bounded sequence .

Exercise 1.4 Show that a sequence (an) is bounded if and only ifthere exists M > 0 such that |an| ≤M for all n ∈ N.

Theorem 1.3 Every convergent sequence is bounded. The conversein not true.

Proof. Suppose (an) converges to x. Then there exists N ∈ Nsuch that |an − x| ≤ 1 for all n ≥ N . Hence

|an| = |(an − a) + a| ≤ |an − a|+ |x| ≤ 1 + |a| ∀n ≥ N,

so that

|an| ≤ max1 + |a|, |a1|, |a2|, . . . , |aN−1| ∀n ∈ N.

To see that the converse of the theorem is not true, consider thesequence ((−1)n). It is a bounded sequence, but not convergent.

The above theorem can be used to show that certain sequence isnot convergent, as in the following example.

EXAMPLE 1.4 For n ∈ N, let

an = 1 +12

+13

+ . . .+1n.


Then (an) diverges: To see this, observe that

a2n = 1 +12

+13

+ . . .+12n

= 1 +12

+(

13

+14

)+(

15

+16

+17

+18

)+

. . .+(

12n − 1

+ . . .+12n

)≥ 1 +

n

2.

Hence, (an) is not a bounded sequence, so that it diverges.

Exercise 1.5 If (an) converges to a and a 6= 0, then show that thereexists N ∈ N such that |an| ≥ |a|/2 for all n ≥ N .

Using the Theorem 1.3, the following result can be deduced

Theorem 1.4 Suppose an → a and bn → b as n → ∞. Then wehave the following.

(i) anbn → ab as n→∞.

(ii) If bn 6= 0 for all n ∈ N and b 6= 0, then an/bn → a/b asn→∞.


Exercise 1.7 Suppose an → a and bn → b as n→∞. If b 6= 0, thenshow that there exists k ∈ N such that bn 6= 0 for all n ≥ k, andan+k/bn+k → a/b as n→∞.

The following theorem is useful for asserting convergence of se-quences.

Theorem 1.5 Suppose that f is a continuous function defined onan interval J and a ∈ J . If (an) is a sequence in J such that an → a,then f(an) → f(a).

Proof. Suppose (an) is a sequence in J such that an → a. Letε > 0 be given. Since f is continuous at a, there exists δ > 0 suchthat

x ∈ J, |x− a| < δ ⇒ |f(x)− f(a)| < ε. (∗)


Now, since (an) converges to a, there exists N ∈ N such that

n ≥ N ⇒ |an − a| < δ.

Hence, from (∗) above, |f(an)− f(a)| < ε for all n ≥ N .

Remark 1.1 The converse of Theorem 1.5 also holds, i.e., if for everysequence (an) in J which converges to a ∈ J , we have limn→∞ f(an) =f(a), then f is continuous at a.

1.1.2 Monotonic Sequences

We can infer the convergence or divergence of a sequence in certaincases by observing the way the terms of the sequence varies.

Definition 1.7 A sequence (an) is said to be monotonically in-creasing if an ≤ an+1 for all n ∈ N; and is said to be monotonicallydecreasing if an ≥ an+1 for all n ∈ N. In the above definition, if strictinequality occur, then we say that the sequence is strictly increasingand strictly decreasing, respectively.

In the proof of the next theorem we shall make use of an impor-tant property of the set R of real numbers:

Definition 1.8 A subset S of R is said to be bounded above if thereexists c ∈ R such that x ≤ c for all x ∈ S, and in that case c is calledan upper bound of S.

Similarly we can define lower bound for a subset of R.

Definition 1.9 A number b0 ∈ R is called a least upper bound orsupremum of S ⊆ R if for every a < b0, there exists x ∈ S such thatx > a.

A number a0 ∈ R is called a greatest lower bound or infimum ofS ⊆ R if for every b > a0, there exists x ∈ S such that x < b.

It can be seen that supremum (respectively, infimum) if exists isunique.

Least upper bound property: If S ⊆ R is bounded above, thenS has a least upper bound. We may write this least upper bound aslub(S) or sup(S).


Greatest lower bound property: If S ⊆ R is bounded below,then S has a greatest lower bound.

Theorem 1.6 Every sequence which is monotonically increasing andbounded above is convergent. Also, every sequence which is mono-tonically decreasing and bounded below is convergent.

Proof. Suppose (an) is a monotonically increasing sequence ofreal numbers which is bounded above. Then the set S := an : n ∈ Nis bounded above. Hence, by the least upper bound property of R,S has a least upper bound, say b. Now, let ε > 0 be give. Then, bythe definition of the least upper bound, there exists N ∈ N such thataN > b− ε. Since an ≥ aN for every n ≥ N , we get

b− ε < aN ≤ an ≤ b < b+ ε ∀n ≥ N.

Thus we have proved that an → b as n→∞.

To see the last part, suppose that (bn) is a monotonically de-creasing sequence which is bounded below. Then, it is seen that thesequence (an) defined by an = −bn for all n ∈ N is monotonically in-creasing and bounded above. Hence, by the first part of the theorem,an → a for some a ∈ R. Then, bn → b := −a.

Note that a convergent sequence need not be monotonically in-creasing or monotonically decreasing. For example, look at the se-quence ((−1)n/n).

1.1.3 Subsequence

Definition 1.10 A sequence (bn) is called a subsequence of a se-quence (an) if there is a strictly increasing sequence (kn) of naturalnumbers such that bn = akn for all n ∈ N.

Thus, subsequences of a real sequence (an) are of the form (akn),where (kn) is a strictly increasing sequence natural numbers.

For example, given a sequence (an), the sequences (a2n), (a2n+1),(an2), (a2n) are some of its subsequences. As concrete examples,(1/2n), and (1/(2n+ 1)), (1/2n) are subsequences of (1/n).


A sequence may not converge, but it can have convergent subse-quences. For example, we know that the sequence ((−1)n) diverges,but the subsequences (an) and (bn) defined by an = 1, bn = −1 forall n ∈ N are convergent subsequences of ((−1)n).

However, we have the following result.

Theorem 1.7 If a sequence (an) converges to x, then all its subse-quences converge to the same limit x.


What about the converse of the above theorem? Obviously, if allsubsequences of a sequence (an) converge to the same limit x, then(an) also has to converge to x, as (an) is a subsequence of itself.

Suppose every subsequence of (an) has at least one subsequencewhich converges to x. Does the sequence (an) converges to x? Theanswer is affirmative, as the following theorem shows. Its proof isleft as an exercise.

Theorem 1.8 If every subsequence of (an) has at least one subse-quence which converges to x, then (an) also converges to x.

1.1.4 Further Examples

EXAMPLE 1.5 Let a sequence (an) be defined as follows :

a1 = 1, an+1 =2an + 3

4, n = 1, 2, . . . .

We show that (an) is monotonically increasing and bounded above.

Note that

an+1 =2an + 3

4=an

2+

34≥ an ⇐⇒ an ≤

32.

Thus it is enough to show that an ≤ 3/2 for all n ∈ N.

Clearly, a1 ≤ 3/2. If an ≤ 3/2, then an+1 = an/2 + 3/4 <3/4 + 3/4 = 3/2. Thus, we have proved that an ≤ 3/2 for all n ∈ N.Hence, by Theorem 1.6, (an) converges. Let its limit be a. Thentaking limit on both sides of an+1 = 2an+3

4 we have

a =2a+ 3

4i.e., 4a = 2a+ 3 so that a =

32.


EXAMPLE 1.6 Let a sequence (an) be defined as follows :

a1 = 2, an+1 =12

(an +

2an

), n = 1, 2, . . . .

It is seen that, if the sequence converges, then its limit would be√

2.

Since a1 = 2, one may try to show that (an) is monotonicallydecreasing and bounded below.

Note that

an+1 :=12

(an +

2an

)≤ an ⇐⇒ a2

n ≥ 2, i.e., an ≥√

2.

Clearly a1 ≥√

2. Now,

an+1 :=12

(an +

2an

)≥√

2 ⇐⇒ a2n − 2

√2an + 2 ≥ 0,

which is always true. Thus, (an) is monotonically decreasing andbounded below, so that by Theorem 1.6, (an) converges.

EXAMPLE 1.7 Consider the sequence (an) with an = (1 + 1/n)n

for all n ∈ N. We show that (an) is monotonically increasing andbounded above. Hence, by Theorem 1.6, (an) converges.

Note that

an =(

1 +1n

)n

= 1 + n.1n

+n(n− 1)

2!1n2

+ · · ·+ n(n− 1) . . . 2.1n!

1nn

= 1 + 1 +12!

(1− 1

n

)+

13!

(1− 1

n

)(1− 2

n

)+

· · ·+ 1n!

(1− 1

n

)· · ·(

1− n− 1n

)≤ an+1.

Also

an = 1 + 1 +12!n(n− 1)

n2+ · · ·+ 1

n!n(n− 1) . . . 2.1

nn

≤ 1 + 1 +12!

+13!

+ · · ·+ 1n!

≤ 1 + 1 +12

+122

+ · · ·+ 12n−1

< 3.


In the four examples that follow, we shall be making use of The-orem 1.2 without mentioning it explicitly.

EXAMPLE 1.8 Consider the sequence (bn) with

bn = 1 +11!

+12!

+13!

+ · · ·+ 1n!

∀n ∈ N.

Clearly, (bn) is monotonically increasing, and we have noticed in thelast example that it is bounded above by 3. Hence, by Theorem 1.6,it converges.

Let (an) and (bn) be as in Examples 1.7 and 1.8 respectively, andlet a and b their limits. We show that a = b.

We have observed in last example that 2 ≤ an ≤ bn ≤ 3. Hence,taking limits, it follows that a ≤ b. Notice that

an = 1 + 1 +12!

(1− 1

n

)+

13!

(1− 1

n

)(1− 2

n

)+

. . .+1n!

(1− 1

n

). . .

(1− n− 1

n

).

Hence, for m,n with m ≤ n, we have

an ≥ 1 + 1 +12!

(1− 1

n

)+

13!

(1− 1

n

)(1− 2

n

)+

· · ·+ 1m!

(1− 1

n

). . .

(1− m− 1

n

).

Taking limit as n→∞, we get (cf. Theorem 1.2 (c))

x ≥ 1 +11!

+12!

+13!

+ · · ·+ 1m!

= bm.

Now, taking limit as m → ∞, we get a ≥ b. Thus we have proveda = b.

The common limit in the above two examples is denoted by theletter e.

EXAMPLE 1.9 Let a > 0. We show that, if 0 < a < 1, then thesequence (an) converges to 0, and if a > 1, then (an) diverges toinfinity.


Suppose 0 < a < 1. Then we can write a = 1/(1 + r), r > 0, andwe have

an = 1/(1+r)n = 1/ (1 + nr + · · ·+ rn) < 1/(1+nr) → 0 as n→∞.

Hence, an → 0 as n→∞.

An alternate way: Let xn = an. Then (xn) is monotonically decreas-ing and bounded below by 0. Hence (xn) converges, to say x. Thenxn+1 = an+1 = axn → ax. Hence, x = ax. This shows that x = 0.

Next suppose a > 1. Then, since 0 < 1/a < 1, the sequence(1/an) converges to 0, so that (an) diverges to infinity. (Why ?)

EXAMPLE 1.10 The sequence (n1/n) converges to 1.

Note that n1/n = 1 + rn for some sequence (rn) of positive reals.Then we have

n = (1 + rn)n ≥ n(n− 1)2

r2n,

so that r2n ≤ 2/(n−1) for all n ≥ 2. From this it follows, by Theorem1.2(c), that rn → 0, and hence n1/n = 1 + rn → 1.

EXAMPLE 1.11 For any a > 0, (a1/n) converges to 1.

If a > 1, then we can write a1/n = 1 + rn for some sequence (rn)of positive reals. Then we have

a = (1 + rn)n ≥ nrn so that rn ≤ a/n→ 0,

and hence a1/n = 1 + rn → 1.

In case 0 < a < 1, then 1/a > 1, so that 1/a1/n = (1/a)1/n → 1.Hence, an → 1.

EXAMPLE 1.12 Let (an) be a bounded sequence of non-negativereal numbers. Then (1 + an)1/n → 1 as n→∞:

This is seen as follows: Let M > 0 be such that 0 ≤ an ≤ M forall n ∈ N. Then, 1 ≤ (1 + an)1/n ≤ (1 + M)1/n for all n ∈ N. ByExample 1.11, (1 +M)1/n → 0. Hence the result follows by makinguse of part (d) of Theorem 1.2.

EXAMPLE 1.13 As an application some of the results discussedabove, consider the sequence (an) with an = (1+1/n)1/n, n ∈ N. Wealready know that lim

n→∞an = 1. Now, two more proofs for the same.


(i) Note that 1 ≤ an ≤ 21/n, and 21/n → 1 as n→∞.

(ii) Also, note that an = eln(1+1/n)1/n= e

1n

ln(1+1/n). Since 1/n→0 and ln(1 + 1/n) → log 1 = 0, it follows that an = eln(1+1/n)1/n →e0 = 1. Here, we used the fact that the function f(x) := ex ln(1+x) iscontinuous at 0 and used Theorem 1.5.

EXAMPLE 1.14 Consider the sequence (an) with an = (1+n)1/n.Then an → 1 as n→∞. We give two proofs for this result.

(i) Observe that an = n1/n (1 + 1/n)1/n. We already know thatn1/n → 1, and (1 + 1/n)1/n → 1 as n→∞.

(ii) Observe that n1/n ≤ (1 + n)1/n ≤ (2n)1/n = 21/nn1/n, wheren1/n → 1 and 21/n → 1 as n→∞.

1.1.5 Cauchy sequence

Theorem 1.9 If a real sequence (an) converges, then for every ε > 0,there exists N ∈ N such that

|an − am| < ε ∀n,m ≥ N.

Proof. Suppose an → a as n→∞, and let ε > 0 be given. Thenwe know that there exists N ∈ N such that |an − a| < ε/2 for alln ≥ N . Hence, we have

|an − am| ≤ |an − a|+ |a− am < ε ∀n,m ≥ N.

This completes the proof.

If a sequence satisfies the conclusion of the above theorem thenwe say that it is a Cauchy sequence.

Thus what we have shown is that every convergent sequence is aCauchy sequence. In fact, the converse of this statement is also true:

Theorem 1.10 Every Cauchy sequence of real numbers converges.

The proof of the above theorem requires more ideas from math-ematical analysis, which are beyond the scope of this course.

Here is an example of a general nature.


EXAMPLE 1.15 Let (an) be a sequence of real numbers. Supposethere exists a positive real number ρ < 1 such that

|an+1 − an| ≤ ρ|an − an−1| ∀n ∈ N, n ≥ 2.

Then (an) is a cauchy sequence. To see this first we observe that

|an+1 − an| ≤ ρn−1|a2 − a1| ∀n ∈ N, n ≥ 2.

Hence, for n > m,

|an − am| ≤ |an − an−1|+ . . .+ |am+1 − am|≤ (ρn−2 + . . .+ ρm−1)|a2 − a1|≤ ρm−1(1 + ρ+ . . .+ ρn−m−3)|a2 − a1|

≤ ρm−1

1− ρ|a2 − a1|.

Since ρm−1 → 0 as m → ∞, given ε > 0, there exists N ∈ N suchthat |an − am| < ε for all n,m ≥ N .

Exercise 1.9 Given a, b ∈ R and 0 < ρ < 1, let (an) be a sequenceof real numbers defined by a1 = a, a2 = b and

an+1 = (1 + ρ)an − ρan−1 ∀n ∈ N, n ≥ 2.

Show that (an) is a Cauchy sequence. Also, show that (an) convergesto (b+ ρa)/(1− ρ).

Exercise 1.10 Suppose f is a continuous function defined on aninterval J . If there exists 0 < ρ < 1 such that

|f(x)− f(y)| ≤ ρ|x− y| ∀x, y ∈ J,

then the for any a ∈ J , the sequence (an) defined by

a1 = f(a), an+1 := f(an) ∀n ∈ N,

is a Cauchy sequence. Show also that the limit of the sequence (an)is independent of the choice of a.

Exercise 1.11 Let sn = 1 + 12 + . . .+ 1

n for n ∈ N. Then show that(sn) is not a Cauchy sequence.

Hint: For any n ∈ N, note that s2n − sn ≥ 12 .


1.1.6 Additional Exercises

1. Establish convergence or divergence of the following sequences:

(i)(

n

n+ 1

)(ii)

((−1)nn

n+ 1

)(iii)

(2n

3n2 + 1

),

(iv)(

2n2 + 33n2 + 1

).

2. Suppose (an) is a real sequence such that an → 0 as n → ∞.Show the following:

(i) The sequence (a2n) converges to 0.

(ii) If an > 0 for all n, then the sequence (1/an) diverges toinfinity.

3. If (an) converges and an ≥ x for all n ∈ N, then show thatx ≥ 0 and (

√an) converges to

√x.

4. Suppose (an) is a sequence of positive real numbers. Show thefollowing:

(i) If limn→∞ (an+1/an) < 1, then (an) converges to 0.

(ii) If limn→∞ (an+1/an) > 1, then (an) diverges.

5. Let a1 = 1, an+1 =√

2 + an for all n ∈ N. Show that (an)converges. Also, find its limit.

6. Prove that if (an) is a Cauchy sequence having a subsequencewhich converges to a, then (an) itself converges to a.

7. Suppose (an) is a sequence such that the subsequences (a2n−1)and (a2n) converge to the same limit, say a. Show that (an)also converges to a.

8. Show that, if a sequence (an) has the property that an+1−an →0 as n→∞, then (an) need not converge.

9. If 0 < a < 1, then show that the sequence (nan) converges to0.

Series of Real Numbers 15

10. If 0 < a < b and an = (an + bn)1/n for all n ∈ N, then showthat (an) converges to b.

Hint: Note that (an + bn)1/n = b(1 +(

ab

)n)1/n.

11. Let a1 = 1 and an+1 = 14(2an + 3) for all n ∈ N. Show that

(an) is monotonically increasing and bounded above. Find itslimit.

12. Let a1 = 1 and an+1 = an1+an

for all n ∈ N. Show that (an)converges. Find its limit.

13. If an =√n+ 1 −

√n for all n ∈ N,,then show that (an) and

(nan) converge. Find their limits.

14. If 0 < a < b and an+1 = (annb

n)1/2 and bn+1 = an+bn2 for all

n ∈ N with a1 = a, b1 = b, then show that (an) and (bn)converge to the same limit.

1.2 Series of Real Numbers

Definition 1.11 A series of real numbers is an expression of the

form∞∑

n=1

an where (an) is a sequence of real numbers. We may write

a series∞∑

n=1

an also as a1 + a2 + a3 + . . ..

The number an is called the n-th term of the series∑∞

n=1 an, and

the sequence (sn) defined by sn :=n∑

i=1

ai is called the n-th partial

sum of the series∑∞

n=1 an.

1.2.1 Convergence and Divergence of Series

Definition 1.12 A series∑∞

n=1 an is said to converge (to s ∈ R) ifthe sequence sn of partial sums of the series converge (to s ∈ R).

If∑∞

n=1 an converges to s, then we write∑∞

n=1 an = s.

A series which does not converge is called a divergent series.


A necessary condition

Theorem 1.11 If∑∞

n=1 an converges, then an → 0 as n → ∞.Converse does not hold.

Proof. Clearly, if sn is the n-th partial sum of the convergentseries

∑∞n=1 an, then

an = sn − sn−1 → 0 as n→∞.

To see that the converse does not hold it is enough to observethat the series

∑∞n=1 an with an = 1

n for all n ∈ N diverges whereasan → 0.

The proof of the following corollary is immediate from the abovetheorem.

Corollary 1.12 Suppose (an) is a sequence of positive terms suchthat an+1 > an for all n ∈ N. Then the series

∑∞n=1 an diverges.

The above theorem and corollary shows, for example, that theseries

∑∞n=1

nn+1 diverges.

EXAMPLE 1.16 We have seen that the sequence (sn) with sn =n∑

k=1

1k!

converges. Thus, the series∞∑

n=1

1n!

converges. Also, we have seen that

the sequence (σn) with σn =n∑

k=1

1k

diverges. Hence,∞∑

n=1

1n

diverges.

EXAMPLE 1.17 Consider the geometric series series∑∞

n=1 aqn−1,

where a, q ∈ R. Note that sn = a + aq + . . . + aqn−1 for n ∈ N.Clearly, if a = 0, then sn = 0 for all n ∈ N. Hence, assume thata 6= 0. Then we have

sn =

na if q = 1,a(1−qn)

1−q if q 6= 1.


Thus, if q = 1, then (sn) is not bounded; hence not convergent. Ifq = −1, then we have

sn =a if n odd,0 if n even.

Thus, (sn) diverges for q = −1 as well. Now, assume that |q| 6= 1. Inthis case, we have ∣∣∣∣sn −

a

1− q

∣∣∣∣ = |a||1− q|

|q|n.

This shows that, if |q| < 1, then (sn) converges to a1−q , and if |q| > 1,

then (sn) is not bounded, hence diverges.

Theorem 1.13 Suppose (an) and (bn) are sequences such that forsome k ∈ N, an = bn for all n ≥ k. Then

∑∞n=1 an converges if and

only if∑∞

n=1 bn converges.

Proof. Suppose sn and σn be the n-th partial sums of the se-ries

∑∞n=1 an and

∑∞n=1 bn respectively. Let α =

∑ki=1 ai and β =∑k

i=1 bi. Then we have

sn − α =n∑

i=k+1

ai =n∑

i=k+1

bi = σn − β ∀n ≥ k.

From this it follows that the sequence (sn) converges if and only if(σn) converges.

From the above theorem it follows if∑∞

n=1 bn is obtained from∑∞n=1 an by omitting or adding a finite number of terms, then

∑∞n=1 bn

converges if and only if∑∞

n=1 an converges.

The proof of the following theorem is left as an exercise.

Theorem 1.14 Suppose∑∞

n=1 an converges to s and∑∞

n=1 bn con-verges to σ. Then for every α, β ∈ R,

∑∞n=1(αan + β bn) converges

to α s+ β σ.


1.2.2 Some Tests for Convergence

Theorem 1.15 (Comparison test) Suppose (an) and (bn) are se-quences of non-negative terms, and an ≤ bn for all n ∈ N. Then,

(i)∞∑

n=1

bn converges =⇒∞∑

n=1

an converges,

(ii)∞∑

n=1

an diverges =⇒∞∑

n=1

bn diverges.

Proof. Suppose sn and σn be the n-th partial sums of the series∑∞n=1 an and

∑∞n=1 bn respectively. By the assumption, we get 0 ≤

sn ≤ σn for all n ∈ N, and both (sn) and (σn) are monotonicallyincreasing.

(i) Since (σn) converges, it is bounded. Let M > 0 be such thatσn ≤ M for all n ∈ N. Then we have sn ≤ M for all n ∈ N. Since(sn) are monotonically increasing, it follows that (sn) converges.

(ii) Proof of this part follows from (i) (How?).

Corollary 1.16 Suppose (an) and (bn) are sequences of positiveterms.

(a) Suppose ` := limn→∞

an

bnexists. Then we have the following:

(i) If ` > 0, then∞∑

n=1

bn converges ⇐⇒∞∑

n=1

an converges.

(ii) If ` = 0, then∞∑

n=1

bn converges ⇒∞∑

n=1

an converges.

(b) Suppose limn→∞

an

bn= ∞. Then

∞∑n=1

an converges ⇒∞∑

n=1

bn con-

verges.

Proof. (a) Suppose limn→∞

an

bn= ` = `.

(i) Let ` > 0. Then for any ε > 0 there exists n ∈ N such that


`− ε <an

bn< `+ ε for all n ≥ N . Equivalently, (`− ε)bn < an < (`+ ε)bn

for all n ≥ N . Had we taken ε = `/2, we would get`

2bn < an <

3`2bn

for all n ≥ N . Hence, the result follows by comparison test.

(ii) Suppose ` = 0. Then for ε > 0, there exists n ∈ N such that−ε < an

bn< ε for all n ≥ N . In particular, an < εbn for all n ≥ N .

Hence, we get the result by using comparison test.

(b) By assumption, there exists N ∈ N such thatan

bn≥ 1 for all

n ≥ N . Hence the result follows by comparison test.

EXAMPLE 1.18 We have already seen that the sequence (sn) with

sn =n∑

k=1

1k!

converges. Here is another proof for the same fact: Note

that1n!≤ 1

2n−1for all n ∈ N. Since

∞∑n=1

12n−1

converges, it follows

from the above theorem that∞∑

n=1

1n!

converges.

EXAMPLE 1.19 Since1√n≥ 1n

for all n ∈ N, and since the series∞∑

n=1

1n

diverges, it follows from the above theorem that the series

∞∑n=1

1√n

also diverges.

Theorem 1.17 (de’Alembert’s test) Suppose (an) is a sequenceof positive terms such that lim

n→∞

an+1

an= ` exists. Then we have the

following:

(i) If ` < 1, then the series∑∞

n=1 an converges.

(ii) If ` > 1, then the series∑∞

n=1 an diverges.


Proof. (i) Suppose ` < q < 1. Then there exists N ∈ N such thatan+1

an< q ∀n ≥ N.

In particular,

an+1 < q an < q2an−1 < . . . < qn a1,∀n ≥ N.

Since∑∞

n=1 qn converges, by comparison test,

∑∞n=1 an also con-

verges.

(ii) Since 1 < p < `. Then there exists N ∈ N such thatan+1

an> p > 1n ≥ N.

From this it follows that (an) does not converge to 0. Hence∑∞

n=1 an

diverges.

Theorem 1.18 (Cauchy’s test) Suppose (an) is a sequence of pos-itive terms such that limn→∞ an

1/n = ` exists. Then we have thefollowing:

(i) If ` < 1, then the series∑∞

n=1 an converges.

(ii) If ` > 1, then the series∑∞

n=1 an diverges.

Proof. (i) Suppose ` < q < 1. Then there exists N ∈ N such that

an1/n < q ∀n ≥ N.

Hence, an < qn for all n ≥ N . Since the∑∞

n=1 qn converges, by

comparison test,∑∞

n=1 an also converges.

(ii) Since 1 < p < `. Then there exists N ∈ N such that

an1/n > p > 1 ∀n ≥ N.

Hence, an ≥ 1 for all n ≥ N . Thus, (an) does not converge to 0.Hence,

∑∞n=1 an also diverges.

We remark that both d’Alembert’s test and Cauchy test are silentfor the case lim

n→∞

an+1

an= 1. But, for such case, we may be able infer

the convergence or divergence by some other means.


EXAMPLE 1.20 For every x ∈ R, the series∞∑

n=1

xn

n!converges:

Here, an =xn

n!. Hence

an+1

an=

x

n+ 1∀n ∈ N.

Hence, it follows that limn→∞

an+1

an= 0, so that by d’Alemberts test,

the series converges.

EXAMPLE 1.21 The series∞∑

n=1

(n

2n+ 1

)n

converges: Here

an1/n =

n

2n+ 1→ 1

2< 1.

Hence, by Cauchy’s test, the series converges.

EXAMPLE 1.22 Consider the series∞∑

n=1

1n(n+ 1)

. In this series,

we see that limn→∞

an+1

an= 1 = lim

n→∞an

1/n. However, the n-th partial

sum sn is given by

sn =n∑

k=1

1k(k + 1)

=n∑

k=1

(1k− 1k + 1

)= 1− 1

n+ 1.

Hence sn converges to 1.


n=1

1n2

. In this case, we see

that1

(n+ 1)2=

1n(n+ 1)

∀n ∈ N.

Since∞∑

n=1

1n(n+ 1)

converges, by comparison test, the given series

also converges.


n=1

1np

for p 6= 1. It is easily


seen from the graph of the function f(x) := 1/xp, x ∈ R, that

n∑k=2

1np

≤∫ n

1

1xp

dx =n1−p − 1

1− p= σn, say,

andn∑

k=1

1np

≥∫ n+1

1

1xp

dx =n1−p − 1

1− p= σn+1.

Note that, for p > 1, σn converges to 1/(p − 1), and for p < 1,σn is unbounded. Hence, we can conclude that the given seriesconverges if p > 1, and diverges if p ≤ 1.

1.2.3 Alternating series

Definition 1.13 A series of the form∑∞

n=1(−1)n+1un where unis a sequence of positive terms is called an alternating series.

Theorem 1.19 (Leibniz’s theorem) Suppose un is a sequence ofpositive terms such that un ≥ un+1 for all n ∈ N, and un → 0 asn→∞. Then the alternating series

∑∞n=1(−1)n+1un converges.

Proof. Let sn be the n-th partial sum of the alternating series∑∞n=1(−1)n+1un. We observe that

s2n+1 = s2n + u2n+1 ∀n ∈ N.

Since un → 0 as n → ∞, from the above relation it follows thats2n converges if and only if s2n+1 converges, and in that case,sn converges. We show that s2n converges. Note that

s2n = (u1 − u2) + (u3 − u4) + . . .+ (u2n−1 − u2n),

s2n = u1 − (u2 − u3)− . . . (u2n−2 − u2n−1)− u2n

for all n ∈ N. Hence, it follows that s2n is monotonically increasingand bounded above. Therefore s2n converges.

By the above theorem, it follows that the series∞∑

n=1

(−1)n+1

ncon-

verges.


Consider an alternating series∑∞

n=1(−1)n+1un, where un ≥ un+1

for all n ∈ N, and un → 0 as n→∞. Let sn be the n-th partial sumof the alternating series

∑∞n=1(−1)n+1un. By the above theorem,

s := limn→∞sn exists.

We have shown that s2n is an increasing sequence. Similarly,it can be shown that s2n−1 is a decreasing sequence. Note that

s2n = s2n−1 − u2n, s2n+1 = s2n + u2n+1 ∀n ∈ N.

From the above observations, it follows that for m ≥ n,

s2n−1 = s2n + u2n ≤ s2m + u2n, s2m+1 ≤ s2n+1 = s2n + u2n+1.

Hence, taking limit as m→∞, we have

s2n−1 ≤ s+ u2n, s ≤ s2n + u2n+1

for all n ∈ N, i.e.,

s2n−1 − s ≤ u2n, s− s2n ≤ u2n+1 ∀n ∈ N.

Since s2n−1 ≤ s, s ≤ s2n for all n ∈ N, we get

|s− sn| ≤ un+1 ∀n ∈ N.

1.2.4 Absolute convergence


n=1 an is said to converge absolutely, if∑∞n=1 |an| converges.

Theorem 1.20 Every absolutely convergent series converges.

Proof. Suppose∑∞

n=1 an is an absolutely convergent series. Letsn and σn be the n-th partial sums of the series

∑∞n=1 an and

∑∞n=1 |an|

respectively. Then, for n > m, we have

|sn − sm| =

∣∣∣∣∣∣n∑

j=m+1

an

∣∣∣∣∣∣ ≤n∑

j=m+1

|an| = |σn − σm|.

Since, σn converges, it is a Cauchy sequence. Hence, form theabove relation it follows that sn is also a Cauchy sequence. There-fore, by the Cauchy criterion, it converges.


Another proof without using Cauchy criterion. Suppose∑∞

n=1 an

is an absolutely convergent series. Let sn and σn be the n-th par-tial sums of the series

∑∞n=1 an and

∑∞n=1 |an| respectively. Then it

follows thatsn + σn = 2pn,

where pn is the sum of all positive terms from a1, . . . , an. Sinceσn converges, it is bounded, and since pn ≤ σn for all n ∈ N,the sequence pn is also bounded. Moreover, pn is monotonicallyincreasing. Hence pn converge as well. Thus, both sn, pnconverges. Now, since sn = 2pn−σn for all n ∈ N, the sequence snalso converges.


n=1 an is said to converge conditionallyif∑∞

n=1 an converges, but∑∞

n=1 |an|diverges.

EXAMPLE 1.25 The series∑∞

n=1(−1)n+1

n is conditionally conver-gent.


n=1(−1)n+1

n2 is absolutely convergent.


n=1(−1)n+1

n! is absolutely convergent.

EXAMPLE 1.28 For any α ∈ R, the series∑∞

n=1sin(nα)

n2 is abso-lutely convergent: Note that∣∣∣∣sin(nα)

n2

∣∣∣∣ ≤ 1n2

∀n ∈ N.

Since∑∞

n=11n2 converges, by comparison test,

∑∞n=1

∣∣∣ sin(nα)n2

∣∣∣ alsoconverges.

Here are two more results whose proofs are based on some ad-vanced topics in analysis


n=1 an is an absolutely convergent se-ries and (bn) is a sequence obtained by rearranging the terms of(an). Then

∑∞n=1 bn is also absolutely convergent, and

∑∞n=1 an =∑∞

n=1 bn.


n=1 an is a conditionally convergent se-ries. The for every α ∈ R, there exists a sequence (bn) whose termsare obtained by rearranging the terms of (an) such that

∑∞n=1 bn = α.


To illustrate the last theorem consider the conditionally conver-gent series

∑∞n=1

(−1)n+1

n . Consider the following rearrangement ofthis series:

1− 12− 1

4+

13− 1

6− 1

8+ · · ·+ 1

2k − 1− 1

4k − 2− 1

4k+ · · · .

Thus, if an = (−1)n+1

n for all n ∈ N, the rearranged series is∑∞

n=1 bn,where

b3k−2 =1

2k − 1, b3k−1 =

14k − 2

, b3k =14k

for k = 1, 2, . . .. Let sn and σn be the n-th partial sums of the series∑∞n=1 an and

∑∞n=1 bn respectively. Then we see that

σ3k = 1− 12− 1

4+

13− 1

6− 1

8+ · · ·+ 1

2k − 1− 1

4k − 2− 1

4k

=(

1− 12− 1

4

)+(

13− 1

6− 1

8

)+ · · ·+

(1

2k − 1− 1

4k − 2− 1

4k

)=

(12− 1

4

)+(

16− 1

8

)+ · · ·+

(1

4k − 2− 1

4k

)=

12

[(1− 1

2

)+(

13− 1

4

)+ · · ·+

(1

2k − 1− 1

2k

)]=

12s2k.

Also, we have

σ3k+1 = σ3k +1

2k + 1, σ3k+2 = σ3k +

12k + 1

− 14k + 2

.

We know that sn converge. Let limn→∞ sn = s. Since, an → 0 asn→∞, it then follows that

limk→∞

σ3k =s

2, lim

k→∞σ3k+1 =

s

2, lim

k→∞σ3k+2 =

s

2.

Hence, we can infer that σn → s/2 as n→∞.

2

Definite Integral

2.1 Introduction

In calculus we define integral of a function f : [a, b] → R between thelimits a and b, i.e., ∫ b

af(x)dx,

to be the number g(b) − g(a), where g : [a, b] → R is such that itsderivative of f . One immediate question one raises would be:

(i) Given any function f : [a, b] → R, does there exist a differen-tiable function g : [a, b] → R such that g′(x) = f(x)?

Obviously, it is not necessary to have such a function g.

Another point one recalls is that if g : [a, b] → R is a differentiablefunction, then g′(x) has a geometric meaning, namely, it representsthe slope of the tangent to the graph of g at the point x.

(ii) Do we have a geometric meaning to the integral∫ ba f(t)dt?

We answer both the above questions affirmatively for a certainclass of functions f by giving a geometric definition of the conceptof integral.

Suppose f is a bounded (real valued) function defined on a closedinterval [a, b], i.e., f : [a, b] → R, and there exist m > 0, M > 0 suchthat m ≤ f(x) ≤ M for all x ∈ [a, b]. Our attempt is to associate anumber γ to such a function such that, in case f(x) ≥ 0 for x ∈ [a, b],then γ is the area of the region Rf bounded by the graph of f , x-axis,and the ordinates at a and b. We may not succeed to do this for allbounded functions f .

26

Introduction 27

Suppose, for a moment, f(x) ≥ 0 for all x ∈ [a, b]. Let us agreethat we have some idea about what the area under the graph of f .Then it is clear that

m(b− a) ≤ area(Rf ) ≤M(b− a). (2.1.1)

Thus, we get estimates for area(Rf ). To get better estimates, let usconsider a point c such that a < c < b, and let m1,m2,M1,M2 besuch that

m1 ≤ f(x) ≤M1 ∀x ∈ [a, c]; m2 ≤ f(x) ≤M2 ∀x ∈ [c, b].

Then it is obvious that

m1(c− a) +m2(b− c) ≤ area(Rf ) ≤M1(c− a) +M2(b− c). (2.1.2)

Since

m(b− a) = m(c− a) +m(b− c) ≤ m1(c− a) +m2(b− c),

M(b− a) = M(c− a) +M(b− c) ≥M1(c− a) +M2(b− c),

we can infer that the estimates in (2.1.2) are better than those in(2.1.1). We would have got still better estimates if we had takenm1,m2,M1,M2 as

m1 = inff(x) : a ≤ x ≤ c, m2 = inff(x) : c ≤ x ≤ b,

M1 = supf(x) : a ≤ x ≤ c, M2 = supf(x) : c ≤ x ≤ b,

We can improve these bounds by taking more and more points in[a, b]. This is the basic idea of Riemann integration.

In the above, for S ⊆ R, the expressions inf S and supS denoterespectively the greatest lower bound and least upper bound of S. Letus recall the definitions of these concepts.

Let S be a set of real numbers.

A number β is called an upper bound of S if s ≤ β for all s ∈ S,and a number α is called a lower bound of S if s ≥ α for all s ∈ S.

By a least upper bound (lub) or supremum of S we meana number β0 which is an upper bound of S, and every other upperbound of S is strictly bigger than β0. Similarly, a greatest lower

28 Definite Integral

bound (glb) or infimum of S we mean a number α0 which is alower bound of S, and every other lower bound of S is strictly lessthan α0.

Alternatively, β0 ∈ R is the supremum of S iff for every ε > 0,there exists s ∈ S such that β0 − ε ≤ s, and α0 ∈ R is the infimumof S iff for every ε > 0, there exists t ∈ S such that t ≤ α0 + ε.

2.2 Upper and Lower Sums

Let f : [a, b] → R be a bounded function, and let P be a partition of[a, b], i.e., a finite set P = x0, x1, x2, . . . , xk of points in [a, b] suchthat

P : a = x0 < x1 < x2 < . . . < xk = b.

Corresponding to this partition and the function f , we associate twonumbers:

L(P, f) :=k∑

i=1

mi∆xi, U(P, f) :=k∑

i=1

Mi∆xi,

where, for i = 1, . . . , k, ∆xi = xi − xi−1, and

mi = inff(x) : xi−1 ≤ x ≤ xi, Mi = supf(x) : xi−1 ≤ x ≤ xi.

Definition 2.1 The quantities U(P, f) and L(P, f) are called uppersum and lower sum, respectively, of the function f associated withthe partition P .

Note that if f(x) ≥ 0 for all x ∈ [a, b], then L(P, f) and is thetotal area of the rectangles with lengths mi and widths xi − xi−1,and U(P, f) is the total area of the rectangle with lengths Mi andwidths xi−xi−1, for i = 1, . . . , k. Thus, it is intuitively clear that therequired area, say γ, under the graph of f must satisfy the relation:

L(P, f) ≤ γ ≤ U(P, f)

for all partitions P of [a, b].

Now, let P be the set of all partitions of [a, b]. Then we have

L(P, f) ≤ U(P, f) ∀P ∈ P.

Upper and Lower Sums 29

Now, let

m = inff(x) : a ≤ x ≤ b, M = supf(x) : a ≤ x ≤ b.

Sincem ≤ mi ≤Mi ≤M ∀ i = 1, . . . , k,

and since

k∑i=1

m∆xi = m(b− a),k∑

i=1

M∆xi = M(b− a),

it follows that

m(b− a) ≤ L(P, f) ≤ U(P, f) ≤M(b− a) ∀P,Q ∈ P.

Hence, the set L(P, f) : P ∈ P is bounded above by M(b−a), andthe set U(Q, f) : Q ∈ P is bounded below by m(b− a). Thus

α(f) := supL(P, f) : P ∈ P ≤M(b− a),β(f) := infU(P, f) : P ∈ P ≥ m(b− a)

exist. The quantities α(f) and β(f) are known as lower integral andupper integral respectively.

In defining the quantities α(f) and β(f), we have assume animportant property of the set R of real numbers. That reads asfollows:

Least Upper Bound Property: If S is a subset of R which isbounded above, then supS exists.

Using the above property the following property can be proved:

Geast Lower Bound Property: If S is a subset of R which isbounded below, then inf S exists.

Although we know that L(P, f) ≤ U(P, f) for all P ∈ P, it isnot obvious whether α(f) ≤ β(f). In fact, it is true. To see this, wemake use of the following easily verifiable result.

Lemma 2.1 For P,Q ∈ P, let P = P∪Q, i.e., the partition obtainedby taking all the points in P ∪Q. Then

L(P, f) ≤ L(P , f), U(P , f) ≤ U(Q, f).


By the above Lemma 2.1, we have

L(P, f) ≤ U(Q, f) ∀P,Q ∈ P.

Therefore, it follows that,

α(f) ≤ β(f).

As a consequence, we also have

β(f)− α(f) ≤ U(P, f)− L(P, f),

m(b− a) ≤ L(P, f) ≤ α(f) ≤ β(f) ≤ U(Q, f) ≤M(b− a)

for all P,Q ∈ P.

2.3 Integrability and integral

Definition 2.2 If α(f) = β(f), then we say that f is integrable on[a, b], and the common value is called the integral of f . The integralof f , if exists, is denoted by∫ b

af(x) dx.

Note that if∫ ba f(x) dx exists, then

L(P, f) ≤∫ b

af(x) dx ≤ U(Q, f) ∀P,Q ∈ P.

Theorem 2.2 Suppose f is integrable on [a, b], and m, M are suchthat m ≤ f(x) ≤M for all x ∈ [a, b]. Then

m(b− a) ≤∫ b

af(x) dx ≤M(b− a).

In particular, if M0 > 0 is such that |f(x)| ≤ M0 for all x ∈ [a, b],then ∣∣∣∣∫ b

af(x) dx

∣∣∣∣ ≤M0(b− a).

Integrability and integral 31

Proof. We know that for any partition P on [a, b],

m(b− a) ≤ L(P, f) ≤∫ b

af(x) dx ≤ U(P, f) ≤M(b− a).

Hence the result.

Remark 2.1 Suppose f : [a, b] → R is integrable. Then we define∫ a

bf(x) dx := −

∫ b

af(x) dx.

Also, for any function function f : [a, b] → R, we define∫ τ

τf(x) dx := 0 ∀ τ ∈ [a, b].

EXAMPLE 2.1 Let f(x) = c for all x ∈ [a, b], for some c ∈ R.Then we have

L(P, f) = c(b− a), U(P, f) = c(b− a)

Hence

c(b− a) = L(P, f) ≤ α(f) ≤ β(f) ≤ U(P, f) = c(b− a),

showing the integrability of f , and∫ ba f(x)dx = c(b− b).

EXAMPLE 2.2 Let f(x) = x for all x ∈ [a, b] for some c ∈ R.Let xi = a + i (b−a)

n , for i = 1, . . . , n. Then with Pn = xini=1 is a

partition of [a, b]. We note that mi = xi−1, Mi = xi for i = 1, . . . , n,and hence

L(Pn, f) =n∑

i=1

xi−1∆xi =n∑

i=1

[a+ (i− 1)

(b− a)n

](b− a)n

,

U(Pn, f) =n∑

i=1

xi∆xi =n∑

i=1

[a+ i

(b− a)n

](b− a)n

.

It is see that

L(Pn, f) = a(b− a)− (b− a)2

n+ (b− a)2

n(n+ 1)2n2

→ b2 − a2

2,


U(Pn, f) = a(b− a) +(b− a)2

n2

n(n+ 1)2

→ b2 − a2

2as n→∞. Now, since

L(Pn, f) ≤ α(f) ≤ β(f) ≤ U(Pn, f) ∀n ∈ N,

it follows by taking limit as n→∞ that

α(f) = β(f) =b2 − a2

2.

In the above example what we have showed is that for a sequence(Pn) of partitions, the sequences U(Pn, f) and L(Pn, f) convergeto the same point. This is true in general as well as well as thefollowing theorem shows,

Theorem 2.3 If there is a sequence (Pn) of partitions such that boththe sequences U(Pn, f) and L(Pn, f) converge to the same point,say γ, then f is Riemann integrable, and γ =

∫ ba f(x)dx.

Proof. Suppose (Pn) is a sequence of partitions of [a, b] such thatboth the sequences U(Pn, f) and L(Pn, f) converge to the samepoint, say γ. We know that

L(Pn, f) ≤ α(f) ≤ β(f) ≤ U(Pn, f) ∀n ∈ N.

By hypothesis, it then follows that α(f) = β(f) = γ.

In fact, we have the following.

Theorem 2.4 Let f : [a, b] → R be a bounded function. If for everyε > 0, there exists a partition P such that U(P, f) − L(P, f) < ε,then f is integrable. Conversely, if f is integrable, then for everyε > 0, there exists a partition P such that U(P, f)− L(P, f) < ε.

Proof. Suppose that for every ε > 0, there exists a partition Psuch that U(P, f)− L(P, f) < ε. So, let ε > 0, and P be a partitionsuch that U(P, f)− L(P, f) < ε. Since

β(f)− α(f) ≤ U(P, f)− L(P, f),

Integrability and integral 33

it follows that β − α < ε for all ε > 0. Consequently, α = β, andhence f is integrable.

Conversely, suppose f is integrable on [a, b], i.e., γ :=∫ ba f(x) dx

exists. Then we know that

supL(P, f) : P ∈ P = γ = infU(P, f) : P ∈ P.

Now, let ε > 0. Then there exist partitions P1 and P2 such that

U(P1, f) < γ +ε

2, γ − ε

2< L(P2, f).

ThusU(P1, f)− ε

2< γ < L(P2, f) +

ε

2.

Now, let P = P1 ∪ P2. Then, by Lemma 2.1, we have

U(P, f) ≤ U(P1, f), L(P2, f) ≤ L(P, f).

Hence,U(P, f)− ε

2< γ < L(P, f) +

ε

2.

Thus,U(P, f)− L(P, f) < ε.


For computation of∫ ba f(x) dx, the following result is sometimes

useful.

Theorem 2.5 Suppose f : [a, b] → R is a bounded function. Supposethere exits a sequence Pn of partitions of [a, b] such that

U(Pn, f)− L(Pn, f) → 0 as n→∞.

Then f is integrable. Conversely, suppose f is integrable. Then thereexits a sequence Pn of partitions of [a, b] such that

U(Pn, f)− L(Pn, f) → 0 as n→∞.

In such case,

U(Pn, f) →∫ b

af(x) dx, L(Pn, f) →

∫ b

af(x) dx as n→∞.


Proof. Suppose there exits a sequence Pn of partitions of [a, b]such that

U(Pn, f)− L(Pn, f) → 0 as n→∞.

Then,

0 ≤ β(f)− α(f) ≤ U(Pn, f)− L(Pn, f) → 0 as n→∞.

Hence, α(f) = β(f), and hence f is integrable.

Conversely, suppose f is is integrable. Then, by Theorem 2.4, foreach n ∈ N, there exists a partition Pn on [a, b] such that

U(Pn, f)− L(Pn, f) <1n.

Hence, U(Pn, f)− L(Pn, f) → 0 as n→∞.

Since

L(Pn, f) ≤ α(f) ≤ β(f) ≤ U(Pn, f) ∀n ∈ N

it also follows that

U(Pn, f) →∫ b

af(x) dx, L(Pn, f) →

∫ b

af(x) dx as n→∞.


Can we impose some conditions on (Pn) which guarantees theconvergence U(Pn, f) − L(Pn, f) → 0 as n → ∞? We shall takeup this issue in the next section. Before doing that let us introduceanother concept, the Riemann sums of f .

2.3.1 Riemann sum

Definition 2.3 Let f : [a, b] → R be a function, P = xi : i =1, . . . , k be a partition of [a, b], and T = ti : i = 1, . . . , k be suchthat xi−1 ≤ ti ≤ xi for all i = 1, . . . , k. Such a set T is called a setof tags on P . The quantity

S(P, f, T ) :=k∑

i=1

f(ti)∆xi

is called the Riemann sum for f associated with the partition P andthe set T of tags on P .

Integral of Continuous Functions 35

We may observe that for any partition P and for any set T oftags on P ,

L(P, f) ≤ S(P, f, T ) ≤ U(P, f).

This observation together with Theorem 2.5 gives the following re-sult.

Theorem 2.6 Suppose f : [a, b] → R is a bounded function. If Pnis a sequence of partitions of [a, b] such that

U(Pn, f)− L(Pn, f) → 0 as n→∞,

then f is integrable, and

S(Pn, f, Tn)) →∫ b

af(x) dx,

where Tn is any set of tags on Pn, n ∈ N.

Remark 2.2 in the Appendix, we have given a characterization ofintegrability using Riemann sums (see Theorem 9.1).

2.4 Integral of Continuous Functions

Next we prove that every continuous function defined on a closedinterval [a, b] is integrable. First we recall some definitions.

Definition 2.4 Let J be an interval (open or closed or semi-open).A function f : J → R is said to be continuous at a point x0 ∈ Jif for every ε > 0, there exists δ > 0 (depends in general on ε and onthe point x0) such that

|f(x)− f(x0)| < ε whenever |x− x0| < δ.

Definition 2.5 Let J be an interval (open or closed or semi-open).A function f : J → R is said to be continuous on J if it is contin-uous at every point in J .

Definition 2.6 definition1 Let J be an interval (open or closedor semi-open). A function f : J → R is said to be uniformlycontinuous on J if for every ε > 0, there exists δ > 0 (depends ingeneral on ε) such that

|f(x)− f(y)| < ε whenever |x− y| < δ.


Now we list a few important properties of continuous functions.

Suppose f : [a, b] → R is continuous. The we have the following:

• f is bounded and uniformly continuous.

• Intermediate-value property: Suppose α, β ∈ R such thatα < β. Suppose x1, x2 ∈ [a, b] are such that f(x1) = α, f(x2) = β. Ifc1 < γ < c2, then then there exists x0 in the interval with end-pointsx1, x2 such that f(x0) = γ.

• For every α, β ∈ [a, b] with α < β, there exist ξ, η ∈ [α, β] suchthat

f(ξ) = infα≤x≤β

f(x), f(η) = supα≤x≤β

f(x).

Definition 2.7 Let J be an interval. A function f : J → R is saidto be piece-wise continuous if f is not continuous only at a finitenumber of points in J .

Note: A piece-wise continuous function need not be bounded. Forexample, the function f : [−1, 1] → R defined by f(x) = 1/x forx 6= 0 and f(0) = 0 is piece-wise continuous, but not bounded.

In the following we shall make use of the following notation:

For a partition P := xi : i = 1, . . . , k of [a, b], we denote

µ(P ) := maxxi − xi−1 : i = 1, . . . , kn,

and it is called the mesh of the partition P .

Theorem 2.7 Suppose f : [a, b] → R is a continuous function. Thenf is integrable. In fact, we have the following:

(i) For every ε > 0 there exists a δ > 0 such that for everypartition P of [a, b] with µ(P ) < δ, we have

U(P, f)− L(P, f) < ε.

(ii) Suppose Pn is a sequence of partitions of [a, b] such thatµ(Pn) → 0 as n → ∞. Then the sequences U(Pn, f), L(Pn, f)S(Pn, f, Tn) converge to the same limit

∫ ba f(x) dx.

Integral of Continuous Functions 37

Proof. Let f : [a, b] → R be a continuous function, and let ε > 0be given. Let P : a = x0 < x1 < x2 . . . < xk = b be a partition of[a, b]. Then

U(P, f)− L(P, f) =k∑

i=1

(Mi −mi)(xi − xi−1).

Since f is continuous on the closed interval [a, b], there exists ξi, ηi in[xi−1, xi] such that Mi = f(ξi), mi = f(ηi) for i = 1, . . . , k. Hence,

U(P, f)− L(P, f) =k∑

i=1

[f(ξi)− f(ηi)](xi − xi−1).

Again, since f is uniformly continuous on [a, b], there exists δ > 0such that |f(t)− f(s)| < ε/(b− a) whenever |t− s| < δ. Hence, if wetake P such that µ(P ) < δ, then we have

U(P, f)− L(P, f) =k∑

i=1

[f(ξi)− f(ηi)](xi − xi−1) < ε.

Therefore, by Theorem 2.4, f is integrable.

Next, suppose Pn is a sequence of partitions such that µ(Pn) →0 as n→∞. Let N ∈ N be such that µ(Pn) < δ for all n ≥ N . Then,it follows by (i) above that U(Pn, f) − L(Pn, f) < ε for all n ≥ N ,showing that U(Pn, f)−L(Pn, f) → 0 as n→∞. From this, the lastobservation also follows, since

L(Pn, f) ≤∫ b

af(x)dx ≤ U(Pn, f),

L(Pn, f) ≤ S(Pn, f, Tn) ≤ U(Pn, f)

for all n ∈ N.

EXAMPLE 2.3 Let f(x) = ex for all x ∈ [a, b]. Then, f is contin-uous. Let

hn =(b− a)n

, xi = a+ ihn, ti = xi−1, i = 1, . . . , n.


Then with Pn = xini=1 and T = tin

i=1, we have µ(Pn) → 0, and

S(Pn, f, Tn) = hn

n∑i=1

ea+(i−1)hn = hnea

n∑i=1

α(i−1)n = hne

aαnn − 1αn − 1

,

where αn = ehn . Since αnn = eb−a, we have

S(Pn, f, Tn) = hneaα

nn − 1αn − 1

= ea[eb−a − 1]hn

ehn − 1.

Since limn→∞hn

ehn−1= 1, it follows that

S(Pn, f, Tn) = ea[eb−a − 1]hn

ehn − 1→ ea[eb−a − 1] = eb − ea.

Note: In the above we used the result that, if g is a continuousfunction, then xn → x implies g(xn) → g(x), and also applied thel’Hospital rule: limx→x

ez

ez−1 = limx→x1ez = 1.

Theorem 2.8 Suppose f : [a, b] → R is bounded and piecewise con-tinuous,i.e., there are at most a finite number of points in [a, b] atwhich f is discontinuous. Then f is integrable.

Proof. Let ε > 0 be given. We have to show that there exists apartition P such that U(P, f)− L(P, f) < ε.

Suppose that c ∈ (a, b) such that f is continuous on [a, c) and(c, b]. Let δ > 0 be such that c+ δ < b and c− δ > a. Let f1, f2, f3

be restrictions of f to the intervals [a, c−δ], [c+δ, b] and [c−δ, c+δ],respectively. Since f is continuous on [a, c − δ] and [c + δ, b], f isintegrable on these intervals, so that there exist partitions P1 on[a, c− δ] and P2 on [c+ δ, b] such that

U(P1, f1)− L(P1, f1) <ε

3, U(P2, f2)− L(P2, f2) <

ε

3.

Assume that δ > 0 is so small that (M −m)δ < ε/6. Then for anypartition P3 on [c− δ, c+ δ], we have

U(P3, f3)− L(P3, f3) <ε

3.

Now, for the partition P = P1 ∪ P2 ∪ P3 on [a, b], we have

U(P, f) = U(P1, f1) + U(P2, f2) + U(P3, f3),

Some properties 39

L(P, f) = L(P1, f1) + L(P2, f2) + L(P3, f3).

Hence,U(P, f)− L(P, f) < ε.

Thus, f is integrable. The case of more than one (but finite numberof) points of discontinuity can be handled analogously.

2.5 Some properties

Theorem 2.9 Suppose f is integrable on [a, c] and [c, b]. Then f isintegrable on [a, b], and∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.

Proof. Let f1 = f |[a,c], f2 = f |[c,b]. Let ε > 0 be given. Since f1

and f2 are integrable, there exist partitions P1 and P2 of [a, c] and[c, b] respectively such that

U(P1, f1)− L(P1, f1) <ε

2, U(P2, f2)− L(P2, f2) <

ε

2.

Suppose P = P1 ∪ P2. Then, it can be seen that

L(P, f) = L(P1, f1)+L(P2, f2), U(P, f) = U(P1, f1)+U(P2, f2).

Hence,

U(P, f)−L(P, f) = [U(P1, f1)−L(P1, f1)]+[U(P2, f2)−L(P2, f2)] < ε.

Thus f is integrable. Since

L(P, f) ≤∫ b

af(x)dx ≤ U(P, f),

L(P1, f1)+L(P2, f2) ≤∫ c

af(x)dx+

∫ b

cf(x)dx ≤ U(P1, f1)+U(P2, f2),

it follows that∣∣∣∣∫ c

af(x) dx+

∫ b

cf(x) dx−

∫ b

af(x) dx

∣∣∣∣ < ε.

This is true for all ε > 0. Hence the final result.


Theorem 2.10 (Mean-value theorem) Suppose f is continuouson [a, b]. Then there exists x0 ∈ [a, b] such that

1b− a

∫ b

af(x) dx = f(x0).

Proof. By Theorem 2.2,

m ≤ 1b− a

∫ b

af(x) dx ≤M.

Since there exist ξ, η ∈ [a, b] such that f(ξ) = m, f(η) = M , itfollows from intermediate value property that there exists x0 ∈ [a, b]such that

1b− a

∫ b

af(x) dx = f(x0).

Hence the result.

Theorem 2.11 Suppose f and g are continuous functions on [a, b].Then ∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx∫ b

ac f(x) dx = c

∫ b

af(x) dx ∀ c ∈ R.

Proof. Suppose Pn is a sequence of partitions on [a, b] such thatµ(Pn) → 0 as n → ∞. For each n ∈ N, let Tn be a set of tags onPn. Since f , g are continuous, the functions f + g and c f are alsocontinuous for every c ∈ R, and

S(Pn, f, Tn) →∫ b

af(x) dx, S(Pn, g, Tn) →

∫ b

ag(x) dx,

S(Pn, f + g, Tn) →∫ b

a[f(x) + g(x)] dx.

Now, since

S(Pn, f + g, Tn) = S(Pn, f, Tn) + S(Pn, g, Tn),

S(Pn, c f, Tn) = c S(Pn, f, Tn) ∀ c ∈ R,

Some results 41

it follows that∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx,

∫ b

ac f(x) dx = c

∫ b

af(x) dx ∀ c ∈ R.


Remark 2.3 In the appendix, we have proved the conclusion ofthe above theorem assuming only the integrablility of f and g (seeTheorem 9.2).

Theorem 2.12 Suppose f and g are continuous functions on [a, b]such that f(x) ≤ g(x) for all x ∈ [a, b]. Then∫ b

af(x) dx ≤

∫ b

ag(x) dx.

Proof. By assumption, g(x)− f(x) for all x ∈ [a, b], so that

S(Pn, g, Tn)− S(Pn, f, Tn) = S(Pn, g − f, Tn) ≥ 0 ∀n ∈ N,

so that ∫ b

ag(x) dx−

∫ b

af(x) dx ≥ 0.


2.6 Some results

2.6.1 Indefinite Integral

Theorem 2.13 Suppose f is continuous on [a, b], and for x ∈ [a, b],let

ϕ(x) =∫ x

af(t)dt.

Then ϕ is differentiable and

ϕ′(x) = f(x) ∀x ∈ [a, b].


Proof. Let x ∈ [a, b] and 0 6= h ∈ R be such that x + h ∈ [a, b].Then we have

ϕ(x+ h)− ϕ(x) =∫ x+h

af(t)dt−

∫ x

af(t)dt =

∫ x+h

xf(t)dt.

Hence,

ϕ(x+ h)− ϕ(x)h

− f(x) =1h

∫ x+h

x[f(t)− f(x)]dt.

We know by Theorem 2.2 (applied to the continuous function g(t) :=f(t)− f(x), t ∈ [x, x+ h]) that∣∣∣∣∫ x+h

x[f(t)− f(x)]dt

∣∣∣∣ ≤ h maxt∈[x,x+h]

|f(t)− f(x)|.

Now, let ε > 0 be given and let h be small enough such that

|f(t)− f(x)| < ε whenever t ∈ [x, x+ h].

Hence, it follows that∣∣∣∣ϕ(x+ h)− ϕ(x)h

− f(x)∣∣∣∣ < ε.

Thus limh→0

[ϕ(x+h)−ϕ(x)

h − f(x)]

exists and is equal to f(x). Inother words, ϕ′(x) exists and ϕ′(x) = f(x).

An alternate argument: By mean-value theorem, there exists ξhin the interval with endpoints x, x+ h such that∫ x+h

xf(t)dt = h f(ξh).

Hence,ϕ(x+ h)− ϕ(x)

h= f(ξh).

Since f is continuous at x, it follows that f(ξh) → x as h→ 0. Henceϕ′(x) exists and ϕ′(x) = f(x).

Some results 43

2.6.2 Fundamental Theorem of Integral Calculus

A consequence of Theorem 2.13 is that every continuous function hasan anti-derivative or primitive.

Definition 2.8 A function g is called an anti-derivative of f if g isdifferentiable and g′(x) = f(x) for all x ∈ [a, b].

By the Theorem 2.13, if f is a continuous function, then theindefinite integral ϕ(x) =

∫ xa f(t)dt is an anti-derivative of f .

We may observe that if g1 and g2 are anti-derivatives of f , theng′1(x) = f(x) = g′2(x) for all x ∈ [a, b], i.e., g′1(x) − g′2(x) = 0 for allx ∈ [a, b]. Hence, it follows that g1 − g2 is a constant. Thus, in viewof the above theorem it follows that if g is an anti-derivative of f ,and if ϕ is the indefinite integral of f , then g(x) = ϕ(x) + c for allx ∈ [a, b] and for some constant c ∈ R. Hence, in view of the abovetheorem, we have

g(b)− g(a) = ϕ(b)− ϕ(a) =∫ b

af(t)dt.

Thus we have proved the following theorem.

Theorem 2.14 (Fundamental theorem of integral calculus)Suppose f is continuous on [a, b] and suppose that g is an anti-derivative of f . Then ∫ b

af(t)dt = g(b)− g(a).

The conclusion of the above theorem is also known as Newton-Leibniz formula. The difference g(b) − g(a) is usually written as[g(x)]ba.

Here is another p[roof for the above theorem.

Another Proof. Let g be an antiderivative of f , i.e., g is differ-entiable and g′ = f . Let P : a = x0 < x1 < . . . < xn = b be anypartition of [a, b]. Then by Lagrange’s mean value theorem, thereexists ξi ∈ [xi−1, xi] such that

g(xi)− g(xi−1) = g′(ξi)(xi − xi−1) = f(ξi)(xi − xi−1).


Hence,

g(b)− g(a) =n∑

j=1

[g(xi)− g(xi−1)] =n∑

j=1

f(ξi)(xi − xi−1).

Since the Riemann sum∑n

j=1 f(ξi)(xi − xi−1) is constant for any

partition P , it follows that∑n

j=1 f(ξi)(xi−xi−1) =∫ ba f(x)dx. Thus,∫ b

af(x)dx = g(b)− g(a).


The following result is worth mentioning:

Theorem 2.15 Suppose f is a continuous function.Then for everypartition P of [a, b], there exists a set T of tags on P such that

S(P, f, T ) =∫ b

af(x) dx.

Proof. Let P = xi : i = 1, . . . , k be a partition of [a, b]. Since fis continuous, by mean value theorem (Theorem 2.10), there existsξi ∈ [xi−1, xi] such that∫ xi

xi−1

f(x) dx = f(ξi)(xi − xi−1), i = 1, . . . , k.

Hence, taking T = ξi : i = 1, . . . , k,

S(P, f, T ) =k∑

i=1

f(ξi)(xi − xi−1) =k∑

i=1

∫ xi

xi−1

f(x) dx =∫ b

af(x) dx.


2.6.3 Applications of Fundamental Theorem

Theorem 2.16 (Product formula) Suppose f and g are continu-ous functions on [a, b], and let G be an anti-derivative of g. If f isdifferentiable on [a, b], then∫ b

af(x)g(x) dx = [f(x)G(x)]ba −

∫ b

af ′(x)G(x) dx.

Some results 45

Proof. Recall that if u and v are differentiable, then

(uv)′ = u′v + uv′.

Hence,∫ b

a[u(x)v(x)]′ dx =

∫ b

au′(x)v(x) dx+

∫ b

au(x)v′(x) dx.

Using fundamental theorem,

[u(x)v(x)]ba =∫ b

au′(x)v(x) dx+

∫ b

au(x)v′(x) dx.

Thus, ∫ b

au(x)v′(x) dx = [u(x)v(x)]ba −

∫ b

au′(x)v(x) dx.

Now, taking f(x) = u(x) and v(x) = G(x), we obtain the requiredformula.

Theorem 2.17 (Change of variable formula) Suppose ψ : [α, β] →R is a differentiable function such that ψ(α) = a and ψ(β) = b.Then, for any continuous function f : [a, b] → R,∫ b

af(x) dx =

∫ β

αf(ψ(t))ψ′(t)dt.

Proof. Let F be an anti-derivative of f , i.e., such that F ′(x)f(x).Then taking G(t) = F (ψ(t)) for t ∈ [α, β], we have

G′(t) = F ′(ψ(t))ψ′(t) = f(ψ(t))ψ′(t), t ∈ [α, β].

Hence, by fundamental theorem,∫ β

αf(ψ(t))ψ′(t)dt =

∫ β

αG′(t)dt = G(β)−G(α) = F (ψ(β))−F (ψ(α)).

Hence, ∫ β

αf(ψ(t))ψ′(t)dt = F (b)− F (a) =

∫ b

af(x) dx.



The following examples have been worked out by knowing theantiderivatives of certain functions.

EXAMPLE 2.4 For k ≥ 0,∫ b

axk dx =

[xk+1

k + 1

]b

a

=bk+1 − ak+1

k + 1.

EXAMPLE 2.5 For α 6= 0,∫ b

aeαx dx =

[eαx

α

]b

a

=eαb − eαa

k + 1.

3

Improper Integrals

Recall that we defined definite integral of a function f for the casewhen f is a bounded function defined on a closed interval [a, b]. Incase the assumptions on f are not satisfied, then can we still have anotion of integral? We discuss a few such cases.

3.1 Definitions

Definition 4.1 Suppose f is defined on [a,∞). If ϕ(t) :=∫ ta f(x) dx

exists for every t > a, and if limt→∞ ϕ(t) exists, then we define theimproper integral of f over [a,∞) as∫ ∞

af(x) dx = lim

t→∞

∫ t

af(x) dx.

Definition 4.2. Suppose f is defined on (−∞, b]. If ψ(t) :=∫ bt f(x) dx exists for every t < b, and if limt→∞ ψ(t) exists, then

we define the improper integral of f over (−∞, b] as∫ b

−∞f(x) dx = lim

t→−∞

∫ b

tf(x) dx.

Definition 4.3. Suppose f is defined on R := (−∞,∞). If∫ c−∞ f(x) dx

and∫∞c f(x) dx exists for some c ∈ R, then we define the improper

integral of f over (−∞,∞) as∫ ∞

−∞f(x) dx =

∫ c

−∞f(x) dx+

∫ ∞

cf(x) dx.

47

48 Improper Integrals

We may observe that existence of limt→∞∫ t−t f(x) dx does not,

in general, imply existence of∫∞−∞ f(x) dx. To see this, consider the

following example:

Let f(x) = x for every x ∈ R. Then we have∫ t−t f(x) dx = 0 for

every t ∈ R, but the integrals∫ c−∞ f(x) dx and

∫∞c f(x) dx do not

exist.

Next we consider the case when f is defined on an interval J offinite length, but limx→x0 |f(x)| = ∞, where x0 either belongs to Jor it is an end point of J .

Definition 4.4. Suppose f is defined on (a, b]. If∫ bt f(x) dx exists

for every t ∈ (a, b), and if limδ→0

∫ ba+δ f(x) dx exists, then we define

the improper integral of f over (a, b] as∫ b

af(x) dx = lim

δ→0

∫ b

a+δf(x) dx.

Definition 4.5. Suppose f is defined on [a, b). If∫ ta f(x) dx exists

for every t ∈ (a, b), and if limδ→0

∫ b−δa f(x) dx exists, then we define

the improper integral of f over [a, b) as∫ b

af(x) dx = lim

δ→0

∫ b−δ

af(x) dx.

Definition 4.6. Suppose f is defined on [a, c) and (c, b]. If∫ ca f(x) dx

and∫ bc f(x) dx exist, then we define the improper integral of f

over [a, b] as ∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.

Now we combine the situations in Definitions 4.1, 4.2 with Def-initions 4.4, 4.5, to consider improper integrals over intervals of theform (a,∞) and (−∞, b).

Examples 49

Definition 4.7. Suppose f is defined on (a,∞). If the integrals∫ ta f(x) dx and

∫∞t f(x) dx exist as improper integrals for every t > a,

then we define the improper integral of f over (a,∞) as∫ ∞

af(x) dx =

∫ t

af(x) dx+

∫ ∞

tf(x) dx.

Definition. Suppose f is defined on (−∞, b). If the integrals∫ t−∞ f(x) dx and

∫ bt f(x) dx exist as improper integrals for every

t < b, then we define the improper integral of f over (−∞, b) as∫ b

−∞f(x) dx =

∫ t

−∞f(x) dx+

∫ b

tf(x) dx.

In case an improper integral exists (resp. does not exists), thenwe also say that the improper integral converges (resp. diverges).

3.2 Examples

EXAMPLE 3.1 Consider the improper integral∫ ∞

1

1xdx. Note

that ∫ t

1

1xdx = [lnx]t1 = ln t→∞ as t→∞.

Hence,∫∞1

1x dx diverges.

EXAMPLE 3.2 Consider the improper integral∫ ∞

1

1x2

dx Note

that ∫ t

1

1x2

dx =[−1x

]t

1

= 1− 1t→ 1 as t→∞.

Hence,∫∞1

1x2 dx converges.

EXAMPLE 3.3 For p 6= 1, consider the improper integral∫ ∞

1

1xp

dx.

In this case, we have∫ t

1

1xp

dx =[x−p+1

−p+ 1

]t

1

=t−p+1 − 1−p+ 1

.


Note that,

p > 1 =⇒ t−p+1 − 1−p+ 1

→ 1p− 1

as t→∞,

and

p < 1 =⇒ t−p+1 − 1−p+ 1

→∞ as t→∞,

Hence, ∫ ∞

1

1xp

dx

converges for p > 1,diverges for p ≤ 1.

EXAMPLE 3.4 For p 6= 1, consider the improper integral∫ 1

0

1xp

dx.

In this case, we have∫ 1

δ

1xp

dx =[x−p+1

−p+ 1

]1

δ

=1− δ−p+1

−p+ 1.

Note that,

p > 1 =⇒ δ−p+1 − 1−p+ 1

→∞ as t→∞,

and

p < 1 =⇒ δ−p+1 − 1−p+ 1

→ 11− p

as t→∞,

Hence, ∫ 1

0

1xp

dx

converges for p < 1,diverges for p ≥ 1.

Before giving further examples, let us state a result which willbe useful in asserting the existence of certain improper integral bycomparing it with certain other improper integral.

Suppose J is either an interval (of finite or infinite length) or itis union of two intervals such that J together with some point in Ris an interval. Suppose f defined on J . We denote the improperintegral of f over J by ∫

Jf(x) dx,

and say the the improper integral over J converges whenever it exists,and otherwise, we say that the improper integral

∫J f(x) dx diverges.

Examples 51

Theorem 3.1 Suppose J is above, and f and g are defined on J .

(i) If 0 ≤ f(x) ≤ g(x) for all x ∈ J , and∫J g(x) dx exists, then∫

J f(x) dx exists.

(ii) If∫J |f(x)| dx exists, then

∫J f(x) dx exists.

EXAMPLE 3.5 Since∣∣∣∣sinxxp

∣∣∣∣ ≤ 1xp,

∣∣∣cosxxp

∣∣∣ ≤ 1xp

it follows from Example 3.3 and Theorem 3.1(ii) that the improperintegrals∫ ∞

1

sinxxp

dx,

∫ ∞

1

cosxxp

dx, converge for all p > 1.

EXAMPLE 3.6 Since∣∣∣∣sinxxp

∣∣∣∣ = ∣∣∣∣sinxx∣∣∣∣ 1xp−1

≤ 1xp−1

,∣∣∣cosxxp

∣∣∣ ≤ 1xp

it follows from Example 3.4 above and Theorem ??(ii)that∫ 1

0

sinxxp

dx converges for all p < 2,

∫ 1

0

cosxxp

dx converges for all p < 1.

EXAMPLE 3.7 Observe that

sinxxp

=sinxx

1xp−1

≥ sin 1xp−1

∀x ∈ (0, 1].

Since∫ 10

1xp−1 dx diverges for p− 1 ≥ 1, i.e., for p ≥ 2, it follows that∫ 1

0

sinxxp

dx diverges for all p ≥ 2,


EXAMPLE 3.8 Note that for p > 0, and for β > 1,∫ β

1

sinxxp

dx =[

1xp

(− cosx)]β

1

− p

∫ β

1

1xp+1

cosx dx

=[cos 1− cosβ

βp

]− p

∫ β

1

cosxxp+1

dx.

By the result in Example 3.5,∫ ∞

1

cosxxp+1

dx converges for all p > 0.

Also, cos ββp → 0 as β →∞. Hence,∫ ∞

1

sinxxp

dx converges for all p > 0.

EXAMPLE 3.9 From Examples 3.6, 3.7, 3.8,∫ ∞

0

sinxxp

dx converges for 0 < p < 2.

Now some more results which facilitate the assertion of conver-gence/divergence of improper integrals, whose proofs follow from thedefinition of limits.

Theorem 3.2 Suppose f(x) ≥ 0, g(x) ≥ 0 for all x ∈ [a,∞),∫ ba f(x)dx and

∫ ba g(x)dx exists for every b > a. Suppose further

thatf(x)g(x)

→ ` as x→∞.

(i) If ` 6= 0, then∫∞a f(x)dx converges ⇐⇒

∫∞a g(x)dx con-

verges.

(ii) If ` = 0, then∫∞a g(x)dx converges ⇒

∫∞a f(x)dx converges.

Proof. (i) Suppose ` 6= 0. Then ` > 0, and for ε > 0 with`− ε > 0, there exists x0 ≥ a such that

`− ε <f(x)g(x)

< `+ ε ∀x ≥ x0.

Examples 53

Hence(`− ε)g(x) < f(x) < (`+ ε)g(x) ∀x ≥ x0.

Consequently,∫∞x0f(x)dx converges iff

∫∞x0g(x)dx converges. As∫ x0

a f(x)dx and∫ x0

a g(x)dx exist, the result in (i) follows.

(ii) Suppose ` = 0. Then for ε > 0, there exists x0 ≥ a such that

f(x)g(x)

< ε ∀x ≥ x0.

Thus, f(x) < εg(x) for all x ≥ x0. Hence, convergence of∫∞x0g(x)dx

implies the convergence of∫∞x0f(x)dx. From this the result in (ii)

follows.

Gamma and Beta Functions

Gamma and Beta Functions are certain improper integrals whichappear in many applications.

EXAMPLE 3.10 We show that for x > 0, the improper integral

Γ(x) :=∫ ∞

0tx−1e−t dt

converges. The function Γ(x), x > 0, is called the gamma function.

Note that for tx−1e−t ≤ tx−1 for all t > 0, and∫ 10 t

x−1 dt con-verges for x > 0. Hence, by Theorem 3.1,∫ 1

0tx−1e−t dt converges for x > 0.

Also, we observe thattx−1e−t

t−2→ 0 as t → ∞, and

∫∞1 t−2dt con-

verges. Hence, by Theorem 3.2,∫∞1 tx−1e−t dt converges. Thus,

Γ(x) :=∫ ∞

0tx−1e−t dt =

∫ 1

0tx−1e−t dt+

∫ ∞

1tx−1e−t dt

converges for every x > 0.

EXAMPLE 3.11 We show that for x > 0, y > 0, the improperintegral

β(x, y) :=∫ 1

0tx−1(1− t)y−1 dt


converges. The function β(x, y) for x > 0, y > 0 is called the betafunction.

Clearly, the above integral is proper for x ≥ 1, y ≥ 1. Hence itis enough to consider the case of 0 < x < 1, 0 < y < 1. In this caseboth the points t = 0 and t = 1 are problematic. hence, we considerthe integrals∫ 1/2

0tx−1(1− t)y−1 dt,

∫ 1

1/2tx−1(1− t)y−1 dt.

We note that if 0 < t ≤ 1/2, then (1 − t)y−1 ≤ 21−y so thattx−1(1 − t)y−1 ≤ 21−ytx−1. Since

∫ 1/20 tx−1 dt converges it follows

that∫ 1/20 tx−1(1− t)y−1 dt converges. To deal with the second inte-

gral, consider the change of variable u = 1− t. Then∫ 1

1/2tx−1(1− t)y−1 dt =

∫ 1/2

0uy−1(1− u)x−1 du

which converges by the above argument. Hence,

β(x, y) :=∫ 1

0tx−1(1− t)1−y dt, x > 0, y > 0

converges for every x > 0, y > 0.

3.3 Exercises

Exercise 3.1 Does∫ ∞

1sin(

1x2

)dx converge?

Hint: Note that∣∣∣∣sin( 1

x2

)∣∣∣∣ ≤ 1x2

.


2

cosxx(log x)2

dx converge?

Hint: Observe∣∣∣∣ cosxx(log x)2

∣∣∣∣ ≤ 1x(log x)2

and use the change of vari-

able t = log x.


0e−x2

dx converge?

Exercises 55

Hint: Note that e−x2is continuous on [0, 1], and e−x2 ≤ 1

x2 for1 ≤ x ≤ ∞.


2

sin(log x)x

dx converge?

Hint: Use the change of variable t = log x, and the fact that∫∞log 2 sin t dt diverges.

Exercise 3.5 Does∫ 1

0lnxdx converge?

Hint: Use the change of variable t = log x.

4

Geometric and MechanicalApplications of Integrals

4.1 Computing Area

4.1.1 Using Cartesian Coordinates

Suppose a curve is given by an equation

y = f(x), a ≤ x ≤ b,

where f : [a, b] → R is a continuous function such that f(x) ≥ 0 forall x ∈ [a, b]. Then

limµ(P )→0

k∑j=1

f(ξi)∆xi =∫ b

ay dx

is the area of the region bounded by the graph of f , the x-axis, andthe ordinates at x = a and x = b.

Suppose the curve is given in parametric form:

x = ϕ(t), y = ψ(t), , α ≤ t ≤ β,

such that a = ϕ9α), b = ψ(β). Then the length of the curve talesthe form ∫ β

αψ(t)ϕ′(t)dt.

If f takes both positive and negative values, but changes signonly at a finite number of points, then the area is given by∫ b

a|f(x)| dx.

56

Computing Area 57

• Suppose f : [a, b] → R and g : [a, b] → R are continuous functionssuch that f(x) ≤ g(x) for all x ∈ [a, b]. Then the area of the regionbounded by the graphs of f and g, and the ordinates at x = a andx = b is given by

limµ(P )→0

k∑j=1

[g(ξi)− f(ξi)]∆xi =∫ b

a[g(x)− f(x)] dx.

4.1.2 Using Polar Coordinates

Suppose a curve is given in polar coordinates as

ρ = ϕ(θ), α ≤ θ ≤ β,

where ϕ : [α, β] → R is a continuous function. Then the the area ofthe region bounded by the graph of ϕ and the rays θ = α and θ = βis is given by

limµ(P )→0

k∑j=1

12[ϕ(ξi)∆θi]ϕ(ξi) = lim

µ(P )→0

k∑j=1

12[ϕ(ξi)]2∆θi =

12

∫ β

αρ2 dθ.

4.1.3 Examples

Using cartesian coordinates and parametrization

EXAMPLE 4.1 We find the area bounded by the cures defined by

y =√x, y = x2, x ≥ 0 :

Note that the points of intersection of the curves are at x = 0 andx = 1. Also,

√x ≥ x2 for 0 ≤ x ≤ 1. Hence, the required area is∫ 1

0

(√x− x2

)dx =

[x3/2

3/2− x3

3

]1

0

=13.

EXAMPLE 4.2 We find the area bounded by the ellipse

x = a cos t, y = b sin t, 0 ≤ t ≤ 2π :

The required area is

4∫ a

0y dx = 4

∫ 0

π/2(b sin t)(−asint) dt = 2ab

∫ π/2

0(1−cos 2t dt = πab.

58 Geometric and Mechanical Applications of Integrals

EXAMPLE 4.3 We find the area bounded by one arch of the cy-cloid

x = a(t− sin t), y = a(1− cos t).

One arch of the cycloid is obtained by varying t over the interval[0, 2π]. Thus, the required area is∫ 2πa

0y dx =

∫ 2π

0y(t)x′(t) dt =

∫ 2π

0a2(1− cos t)2 dt = 3πa2.

Using polar coordinates

EXAMPLE 4.4 We find the area bounded by a circle of radius a.Without loss of generality assume that the centre of the circle is theorigin. Then, the circle can be represented in polar coordinates as

ρ = a.

Hence the required area is

12

∫ 2π

0ρ2 dθ = πa2.

EXAMPLE 4.5 We find the area bounded by the lemniscate

ρ = a√

cos 2θ.


2

[12

∫ π/4

−π/4ρ2 dθ

]= a2

∫ π/4

−π/4cos 2θdθ = a2.

4.2 Computing Arc Length

4.2.1 Using Cartesian Coordinates

Suppose a curve is given by and equation

y = f(x), a ≤ x ≤ b,

Computing Arc Length 59

where f : [a, b] → R is a continuous function. Then the length of thecurve is given by

A := limµ(P )→0

k∑j=1

√(xi − xi−1)2 + (yi − yi−1)2,

whereyi − yi−1 = f(xi)− f(xi−1) = f ′(ξi)∆xi,

for some ξi ∈ [xi−1, xi], i = 1, . . . , k. Hence,

A := limµ(P )→0

k∑j=1

√(xi − xi−1)2 + (yi − yi−1)2

= limµ(P )→0

k∑j=1

√(∆xi)2 + [f ′(ξi)∆xi]2

= limµ(P )→0

k∑j=1

√1 + [f ′(ξi)]2∆xi

=∫ b

a

√1 +

(dy

dx

)2

dx.

• If the curve y = f(x), a ≤ x ≤ b, is given in parametric form:

x = φ(t), y = ψ(t), c ≤ t ≤ d,

then

A =∫ b

a

√1 +

(dy

dx

)2

dx =∫ d

c

√(dφ

dt

)2

+(dψ

dt

)2

dt.

4.2.2 Using Polar Coordinates

Suppose a curve is given in polar coordinates as

ρ = ϕ(θ), α ≤ θ ≤ β,

where ϕ : [α, β] → R is a continuous function. Since

x = ρ cos θ, y = ρ sin θ, α ≤ θ ≤ β,


we have

A =∫ β

α

√(dx

dθ

)2

+(dy

dθ

)2

dθ.

Note that

dx

dθ= ρ′ cos θ + ρ(− sin θ),

dy

dθ= ρ′ sin θ + ρ cos θ.

Hence, it follows that

A =∫ β

α

√(dx

dθ

)2

+(dy

dθ

)2

dθ

=∫ β

α

√ρ2 + ρ′2 dθ.

4.2.3 Examples

Using cartesian coordinates

EXAMPLE 4.6 We find the length of the circumference of a circleof radius a.

Without loss of generality assume that the centre of the circle isthe origin,i.e., the circle is given by x2+y2 = a2. The required lengthis

L := 4∫ a

0

√1 +

(dy

dx

)2

dx, y =√a2 − x2.

Thus,

L := 4a∫ a

0

dx√a2 − x2

= 2πa.

Using parametric form

EXAMPLE 4.7 Now we find the length of the circle when it isrepresented by the equations

x = a cos θ, y = a sin θ, 0 ≤ θ ≤ 2π.

Computing Arc Length 61

The required length is

L := 4∫ π/2

0

√(dx

dθ

)2

+(dy

dθ

)2

dθ

= 4∫ π/2

0

√a2 sin2 θ + a2 cos2 θ dθ = 2πa.

EXAMPLE 4.8 Let us find the length of the ellipse

x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π.


L := 4∫ π/2

0

√(dx

dθ

)2

+(dy

dθ

)2

dθ

= 4∫ π/2

0

√a2 sin2 θ + b2 cos2 θ dθ

= 4∫ π/2

0

√a2(1− cos2 θ) + b2 cos2 θ dθ

= 4∫ π/2

0

√a2 − (a2 − b2) cos2 θ dθ

= 4a∫ π/2

0

√1− β2 cos2 θ dθ,

where β =√

a2−b2

a . The above integral is not expressible in standardform unless β = 1, i.e., unles b = a in which case the ellipse is thecircle. But, the integral can be approximately computed numerically.

EXAMPLE 4.9 We find the length of the astroid: x = a cos3 t,y = a sin3 t.


L := 4∫ π/2

0

√(dx

dθ

)2

+(dy

dθ

)2

dθ

= 4∫ π/2

0

√9a2 cos4 t sin2 t+ 9a2 sin4 t cos3 t dθ

= 12a∫ π/2

0

√cos2 t sin2 dt = 6a.


Using polar coordinates

EXAMPLE 4.10 We find the length of the cardioid ρ = a(1+cos θ).

The required length is L :=∫ 2π0

√ρ2 + ρ′2 dθ. Since

ρ2 = a2(1 + cos θ)2, ρ′2 = a2 sin2 θ,

we have

L =√

2a∫ 2π

0

√1 + cos θ dθ = 4a

∫ 2π

0

∣∣∣∣cosθ

2

∣∣∣∣ dθ = 8a.

4.3 Computing Volume of a Solid

Suppose that a three dimensional object, a solid, lies between twoparallel planes x = a and x = b. Let α(x) be the area of the crosssection of the solid at the point x, with cross section being parallelto the yz-plane. We assume that the function α(x), x ∈ [a, b] iscontinuous. Now, consider a partition P : a = x0 < x1 < . . . < xk =b of the interval [a, b]. Then the volume of the solid is given by

limµ(P )→0

k∑j=1

α(ξi)∆xi =∫ b

aα(x) dx.

EXAMPLE 4.11 Let us compute the volume of the solid enclosedby the ellipsoid

x2

a2+y2

b2+z2

c2= 1.

For a fixed x ∈ [−a, a], the boundary of the cross section at x is givenby the equation

y2

b2+z2

c2= 1− x2

a2,

i.e.,

y2

φ(x)2+

z2

ψ(x)2= 1, where φ(x) = b

√1− x2

a2, ψ(x) = c

√1− x2

a2.

Hence,

α(x) = πφ(x)ψ(x) = πbc

(1− x2

a2

),

Computing Volume of a Solid of Revolution 63

and the required volume is

V :=∫ a

−aα(x) dx = πbc

∫ a

−a

(1− x2

a2

)dx =

43πabc.

In particular, volume of the solid bounded by the sphere x2+y2+z2 =a2 is 4

3πa3.

4.4 Computing Volume of a Solid of Revolu-tion

Suppose a solid is obtained by revolving a curve y = f(x), a ≤ x ≤ b,with x-axis as axis of revolution. We would like to find the volumeof the solid.

In this case the area of cross section at x is given by

α(x) = πy2 = π[f(x)]2, a ≤ x ≤ b.

Hence, the volume of the solid of revolution is

V :=∫ b

aα(x) dx = π

∫ b

ay2 dx.

EXAMPLE 4.12 Let us compute the volume of the solid of revolu-tion of the curve y = x2 about x-axis for −a ≤ x ≤ a. The requiredvolume is

V := π

∫ a

−ay2 dx = π

∫ a

−ax4 dx =

25a5.

EXAMPLE 4.13 We compute the volume of the solid of revolutionof the catenary

y =a

2

(ex/a + e−x/a

)about x-axis for 0 ≤ x ≤ b. The required volume is

V := π

∫ b

0y2 dx = π

∫ b

0

a2

4

(ex/a + e−x/a

)2dx =

πa2

4

∫ b

0

(e2x/a + e−2x/a + 2

).

We see that

V :=πa3

8

(e2b/a − e−2b/a

)+πb3

8.


4.5 Computing Area of Surface of Revolution

Suppose a solid is obtained by revolving a curve y = f(x), a ≤ x ≤ b,with x-axis as axis of revolution. We would like to find the area ofthe surface of the solid.


A := limµ(P )→0

k∑j=1

2πf(ξi)∆si,

where P : a = x0 < x1 < . . . < xk = b is a partition of the interval[a, b], and

∆si :=√

1 + [f ′(ξi)]2∆xi, i = 1, . . . , k.

Thus

A = limµ(P )→0

k∑j=1

2πf(ξi)√

1 + [f ′(ξi)]2∆xi = 2π∫ b

ay

√1 +

(dy

dx

)2

dx.

EXAMPLE 4.14 We find the surface of revolution of the parabolay2 = 2px, 0 ≤ x ≤ a for p > 0. The required area is

A = 2π∫ a

0y

√1 +

(dy

dx

)2

dx

= 2π∫ a

0

√2px√

1 +p

2xdx = 2π

√p

∫ a

0

√p+ 2x dx

= 2π√p23

[(2x+ p)3/2 1

2

]a

0

=2π√p

3

[(2a+ p)3/2 − p3/2

].

4.6 Centre of Gravity

Suppose A1, A2, . . . , An are material particles on the plane at coordi-nates (x1, y1), (x2, y2), . . . , (xn, yn) and masses m1,m2, . . .mn respec-tively. Then the centre of gravity of the system of these particlesis at the point A = (xC , yC), where

xC :=∑n

i=1 ximi∑ni=1mi

, yC :=∑n

i=1 yimi∑ni=1mi

.

Centre of Gravity 65

Now we attempt to define the centre of gravity of a material lineand material planar region enclosed by certain curves.

4.6.1 Centre of gravity of a material line in the plane

Suppose a curve L is given by the equation y = f(x), a ≤ x ≤ b. Weassume that this curve is a material line. Suppose the density of thematerial at the point X = (x, y) is γ(X). This density is defined asfollows: Suppose M(X, r) is the mass of an arc of the line containingthe point X with length r. Then the density of the material at thepoint x is defined by

γ(X) := limr→0

M(X, r)r

.

Now, in order to find the centre of gravity of L, we first consider apartition P : a = x0 < x1 < . . . < xk, and take points ξi = [xi−1, xi],i = 1, . . . , n. Then we take the the centre of gravity of the system ofmaterial points at (ξ1, f(ξ1), (ξ2, f(ξ2), . . . , (ξk, f(ξk) as

xC(P ) =∑n

i=1 ξiγi∆si∑ni=1 γi∆si

, yC(P ) :=∑n

i=1 f(ξi)γi∆si∑ni=1 γi∆si

.

Here, ∆si is the length of the arcs joining (xi−1, yi−1) to (xi, yi),and γi is the density at the point (ξi, f(ξi). Here yi = f(xi). Notethat γi∆si is the approximate mass of the arc joining (xi−1, yi−1) to(xi, yxi). Now, the centre of gravity of L is at (xC , yC), where

xC = limµ(P )→0

∑ni=1 ξiγi∆si∑ni=1 γi∆si

, yC := limµ(P )→0

∑ni=1 f(ξi)γi∆si∑n

i=1 γi∆si.

Assuming that the function γ(X) := γ(x, f(x)) is continuous on [a, b],we see that

xC =

∫ ba xγ(x, y)

√1 +

(dydx

)2dx

∫ ba γ(x, y)

√1 +

(dydx

)2dx

, yC =

∫ ba yγ(x, y)

√1 +

(dydx

)2dx

∫ ba γ(x, y)

√1 +

(dydx

)2dx

.

Example. We find the centre of gravity of the semi-circlular arcx2 + y2 = a2, y ≥ 0, assuming that the density of the material is


constant. In this case, y = f(x) :=√a2 − x2, so that it follows that√

1 +(dy

dx

)2

=a√

a2 − x2.

Hence, since γ(x, y) is constant,

xC = 0 yC =

∫ a−a y

√1 +

(dydx

)2dx

∫ a−a

√1 +

(dydx

)2dx

=2aπ.

4.6.2 Centre of gravity of a material planar region

Next we consider the centre of gravity of a material planar region Ωbounded by two curves

y = f(x), y = g(x), with f(x) ≤ g(x) a ≤ x ≤ b.

Suppose that the density of the material at the point X is γ(X).This density is defined as follows: Suppose M(X, r) is the mass ofthe circular region S(X, r) ⊆ Ω with centre at x and radius r > 0,and α(X, r) is the area of the same circular region. Then the densityof the material at the point x is defined by

γ(X) := limr→0

M(X, r)α(X, r)

.

Now, in order to find the centre of gravity of Ω, we first look at thefollowing special case: Suppose Ω is a rectangle given by a1 ≤ x ≤ b1,a2 ≤ y ≤ b2. Then we can infer that the centre of gravity of suchrectangle is located at the point(

a1 + b12

,a2 + b2

2

).

Taking the above obervation into account, we consider a partitionP : x0 < x1 < . . . < xk of the interval [a, b], and consider therectangular strips:

Ri : xi−1 ≤ x ≤ xi, f(ξi) ≤ y ≤ g(ξi), i = 1, . . . , k,

Centre of Gravity 67

where ξi = xi−1+xi

2 , i = 1, . . . , k. If γ is the (constant) density of thematerial, then the mass of the rectangular strip Ri is

mi = γ[g(ξi)− f(ξi)]∆xi, i = 1, . . . , k.

Assuming that the mass of the rectangular strip Ri is concentratedat its mid-point:

Xi :(ξi,f(ξi) + g(ξi)

2

),

we consider the centre of gravity of the system of material points atXi as

xC,P :=∑n

i=1 ξimi∑ni=1mi

, yC,P :=∑n

i=1f(ξi)+g(ξi)

2 mi∑ni=1mi

.

Now the centre of gravity of Ω is defined as

xC = limµ(P )→0

xC,P , yC = limµ(P )→0

yC,P ,

i.e.,

xC = limµ(P )→0

∑ni=1 ξiγ[g(ξi)− f(ξi)]∆xi∑ni=1 γ[g(ξi)− f(ξi)]∆xi

=

∫ ba x[g(x)− f(x)] dx∫ ba [g(x)− f(x)] dx

yC = limµ(P )→0

∑ni=1

12 [f(ξi + g(ξi)]γ[g(ξi)− f(ξi)]∆xi∑n

i=1 γ[g(ξi)− f(ξi)]∆xi

=12

∫ ba [f(x+ g(x)][g(x)− f(x)] dx∫ b

a [g(x)− f(x)] dx

EXAMPLE 4.15 We find the coordinates of the centre of gravityof a segment of a parabola y2 = a x cut off by the straight line x = a.

In this case

f(x) = −√a x, g(x) =

√a x, 0 ≤ x ≤ a.

Hence the coordinates of the centre of gravity are


xC =

∫ ba x[g(x)− f(x)] dx∫ ba [g(x)− f(x)] dx

=2∫ ba x√a x dx∫ b

a 2√a x dx

=35a.

yC =12

∫ ba [f(x+ g(x)][g(x)− f(x)] dx∫ b

a [g(x)− f(x)] dx= 0.

4.7 Moment of Inertia

Suppose there are nmaterial points in the plane with massesm1,m2, . . .mn

respectively. Suppose that these points are at distances d1, . . . , dn

from a fixed point O. Then the moment of inertia of the systemof these points with respect to the point O is defined by the quantity:

IO :=n∑

i=1

d2imi.

If O is the origin, and (x1, y1), (x2, y2), . . . , (xn, yn) are the points,then

IO :=n∑

i=1

(x2i + y2

i )mi.

4.7.1 Moment of inertia of a material line in the plane

Suppose a curve L is given by the equation y = f(x), a ≤ x ≤ b. Weassume that this curve is a material line. Suppose the density of thematerial at the point X = (x, y) is γ(X).

Now, in order to find the moment of inertia of L, we first considera partition P : a = x0 < x1 < . . . < xk, and take points ξi =[xi−1, xi], i = 1, . . . , n. Then we consider the moment of inertia of thesystem of material points at (ξ1, ηi), i = 1, . . . , n. Here, ηi = f(ξi),i = 1, . . . , n.

IO,P :=n∑

i=1

(ξ2i + η2i )mi.

Thus,

IO,P :=n∑

i=1

(ξ2i + η2i )γi∆si.

Moment of Inertia 69

Here, ∆si is the length of the arcs joining (xi−1, yi−1) to (ξi, yi), andγi is the density at the point (ξi, ηi). Note that γi∆si is the ap-proximate mass of the arc joining (xi−1, f(xi−1) to (xi, f(xi). Now,assuming that the functions f(x) and γ(x) := γ(x, f(x)) are contin-uous on [a, b], the moment of inertial of L with respect to O is

IO = limµ(P )→0

IO,P

= limµ(P )→0

n∑i=1

(ξ2i + η2i )γi∆si

=∫ b

a(x2 + y2)γ(x, y)

√1 +

(dy

dx

)2

dx.

4.7.2 Moment of inertia of a circular arc with respectto the centre

Suppose the given curve is a circular arc: ρ = a, α ≤ θ ≤ β. Follow-ing the arguments in the above paragraph, we compute the momentof inertia using polar coordinates:

The moment of inertia, in this, case is given by

IO := limµ(P )→0

n∑i=1

d2imi,

where di = a, mi = γia∆θi, for i = 1, . . . , n, so that

IO = limµ(P )→0

n∑i=1

a2γi[a∆θi] = a3

∫ β

αγ(θ)dθ.

Here, γ(θ) is the point density. If γ(θ) = γ, a constant, then

IO = a3

∫ β

αγ(θ)dθ = (β − α)γa3.

In particular, M.I of the circle ρ = a, 0 ≤ θ ≤ 2π, is

IO = 2πγa3.


4.7.3 Moment of inertia of a material sector in theplane

The region is R : 0 ≤ ρ ≤ a, α ≤ θ ≤ β with constant density γ. Tofind the M.I. of R, we partition it by rays and circular arcs:

P : α = θ0 < θ1 < θ2 < . . . < θn = β,

Q : 0 = ρ0 < ρ1 < ρ2 < . . . < ρm = a.

Consider the elementary region obtained by the above partition:

Rij : ρj−1 ≤ ρj ≤ a, θi−1 ≤ θi ≤ θi.

Assume that the the mass of this region Rij is concentrated at thepoint (ρj , θi), where ρj ∈ [ρj−1, ρj ], θi ∈ [θi−1, θi]. Then the MI of thematerial point at (ρj , θi) is mijd

2ij where mij is the mass of the region

Rij which is approximately equal to [ρj∆θi∆ρj ]γ, and dij = ρj . Thusthe MI of the sub-sector θi−1 ≤ θ ≤ θi is defined by

limµ(Q)→0

n∑j=1

mijd2ij = lim

µ(Q)→0

n∑j=1

[ρj∆θi∆ρj ]γρ2j

= limµ(Q)→0

n∑j=1

(γρ3

j∆ρj

)∆θi

= γ

(∫ a

0ρ3 dρ

)∆θi

=γa4

4∆θi.

From this, it follows that, the moment of inertia of the sector α ≤θ ≤ β is

limµ(P )→0

m∑i=1

γa4

4∆θi =

(β − α)γa4

4.

In particular, moment of inertia of a circular disc is

πγa4

2=Ma2

2,

where M = πa2γ is the mass of the disc.

Exercise 4.1 If M is the mass of a right circular homogeneous cylin-der with base radius a, then show that its moment of inertia is Ma2

2 .

Additional Exercises 71

4.8 Additional Exercises

1. Find the area of the portion of the circle x2 + y2 = 1 which liesinside the parabola y2 = 1− x.

[Hind: Area enclosed by the circle in the second and thirdquadrant and the area enclosed by the parabola in the first andfourth quadrant. The the required area is π

2 + 2∫ 10

√1− x dx.

Ans: π2 + 4

3 . ]

2. Find the area common to the cardioid ρ = a(1+ cos θ) and thecircle ρ = 3a

2 .

[Hind: The points of intersections of the given curves are givenby 1 + cos θ = 3

2 , i.e., for θ = ±π3 . Hence the required area is

2[

12

∫ π/30

(3a2

)2dθ + 1

2

∫ ππ/3 a

2(1 + cos θ)2 dθ]. Ans: 7

4π−9√

38 . ]

3. For a, b > 0, find the area included betwee the parabolas y2 =4a(x+ a) and y2 = 4b(b− x).

[Hind: Points of intersection of the curves is given by a(x+a) =b(b−x), i.e., x = b2−a2

a+b = b−a; y = 2√ab. The required area is

2×[∫ b−a−a

√4a(x+ a) +

∫ bb−a

√4b(b− x) dx

]. Ans:83

√ab(a+b).]

4. Find the area of the loop of the curve r2 cos θ = a2 sin 3θ

[Hint:r = 0 for θ = 0 and θ = π/3, and r is maximum forθ = π/6. The area is

∫ π/30

r2

2 dθ. ]

5. Find the area of the region bounded by the curves x− y3 = 0and x− y = 0.

[Hint: Points of intersections of the curves are at x = 0, 1,−1.The area is 2

∫ 10 (x1/3 − x)dx. Ans: 1/2 ]

6. find the area of the region that lies inside the circle r = a cos θand outside the cardioid r = a(1− cos θ).

[Hint: Note that the circle is the one with centre at (0, a/2) andradius a/2. The curves intersect at θ = ±π/3. The requiredarea is

∫ π/3−π/3(r

21 − r22)dθ, where r1 = a cos θ, r2 = a(1 − cos θ).

Ans: a3

3 (3√

3− π) ]


7. Find the area of the loop of the curve x = a(1 − t2), y =at(1− t2) for −1 ≤ t ≤ 1.

[Hint: y = 0 for t ∈ −1, 0, 1, and y negative for −1 ≤ t ≤ 0and positive for 0 ≤ t ≤ 1. Also, y2 = x2(a − x)/a so thatthe curve is symmetric w.r.t. the x-axis. Area is 2

∫ a0 ydx =

2∫ 01 y(t)x

′(t)dt. Ans: 8a2/15 ]

8. Find the length of an arch of the cycloid x = a(t − sin t),y = a(1− cos t).

[Hint: The curve cuts the x-axis at x = a and x = 2πa for t = 0and t = 2π respectively. Thus the length is

∫ 2π0

√[x′(t)]2 + [y′(t)]2dt.

Ans: 8a. ]

9. For a > 0, find the length of the loop of the curve 3a y2 =x(x− a)2.

[Hint: The curve cuts the x-axis at x = a, and the curveis symmetric w.r.t. the x-axis. Thus the required area is

2∫ a0

√1 +

(dydx

)2dx. Note that 6ayy′ = (x − a)(3x − a), so

that 1 + y′2 = (3x+a)2

12ax . Ans: 4a√3. ]

10. Find the length of the curve r = 21+cos θ , 0 ≤ θ ≤ π/2.

[Hind: ` :=∫ π/20

√r2 + [r′]2dθ = 2

∫ π/40 sec3 θdθ. Ans:

√2 +

ln(√

2 + 1). ]

11. Find the volume of the solid obtained by revolving the curvey = 4 sin 2x, 0 ≤ x ≤ π/2, about y-axis.

[Hint: writing y = 4 sin 2x, 0 ≤ x ≤ π/4 and y = 4 sin 2u,π/4 ≤ u ≤ π/2, the required volume is

∫ 40 (u2 − x2)dy =

π∫ π/2π/4 u

2(8 cos 2u)du− π∫ π/40 x2(8 cos 2x)dx.

Also, note that the curve is symmetric w.r.t. the line x = π/4.Hence, the required volume is given by π

∫ π/40 [(π

4 −x)2−x2]dy.

Ans: 2π2.]

12. Find the area of the surface obtained by revolving a loop of thecurve 9ax2 = y(3a− y)2 about y-axis.

Additional Exercises 73

[Hind: x = 0 iff y = 0 or y = 3a. The required area is

2π∫ 3a0 x

√1 +

(dxdy

)2dx. Ans: 3πa2. ]

13. Find the area of the surface obtained by revolving about x-axis,an arc of the catenary y = c cosh(x/c) between x = −a andx = a for a > 0.

[Hind: The area is 2π∫ a−a y

√1 + y′2 dx = 2π c

∫ a−a cosh2 x

c dx.Ans: πc [2a+ c sinh 2a

c ]. ]

14. The lemniscate ρ2 = a2 cos 2θ revolves about the line θ = π4 .

Find the area of the surface of the solid generated.

[Hind: The required surface is 2×2π∫ π/4−π/4 h

√ρ2 + ρ′2 dθ, where

h := ρ sin(

π4 − θ

),

ρ = a√

cos 3θ so that ρ2 + ρ′2 = a2

cos 2θ . Ans: 4πa2. ]

15. Find the volume of the solid generated by the cardioid ρ =a(1 + cos θ) about the initial line. [Ans: 8

3 . ]

5

Sequence and Series ofFunctions

5.1 Sequence of Functions

5.1.1 Pointwise Convergence and Uniform Convergence

Let J be an interval in R, and for each x ∈ J , let fn : J → R be afunction.

Definition 5.1 (a) We say that the sequence the sequence (fn)converges pointwise on J if for each x ∈ J , the sequence (fn(x))of real numbers converges.

(b) Suppose (fn) converges pointwise on J . Let f : J → R bedefined by f(x) = limn→∞ fn(x), x ∈ J . Then we say that (fn)converges to f pointwise on J .

Thus, (fn) converges to f pointwise on J if and only if if for everyε > 0 and for each x ∈ J , there exists N ∈ N (depending, in generalon both ε and x) such that |fn(x)− f(x)| < ε for all n ≥ N .

Remark 5.1 As in the case of sequence of real numbers, a sequence(fn) of functions defined on an interval J can b e thought of as afunction ϕ : N → F(J), where F(J) is the set of all real valuedfunctions defined on J .

EXAMPLE 5.1 Consider fn : R → R defined by fn(x) = sin(nx)n ,

x ∈ R and for n ∈ N. Then we see that for each x ∈ R, |fn(x)| ≤ 1/nfor all n ∈ N. Thus, (fn) converges pointwise to f on R, where f isthe zero function on R, i.e., f(x) = 0 foe very x ∈ R.

74

Sequence of Functions 75

Suppose (fn) is a sequence of functions defined on an intervalJ converges pontwise on J . As we have mentioned, it can happenthat for ε > 0, and for each x ∈ J , the number N ∈ N satisfying|fn(x) − f(x)| < ε ∀n ≥ N depends not only on ε but also on thepoint x. For instance, consider the following example.

EXAMPLE 5.2 Let fn(x) = xn for x ∈ [0, 1] and for n ∈ N. Thenwe see that for 0 ≤ x < 1, fn(x) → 0, and fn(1) → 1 as n → ∞.Thus, (fn) converges pointwise to a function f such that f(x) = 0for x ∈ [0, 1) and f(1) = 1. Suppose ε > 0, there exists N ∈ N suchthat |xn−f(x)| < ε for all n ≥ N and for all x ∈ [0, 1]. In particular,|xN | < ε for all for all x ∈ [0, 1), i.e., (1/|x|)N | > 1/ε for all x ∈ [0, 1),i.e.,

N >ln(1/ε)ln(1/|x|)

∀x ∈ [0, 1).

This is impossible, since ln(1/ε)ln(1/|x|) →∞ as x→ 1.

In case, we are able to find an N ∈ N which does not vary as xvaries over J and satisfying |fn(x)− f(x)| < ε ∀n ≥ N , then we saythat (fn) converges uniformly to f on J .

Definition 5.2 Suppose (fn) is a sequence of functions defined on aninterval J . We say that (fn) converges to a function f uniformlyon J if for every ε > 0 there exists N ∈ N (depending only on ε)such that |fn(x)− f(x)| < ε for all n ≥ N and for all x ∈ J .

Clearly, uniform convergence implies pointwise convergence. Butthe converse need not be true, as Example 5.2 shows.

Another way of seeing the nonuniform convergence of the se-quence (fn) in Example 5.2 is as follows: If we have uniform conver-gence, then for any ε > 0, there existsN ∈ N such that |xn| < ε for all

n ≥ N and for all x ∈ [0, 1). In particular, taking xn =(

n

n+ 1

)1/n

we must have |xnn| < ε for all n ≥ N . This is not possible if we had

chosen ε < 1, as |xnn| =

n

n+ 1.

Here is another example to the same effect.

EXAMPLE 5.3 Let fn(x) =nx

1 + n2x2for x ∈ J , where J is an

76 Sequence and Series of Functions

interval containing 0. Note that fn(0) = 0, and for x 6= 0, fn(x) → 0as n → ∞. Hence, we have pointwise convergence. We do not haveuniform convergence, as fn(1/n)) = 1/2 for all n.

EXAMPLE 5.4 Consider the sequence (fn) defined by fn(x) =tan−1(nx). Note that fn(0) = 0, and for x 6= 0, fn(x) → π/2 asn → ∞. Hence, the given sequence does not converge uniformly onany interval containing 0.

Examples 5.3 and 5.4 suggest the following:

Theorem 5.1 Suppose fn and f are functions defined on an intervalJ . If there exists a sequence (xn) in J and c 6= 0 such that an :=fn(xn)−f(xn) → c as n→∞, then (fn) does not converge uniformlyto f on J .

Proof. Suppose (fn) converges uniformly to f on J . Then takingε = |c|/2, there exists N ∈ N such that

‖fn(x)− f(x)| < |c|2

∀n ≥ N, ∀x ∈ J.

In particular,

‖fn(xn)− f(xn)| < |c|2

∀n ≥ N.

Now, taking limit as n→∞, it follows that

|c| = limn→∞

‖fn(xn)− f(xn)| < |c|2

which is a contradiction. Hence our assumption that (fn) convergesuniformly to f on J is wrong.

In the case of Example 5.3, xn = 1/n, and in the case of Example5.4, we may take xn = π/n, so that the assumptions in the abovetheorem are satisfied.

We may observe that in Examples 5.2 and 5.2, the limit functionf is not continuous, although every fn is continuous. This makes usto ask the following:


• Suppose each (fn) is a sequence of continuous function on Jwhich converges to f pointwise. Under what condition can we assertthat f is continuous?

Also, we may want to know the answers to the following questions:

• Suppose each (fn) is a sequence of continuous function on Jwhich converges pointwise to a continuous function f . Do we have∫ b

af(x)dx = lim

n→∞

∫ b

afn(x)dx

for every [a, b] ⊆ J?

• Suppose each (fn) is a sequence of continuously differentiablefunctions on J which converges to a function f . Then, is the functionf differentiable on J? If f is differentiable on J , then do we have therelation

d

dxf(x) = lim

n→∞

d

dxfn(x)dx ?

The answers to the above two questions need not be affirmativeas the following two examples show.

EXAMPLE 5.5 Let fn(x) = nx(1 − x2)n for 0 ≤ x ≤ 1 and forn ∈ N. Then we see that

limn→∞

fn(x) = 0 ∀x ∈ [0, 1].

But, ∫ 1

0fn(x)dx =

n

2n+ 2→ 1

2as n→∞.

EXAMPLE 5.6 Let fn(x) = sin(nx)√n

for x ∈ R and for n ∈ N. Thenwe see that

limn→∞

fn(x) = 0 ∀x ∈ [0, 1].

But,f ′n(x) =√n cos(nx) for all n ∈ N, so that

f ′n(0) =√n→∞ as n→∞.


Continuity of the Limit Function

Theorem 5.2 Suppose (fn) is a sequence of continuous functionsdefined on an interval J which converges uniformly to a function f .Then f is continuous on J .

Proof. Suppose x0 ∈ J . Then for any x ∈ J and for any n ∈ N,

|f(x)− f(x0)| ≤ |f(x)− fn(x)|+ |fn(x)− fn(x0)|+ |fn(x0)− f(x0)|.(∗)

Let ε > 0 be given. Since (fn) converges to f uniformly, there existsN ∈ N such that

|fn(x)− fn(x)| < ε ∀n ≥ N, ∀x ∈ J.

Hence from (∗), we have

|f(x)− f(x0)| ≤ ε+ |fN (x)− fn(x0)|+ ε.

Now, since fN is continuous, there exists δ > 0 such that

|fN (x)− fn(x0)| < ε whenever |x− x0| < δ.

Hence, it follows that

|f(x)− f(x0)| < 3ε whenever |x− x0| < δ.

Thus, f is continuous at x0. This is true for all x0 ∈ J , show is thatf is a continuous function.

Integration and Uniform Convergence

Theorem 5.3 Suppose (fn) is a sequence of continuous functionsdefined on an interval J which converges uniformly to a function f .Then f is continuous and for any [a, b] ⊆ J , and

limn→∞

∫ b

afn(x)dx =

∫ b

af(x)dx.

Proof. We already know by Theorem 5.2 that f is a continuousfunction. Next we note that∣∣∣∣∫ b

afn(x)dx−

∫ b

af(x)dx

∣∣∣∣ ≤ ∫ b

a|fn(x)− f(x)|dx.


let ε > 0 be given. By uniform convergence of (fn) to f , there existsN ∈ N such that

|fn(x)− f(x)| < ε ∀n ≥ N, ∀x ∈ [a, b].

Hence, for all n ≥ N ,∣∣∣∣∫ b

afn(x)dx−

∫ b

af(x)dx

∣∣∣∣ ≤ ∫ b

a|fn(x)− f(x)|dx < ε(b− a).


Differentiation and Uniform Convergence

Theorem 5.4 Suppose (fn) is a sequence of continuously differen-tiable functions defined on an interval J such that

(i) (f ′n) converges uniformly to a function, and

(ii) (fn(a)) converges for some a ∈ J .

Then (fn) converges to a continuously differentiable function f and

limn→∞

f ′n(x) = f ′(x) ∀x ∈ J.

Proof. Let g(x) := limn→∞

f ′n(x) for xinJ , and α := limn→∞

fn(a). Since

the convergence of (f ′n) to g is uniform, by Theorem 5.3, the functiong is continuous and

limn→∞

∫ x

af ′n(t)dt =

∫ x

ag(t)dt.

Let ϕ(x) :=∫ xa g(t)dt, x ∈ J . Then ϕ is differentiable and ϕ′(x) =

g(x) for x ∈ J . But,∫ xa f

′n(t)dt = fn(x)− fn(a). Hence, we have

limn→∞

[fn(x)− fn(a)] = ϕ(x).

Thus, (fn) converges pointwise to a differentiable function f definedby f(x) = ϕ(x) + α, x ∈ J , and (f ′n) converges to f ′.

The following obvious from the above theorem.


Corollary 5.5 Suppose (fn) is a sequence of continuously differ-entiable functions defined on an interval J such that (fn) convergespoitwise to a function f , and (f ′n) converges uniformly on J . Then

d

dx

(lim

n→∞fn(x)

)= lim

n→∞

d

dxfn(x) ∀x ∈ J.

5.2 Series of Functions

Definition 5.3 By a series of functions we mean an expression

of the form∞∑

n=1

fn or∞∑

n=1

fn(x), where (fn) is a sequence of functions

defined on some interval J .

Definition 5.4 Given a series∑∞

n=1 fn(x) of functions, let

sn(x) :=n∑

i=1

fi(x), x ∈ J.

Then sn is called the n-th partial sum of the series∞∑

n=1

fn.

Definition 5.5 (a) We say that a series∑∞

n=1 fn(x) converges ata point x0 ∈ J if the sequence (sn(x0)) converges,

(b) the series∑∞

n=1 fn(x) converges pointwise on J if thesequence (sn) converges pointwise on J , and

(c) the series∑∞

n=1 fn(x) converges uniformly on J if thesequence (sn) converges uniformly on J .

The proof of the following two theorems are obvious from thestatements of Theorems 5.3 and 5.4 respectively.

Theorem 5.6 Suppose (fn) is a sequence of continuous functionson J . If

∑∞n=1 fn(x) converges uniformly on J , say to f(x), then f

is continuous on J , and for [a, b] ⊆ J ,∫ b

af(x)dx =

∞∑n=1

∫ b

afn(x)dx.

Theorem 5.7 Suppose (fn) is a sequence of continuously differen-tiable functions on J . If

∑∞n=1 f

′n(x) converges uniformly on J , and

Series of Functions 81

if∑∞

n=1 fn(x) converges at some point x0 ∈ J , then∑∞

n=1 fn(x)converges to a differentiable function on J , and

d

dx

( ∞∑n=1

fn(x)

)=

∞∑n=1

f ′n(x).

Next we consider a useful sufficient condition to check uniformconvergence. First a definition.

Definition 5.6 We say that∑∞

n=1 fn is a dominated series if thereexists a sequence αn of positive real numbers such that |fn(x)| ≤αn for all n ∈ N and for all x ∈ J , and the series

∑∞n=1 αn converges.

Theorem 5.8 A dominated series converges uniformly.

Proof. Let∑∞

n=1 ϕn be a dominated series defined on an intervalJ , and let (αn) be a sequence of positive reals such that

(i) |fn(x)| ≤ αn for all n ∈ N and for all x ∈ J , and

(ii)∑∞

n=1 αn converges.

Let sn(x) =∑n

i=1 fi(x), n ∈ N. Then for n > m,

|sn(x)−sm(x)| =

∣∣∣∣∣n∑

i=m+1

fi(x)

∣∣∣∣∣ ≤n∑

i=m+1

|fi(x)| ≤n∑

i=m+1

αi = σn−σm,

where σn =∑n

k=1 αk Since∑∞

n=1 αn converges, the sequence σn isa Cauchy sequence. Now, let ε > 0 be given, and let N ∈ N be suchthat

|σn − σm| < ε ∀n,m ≥ N.

Hence, from the relation: |sn(x)− sm(x)| ≤ σn − σm, it follows that

|sn(x)− sm(x)| < ε ∀n,m ≥ N, ∀x ∈ J.

This, in particular implies that sn(x) is also a Cauchy sequenceat each x ∈ J . Hence, sn(x) converges for each x ∈ J . Letf(x) = limn→∞ sn(x), x ∈ J . Then, we have

|f(x)− sm(x)| = limn→∞

|sn(x)− sm(x)| < ε ∀m ≥ N, ∀x ∈ J.

Thus, the series∑∞

n=1 fn converges uniformly f on J .


5.2.1 Examples

EXAMPLE 5.7 The series∞∑

n=1

cosnxn2

and∞∑

n=1

sinnxn2

are domi-

nated series, since∣∣∣cosnxn2

∣∣∣ ≤ 1n2,

∣∣∣∣sinnxn2

∣∣∣∣ ≤ 1n2

∀n ∈ N

and∞∑

n=1

1n2

is convergent.


n=0 xn is a dominated series on [−ρ, ρ]

for 0 < ρ < 1, since |xn| ≤ ρn for all n ∈ N and∑∞

n=0 ρn is conver-

gent.


n=1 xn−1 is not uniformly convergent

on (0, 1); in particular, not dominated on (0, 1). This is seen asfollows: Note that

sn(x) :=n∑

k=1

xk−1 =1− xn

1− x→ f(x) :=

11− x

as n→∞.

Hence, for ε > 0,

|f(x)− sn(x)| < ε ⇐⇒∣∣∣∣ xn

1− x

∣∣∣∣ < ε.

Hence, if there exists N ∈ N such that |f(x) − sn(x)| < ε for alln ≥ N for all x ∈ (0, 1), then we would get

|x|N

|1− x|< ε ∀x ∈ (0, 1).

This is not possible, as |x|N/|1− x| → ∞ as x→ 1.


n=1(1 − x)xn−1 is not uniformlyconvergent on [0, 1]; in particular, not dominated on [0, 1]. This isseen as follows: Note that

sn(x) :=n∑

k=1

(1−x)xk−1 = (1−x)1− xn

1− x=

1− xn if x 6= 10 if x = 1.

Series of Functions 83

Note that

sn(x) → f(x) :=

1 if x 6= 10 if x = 1.

Thus, |sn(x)− f(x)| =xn if x 6= 10 if x = 1.

Hence, for 0 < ε < 1

and x 6= 1,

|f(x)− sn(x)| < ε ⇐⇒ |xn| < ε.

Hence, if there exists N ∈ N such that |f(x) − sn(x)| < ε for alln ≥ N for all x ∈ [0, 1], then we would get |x|N < ε for all x ∈ [0, 1).This is not possible, as |x|N → 1 as x→ 1.

Remark 5.2 Note that if a series∑∞

n=1 fn converges uniformly toa function f on an interval J , then we must have

supx∈J

|sn(x)− f(x)| → 0 as n→∞.

Here, sn is the n-th partial sum of the series. Thus, if∑∞

n=1 fn

converges to a function f on J , and if supx∈J |sn(x) − f(x)| 6→ 0 asn → ∞, then we can infer that the convergence is not uniform. Asan illustration, consider the Example 5.10. There we have

|sn(x)− f(x)| =xn if x 6= 10 if x = 1.

Hence, sup|x|≤1 |sn(x)− f(x)| = 1


n=1

x

n(1 + nx2)on R. Note

thatx

n(1 + nx2)≤ 1n

(1

2√n

),

and∞∑

n=1

1n3/2

converges. Thus, the given series is dominated series,

and hence it converges uniformly on R.


n=1

x

1 + n2x2for x ∈ [c,∞),

c > 0. Note that

x

1 + n2x2≤ x

n2x2=≤ 1

n2x≤ 1n2c


and∞∑

n=1

1n2

converges. Thus, the given series is dominated series, and

hence it converges uniformly on [c,∞).

Next examples shows that in Theorem 5.7, the condition that thederived series converges uniformly is not a necessary condition forthe the conclusion.

EXAMPLE 5.13 Consider the series∑∞

n=0 xn. We know that it

converges to 1/(1− x) for |x| < 1. It can be seen that the derivedseries

∑∞n=1 nx

n−1 converges uniformly for |x| ≤ ρ for any ρ ∈ (0, 1).This follows since

∑∞n=1 nρ

n−1 converges. Hence,

1(1− x)2

=d

dx

11− x

=∞∑

n=1

nxn−1 for |x| ≤ ρ.

The above realtion is true for x in any open interval J ⊆ (−1, 1);because we can choose ρ sucfficieltly close to 1 such that J ⊆ [−ρ, ρ].Hence, we have

1(1− x)2

=∞∑

n=1

nxn−1 for |x| < 1.

Exercises.

1. Let fn(x) = x2

(1+x2)n for x ≥ 0. Show that the series∑∞

n=1 fn(x)does not converge uniformly.

2. Let fn(x) = x1+nx2 , x ∈ R. Show that (fn) converge uni-

formly, whereas (f ′n) does not converge uniformly. Is the re-lation limn→∞ f ′n(x) = (limn→∞ fn(x))′ true for all x ∈ R?

3. Let fn(x) = log(1+n3x2)n2 , and gn(x) = 2nx

1+n3x2 for x ∈ [0, 1].Show that the sequence (gn) converges uniformly to g whereg(x) = 0 for all x ∈ [0, 1]. Using this fact, show that (fn) alsoconverges uniformly to the zero function on [0, 1].

4. Let fn(x) =

n2x, 0 ≤ x ≤ 1/n,−n2x+ 2n, 1/n ≤ x ≤ 2/n,0, 2/n ≤ x ≤ 1.

Show that (fn) does not converge uniformly of [0, 1].

Hint: Use termwise integration.

6

Power Series

6.1 Convergence and Absolute convergence

Power series is a particular case of series of functions.

Definition 6.1 A series of the form∑∞

n=0 anxn is called a power

series. Here, an is a sequence of real numbers.

Note that a power series∑∞

n=0 anxn converges at the point x = 0.

What can we say about its domain of convergence?

Theorem 6.1 (Abel’s Theorem) If∑∞

n=0 anxn converges at a point

x0, then it converges absolutely for every x with |x| < |x0|.

Proof. (i) Suppose∑∞

n=0 anxn converges at a point x0 6= 0. Let

x be such that |x| < |x0|. Then, since we have

|anxn| = |anx

n0 |∣∣∣∣ xx0

∣∣∣∣n ∀n.

Since |anxn0 | → 0 as n →∞, there exists M > 0 such that |anx

n0 | ≤

M for all n, so that we have

|anxn| ≤M

∣∣∣∣ xx0

∣∣∣∣n ∀n.

Now, since∣∣∣ xx0

∣∣∣ < 1, it follows, by comparison test that∑∞

n=0 |anxn|

converges.

The following corollary is an immediate consequence of Abel’sTheorem 6.1.

85

86 Power Series

Corollary 6.2 If∑∞

n=0 |anxn| diverges at a point u0, then the series∑∞

n=0 anxn diverges for every x with |x| > |u0|.

Theorem 6.1 shows that if∑∞

n=0 anxn converges at a non-zero

point, if it diverges at some point, then there exists R > 0 such thatit converges for all x such that |x| < R, and it diverges at all x with|x| > R. In particular, the set of all x such that

∑∞n=0 anx

n convergesis either the singleton set 0 or an an interval. In fact, the abovenumber R is

R = sup|x| :∞∑

n=0

anxn convergesatx.

Definition 6.2 The domain (or interval) of convergence of apower series

∑∞n=0 anx

n is the set

D := x ∈ R :∞∑

n=0

anxn convergesatx.

andR := sup|x| : x ∈ D

is called the radius of convergence of∑∞

n=0 anxn.

Thus, a number R with 0 < R < ∞ is called the radius of con-vergence of

∑∞n=0 anx

n if and only if∑∞

n=0 anxn converges for all x

with |x| < R, and diverges at all x with |x| > R. If the power series∑∞n=0 anx

n converges only at the point 0, then the radius of conver-gence is 0, and if it converges at all points in R, then sup|x| : x ∈ Ddoes not exists, and in that case we say that the radius of convergenceis ∞, i.e., we write R = ∞.

EXAMPLE 6.1 Consider the power series∑∞

n=0 xn. In this case,

we know that the series converges for x with |x| < 1, and divergesfor for x with |x| > 1. Also, the series diverges for x ∈ 1,−1.Hence, its radius of convergence is 1, and its domain (interval) ofconvergence is D = x : −1 < x < 1 = (−1, 1).


n=0xn

n . In this case,we know that the series converges at x = −1 and diverges at x = 1.Hence, its radius of convergence is 1, and its domain (interval) ofconvergence is [−1, 1).

Convergence and Absolute convergence 87


n=0xn

n2 . We know thatthis series converges at x = 1 and x = −1. Since

|xn+1/(n+ 1)2||xn/n2|

= |x| n

n+ 1→ |x| as n→∞,

by ratio test the series∑∞

n=0

∣∣xn

n2

∣∣ diverges for x with |x| > 1. Hence,it cannot converge at any x with |x| > 1. Therefore, the radius ofconvergence is 1, and the domain (interval) of convergence is [−1, 1].


n=0xn

n! . Since

|xn+1/(n+ 1)!||xn/n!|

= |x| 1n+ 1

→ 0 as n→∞,

by ratio test the series converges at every x ∈ R. Hence, the radiusof convergence is ∞, and the domain (interval) of convergence is R.

For finding the radius of convergence and domain of convergence,the following theorem will be useful.

CONVENTION: In the following, we use the following convention:

(i) If an ≥ 0 for all n ∈ N and an →∞ as n→∞, then we writelimn→∞ an = ∞.

(ii) If c = 0, then we write 1/c = ∞, and if c = ∞, then we write1/c = 0.

Theorem 6.3 Consider the power series∑∞

n=0 anxn, and let R be

its radius of convergence.

(a) If limn→∞

∣∣∣an+1

an

∣∣∣ = L ∈ [0,∞], then R = 1/L.

(b) If limn→∞ |an|1/n = ` ∈ [0,∞], then R = 1/`.

Proof. Let un(x) = anxn.

(a) In this case, we have limn→∞

∣∣∣∣un+1(x)un(x)

∣∣∣∣ = L|x|. Now, suppose

that 0 < L < ∞. Then, by d’Alembert’s ratio test, the series∑∞n=0 |anx

n| converges absolutely for |x| < 1/L, and∑∞

n=0 |anxn|

diverges for |x| > 1/L. Since absolute convergence implies conver-gence, we have convergence of

∑∞n=0 anx

n for |x| < 1/L.

88 Power Series

We claim that∑∞

n=0 anxn diverges for |x| > 1/L. Suppose this is

not true. Then there exists x0 such that |x0| > 1/L and∑∞

n=0 anxn

converges at x = x0. Then, taking a point u0 such that 1/L < |u0| <|x0|, it follows by Abel’s Theorem 6.1, that the series

∑∞n=0 |anu

n0 |

converges. This contradicts the fact that∑∞

n=0 |anxn| diverges for

|x| > 1/L. Thus, we justified our claim.

b) In this case, limn→∞

|un+1(x)|1/n = `|x|. Hence, the result followsby making use of Cauchy’s root test, and following the arguments asin (a) above.

Exercise 6.1 Find the interval of convergence of the following powerseries.

(i)∑∞

n=1xn

2n−1 . Ans: [−1, 1) (ii)∑∞

n=1xn

n×4n . Ans: [−4, 4](iii)

∑∞n=0

xn

n! . Ans: R (iv)∑∞

n=0 n!xn Ans: 0.

6.2 Integration and Differentiation

Theorem 6.4 Suppose R > 0 is the radius of convergence of a powerseries

∑∞n=0 anx

n. Then for any ρ with 0 < ρ < R, the series

∞∑n=0

anxn and

∞∑n=1

nanxn−1

converge uniformly on [−ρ, ρ]. Moreover, the function f defined byf(x) :=

∑∞n=0 anx

n, x ∈ (−R,R), is continuous and the radius ofconvergence of

∑∞n=1 nanx

n−1 is R.

Proof. Let 0 < ρ < R, and r such that ρ < r < R. Then forevery x with |x| ≤ ρ, we have

|anxn| ≤ |anr

n|(xr

)n| ≤ |anr

n|(ρr

)n.

Since the series∑∞

n=0 |anρn is convergent, the sequence (anr

n) isbounded, say |anr

n| ≤ M for all n ∈ N, for some M > 0. Also,since ρ

r < 1,∑∞

n=0 anxn is a dominated on [−ρ, ρ]. Hence the series∑∞

n=0 anxn is uniformly convergent on [−ρ, ρ].

Integration and Differentiation 89

By Theorem 6.4, the function f defined by f(x) :=∑∞

n=0 anxn,

x ∈ (−R,R), is continuous on [−ρ, ρ]. Since this is true for any ρwith 0 < ρ < R, it follows that f is continuous on (−R,R).

Next, we note that

|nanxn−1| ≤ n|anρ

n−1| ≤ n|anrn−1|

(ρr

)n−1∀n ∈ N.

Since |anrn−1| converges to 0, there existsM > 0 such that |anr

n−1| ≤M for all n ∈ N. Thus, |nanx

n−1| ≤Mn(ρ

r

)n−1 for all n ∈ N. Sinceρr < 1, it follows that the series

∑∞n=1 nanx

n−1 is dominated on[−ρ, ρ]. In particular, it converges uniformly on [−ρ, ρ].

It remains to show thatR is the radius of convergence of∑∞

n=1 nanxn−1.

Suppose x0 ∈ R such that |x0| < R, and let ρ be such that |x0| <ρ < R. Then we know that

∑∞n=1 nanx

n−1 converges uniformly on[−ρ, ρ], in particular,

∑∞n=1 nanx

n−10 converges. Hence, the radius of

convergence of∑∞

n=1 nanxn−1 is at least R.

Suppose∑∞

n=1 nanxn−1 converges at some point u with |u| >

R. Then taking r with |u| > r > R, we see that∑∞

n=1 nanrn−1

converges. But, |nanrn−1| ≥ |anr

n|/r so that by comparison test∑∞n=1 anr

n converges. This is not possible since r > R. Thus, theradius of convergence of

∑∞n=1 nanx

n−1 is R.

Theorem 6.5 Suppose R > 0 is the radius of convergence of a powerseries

∑∞n=0 anx

n, and let f(x) :=∑∞

n=0 anxn for |x| < R. Then we

have the following:

(a) f(x) is a continuous function for |x| < R, and for [a, b] ⊆(−R,R), ∫ b

af(x)dx =

∞∑n=0

∫ b

a

an

n+ 1[bn+1 − an+1].

(b) f(x) is differentiable for every x ∈ (−R,R), and

d

dxf(x) =

∞∑n=1

nanxn−1.

Proof. (a) This part follows from Theorems 6.4 and 5.6.

90 Power Series

(b) By Theorems 6.4 and 5.7 it follows that

d

dxf(x) =

∞∑n=1

nanxn−1

for every x ∈ [−ρ, ρ] for any ρ with 0 < ρ < R. Now, if x ∈ (−R,R),then we may take ρ such that x ∈ [−ρ, ρ], and the result is valid forsuch x as well.

6.3 Series that can be converted into a powerseries

Some of the series may not be in the standard form∑∞

n=0 anxn, but

can be converted into this form after some change of variable. Forexample, consider the series

(i)∞∑

n=0

an(x−x0)n, (ii)∞∑

n=0

anx3n, (iii)

∞∑n=0

an1xn, (iv)

∞∑n=0

an sinn x.

In each of these cases, we may take a new variable as follows: In (i)y = x− x0, in (ii) y = x3, in (iii) y = 1

x , and in (iv) y = sinx.

Exercises.

1. Find the interval convergence of the following power series.

(i)∑∞

n=1(−1)n

nxn . Ans: (−∞,−1) ∪ [1,∞)

(ii)∑∞

n=1(−1)n3n

(4n−1)xn . Ans: (−∞,−3) ∪ [3,∞),

(iii)∑∞

n=1n(x+5)n

(2n+1)3. Ans: [−6,−4]

(iv)∑∞

n=12n sinn x

n2 . Ans: [−π6 + kπ, π

6 + kπ], k ∈ Z.

2. Find the radius of convergence of∑∞

n=0(n−1)!

nn xn. Ans: e

3. Show that

(i) log(1 + x) =∑∞

n=1(−1)n+1 xn

n for −1 < x ≤ 1,

(ii) tan−1 x =∑∞

n=0(−1)n x2n+1

2n+1 for −1 < x ≤ 1,

(iii) π4 =

∑∞n=0(−1)n 1

2n+1 - the Gregory-Nilakantha series.

7

Fourier Series

While studying the heat conduction problem in the year 1804, Fourierfound it necessary to use a special type of function series associatedwith certain functions f , later known as Fourier series of f . In thischapter we study such series of functions.

7.1 Fourier Series of 2π-Periodic functions

7.1.1 Fourier Series and Fourier Coefficients

Definition 7.1 A series of the form

c0 +∞∑

n=1

(an cosnx+ bn sinnx)

is called a trigonometric series. Here (an) and (bn) are sequencesof real numbers.

We observe that the functions cosnx and sinnx are 2π-periodic.Hence, if c0 +

∑∞n=1 (an cosnx+ bn sinnx) converges to a function

f(x), the f(x) also has to be 2π-periodic. Thus, only a 2π-periodicfunction is expected to have a trigonometric series expansion.

Definition 7.2 A function f : R → R is said to be T -periodic forsome T > 0 if f(x+ T ) = f(x for all x ∈ R.

Now, suppose that f is a 2π-periodic function. We would like toknow whether f can be represented as a Fourier series. Suppose, for

91

92 Fourier Series

a moment, that we can write

f(x) = c0 +∞∑

n=1

(an cosnx+ bn sinnx) ∀x ∈ R.

Then what should be an, bn? To answer this question, let us furtherassume that the series can be termwise integrated. For instance if theabove series is uniformly convergent to f , then termwise integration is

possible; in particular, if (an) and (bn) are such that∞∑

n=0

(|an|+ |bn|)

converges. Observe that∫ π

−πcosnx cosmxdx =

0, if n = mπ, if n 6= m,

∫ π

−πsinnx sinmxdx =

0, if n = mπ, if n 6= m,∫ π

−πcosnx sinmxdx = 0.

Therefore, under the assumption that the series can be integratedtermwise, we get ∫ π

−πf(x)dx = 2πc0,∫ π

−πf(x) cosnxdx = anπ ∀n ∈ N,∫ π

−πf(x) sinnxdx = bnπ ∀n ∈ N.

Thus,

c0 =12π

∫ π

−πf(x)dx

an =1π

∫ π

−πf(x) cosnxdx, n ∈ N,

bn =1π

∫ π

−πf(x) sinnxdx, n ∈ N.

Fourier Series of 2π-Periodic functions 93

Definition 7.3 The Fourier series of a 2π-periodic function f isthe trigonometric series

a0

2+

∞∑n=1

(an cosnx+ bn sinnx) ,

where

an =1π

∫ π

−πf(x) cosnxdx, bn =

1π

∫ π

−πf(x) sinnxdx.

We may write this fact as

f(x) ∼ a0

2+

∞∑n=1

(an cosnx+ bn sinnx) .

The numbers an and bn are called the Fourier coefficients of f .

The following two theorems give sufficient conditions for the con-vergence of the Fourier series of a function f to the function f atcertain points x ∈ R.

Theorem 7.1 Suppose f is a bounded monotonic function on [−π, π).Then the Fourier series of f converges to a function s(x), where

s(x) =f(x) if f continuous at x,12 [f(x−) + f(x+)] if f not continuous at x.

Theorem 7.2 (Dirichlet’s Theorem) Suppose f : R → R is a 2π-periodic function which is piecewise differentiable on (−π, π). Thenthe Fourier series of f converges to a function s(x), where

s(x) =f(x) if f continuous at x,12 [f(x−) + f(x+)] if f not continuous at x.

Remark 7.1 It is known that there are continuous functions f de-fined on [−π, π] whose Fourier series does not converge point wiseto f . Its proof relies on concepts from advanced mathematics (cf.M.T.Nair, Functional Analysis: A First Course, Prentice-Hall of In-dia, new delhi, 2002).

We may observe the following:

94 Fourier Series

• Suppose f is an even function. Then f(x) cosnx is an evenfunction and f(x) sinnx is an odd function. Hence bn = 0 for alln ∈ N, so that in this case the Fourier series of f is

s(x) =a0

2+

∞∑n=1

an cosnx

withan =

2π

∫ π

0f(x) cosnxdx.

In particular,

s(0) =∞∑

n=0

an, s(π) = a0 +∞∑

n=1

(−1)nan.

• Suppose f is an odd function. Then f(x) cosnx is an oddfunction and f(x) sinnx is an even function. Hence an = 0 for alln ∈ N ∪ 0, so that in this case the Fourier series of f is

∞∑n=1

bn sinnx with bn =2π

∫ π

0f(x) sinnxdx. (∗∗)

In particular,

s(π/2) =∞∑

n=0

(−1)nb2n+1.

7.1.2 Even and Odd Expansions

Suppose a function is defined on [0, π). Then we may extend it to[−π, π) in any manner, and then extend to all of R periodically, thatis by defining f(x ± 2π) = f(x). In this fashion we can get manyseries expansions of f all of which coincide on [0, π].

For example, by defining

fodd(x) =f(x) if 0 ≤ x < π,−f(−x) if − π ≤ x < 0,

,

feven(x) =f(x) if 0 ≤ x < π,f(−x) if − π ≤ x < 0,

Fourier Series of 2π-Periodic functions 95

then we may observe that

fodd(−x) = −fodd(x), feven(−x) = feven(x) ∀x ∈ [−π, π],

so that fodd and feven are the odd extension and even extension of frespectively. Therefore,

f(x) ∼ a0

2+

∞∑n=1

an cosnx, x ∈ [0, π), (∗)

and

f(x) ∼∞∑

n=1

bn sinnx, x ∈ [0, π), (∗∗)

with

an =2π

∫ π

0f(x) cosnxdx, bn =

2π

∫ π

0f(x) sinnxdx.

The expansions (∗) and (∗∗) are called, respectively, the even andodd expansions of f on [0, π).

7.1.3 Examples

EXAMPLE 7.1 Consider the 2π-periodic function f with f(x) = xfor x ∈ [−π, π]. Note that the f is an odd function. Hence, an = 0for n = 0, 1, 2, . . ., and the Fourier series is

∞∑n=1

bn sinnx, x ∈ [0, π]

with

bn =2π

∫ π

0x sinnxdx =

2π

[−xcosnx

n

]π0

+∫ π

0

cosnxn

dx

=

2π

−π cosnπ

n

=

(−1)n+12n

.

Thus the Fourier series is

2∞∑

n=1

(−1)n+1

nsinnx.

In particular (using Dirichlet’s theorem), with x = π/2 we have

π

4=

∞∑n=1

(−1)n+1

nsin

nπ

2=

∞∑n=0

(−1)n+1

2n+ 1.

96 Fourier Series

EXAMPLE 7.2 Consider the 2π-periodic function f with f(x) =|x| for x ∈ [−π, π]. Note that the f is an even function. Hence,bn = 0 for n = 1, 2, . . ., and the Fourier series is

a0

2+

∞∑n=1

an cosnx, x ∈ [0, π], an =2π

∫ π

0x cosnxdx.

It can be see that a0 = π, and

a2n = 0, a2n+1 =−4

π(2n+ 1)2, n = 0, 1, 2, . . . .

Thus,

|x| ∼ π

2− 4π

∞∑n=0

cos(2n+ 1)x(2n+ 1)2

, x ∈ [0, π].

Taking x = 0 (Using Dirichlet’s theorem), we have

π2

8=

∞∑n=0

1(2n+ 1)2

.

EXAMPLE 7.3 Consider the 2π-periodic function f with

f(x) =−1, −π ≤ x < 0,

1, 0 ≤ x ≤ π.

Note that the f is an odd function. Hence, an = 0 for n = 0, 1, 2, . . .,and the Fourier series is

∞∑n=1

bn sinnx, x ∈ [0, π]

with

bn =2π

∫ π

0f(x) sinnxdx =

2π

∫ π

0sinnxdx =

2π

(1− cosnπ).

Thus

f(x) ∼ 4π

∞∑n=0

sin(2n+ 1)x2n+ 1

.

Taking x = π/2, we have

π

4=

∞∑n=0

(−1)n

2n+ 1.

Fourier Series of 2`-Periodic functions 97

EXAMPLE 7.4 Consider the 2π-periodic function f with f(x) =x2 for x ∈ [−π, π]. Note that the f is an even function. Hence,bn = 0 for n = 1, 2, . . ., and the Fourier series is

a0

2+

∞∑n=1

an cosnx, x ∈ [0, π], an =2π

∫ π

0x cosnxdx.

It can be see that a0 = 2π2/3, and an = (−1)n4/n2. Thus

x2 ∼ π2

3+ 4

∞∑n=1

(−1)n cosnxn2

, x ∈ .

Taking x = 0 and x = π (Using Dirichlet’s theorem), we have

π2

12=

∞∑n=1

(−1)n+1

n2,

π2

6=

∞∑n=1

1n2

respectively.

7.2 Fourier Series of 2`-Periodic functions

Suppose f is a T -periodic function. We may write T = 2`. Then wemay consider the change of variable t = πx/` so that the functionf(x) = f(`t/π), as a function of t is 2π-periodic. Hence, its Fourierseries is

a0

2+

∞∑n=1

(an cosnt+ bn sinnt)

where

an =1π

∫ π

−πf

(`t

π

)cosntdt =

1`

∫ `

−`f(x) cos

nπx

`dx,

bn =1π

∫ π

−πf

(`t

π

)sinntdt =

1`

∫ `

−`f(x) sin

nπx

`dx.

In particular,

• if f is even, then bn = 0 for all n and

an =2`

∫ `

0f(x) cos

nπx

`dx,

98 Fourier Series

• if f is odd, then an = 0 for all n and

bn =2`

∫ `

0f(x) sin

nπx

`dx,

7.2.1 Fourier series of Functions on Arbitrary intervals

Suppose a function f is defined in an interval [a, b). We can obtainFourier expansion of it on [a, b) as follows:

Method 1: Let us consider a change of variable as y = x− a+b2 . Let

ϕ(y) := f(x) = f(y+ a+b2 ) where −` ≤ y ≤ ` with ` = (b− a)/2. We

can extend ϕ as a 2`-periodic function and obtain its Fourier seriesas

ϕ(y) ∼ a0

2+

∞∑n=1

(an cos

nπ

`y + bn sin

nπ

`y)

where

an =1`

∫ `

−`ϕ(y) cos

nπx

`ydy,

bn =1`

∫ `

−`ϕ(y) sin

nπx

`ydy.

Method 2: Considering the change of variable as y = x − a and` := b − a, we define ϕ(y) := f(x) = f(y + a) where 0 ≤ y < `. Wecan extend ϕ as a 2`-periodic function in any manner and obtain itsFourier series. Here are two specific cases:

(a) For y ∈ [−`, 0], define fe(y) = f(−y). Thus fe on [−`, `] isan even function. In this case,

ϕ(y) ∼ a0

2+

∞∑n=1

an cosnπ

`y

where ` = (b− a)/2 and

an =1`

∫ `

−`ϕ(y) cos

nπx

`ydy.

Fourier Series of 2`-Periodic functions 99

(b) For y ∈ [−`, 0], define fo(y) = −f(−y). Thus fo on [−`, `] isan odd function. In this case,

ϕ(y) ∼∞∑

n=1

bn sinnπ

`y

where

bn =1`

∫ `

−`ϕ(y) sin

nπx

`ydy.

From the series of f we can recover the corresponding series of fon [a, b] by writing y = x− a.

7.2.2 Exercises

Exercise 7.1 Find the Fourier series of the 2π- period function fsuch that:

(a) f(x) =

1, −π2 ≤ x < π

20, π

2 < x < 3π2 .

(b) f(x) =x, −π

2 ≤ x < π2

π − x, π2 < x < 3π

2 .

(c) f(x) =

1 + 2xπ , −π ≤ x ≤ 0

1− 2xπ , 0 ≤ x ≤ π.

(d) f(x) = x2

4 , −π ≤ x ≤ π.

Exercise 7.2 Using the Fourier series in Exercise 7.1, find the sumof the following series:

(a) 1− 13

+15− 1

7+ . . ., (b) 1 +

14

+19

+116

+ . . ..

(c) 1− 14

+19− 1

16+ . . ., (d) 1 +

132

+152

+172

+ . . ..

Exercise 7.3 If f(x) =

sinx, 0 ≤ x ≤ π4

cosx, π4 ≤ x < π

2

, then show that

f(x) ∼ 8π

cosπ

4

[sinx1.3

+sin 3x5.7

+sin 10x9.11

+ . . .

].

100 Fourier Series

Exercise 7.4 Show that for 0 < x < 1,

x− x2 =8π2

[sinxπ

13+

sin 3πx33

+sin 5πx

53+ . . .

].

Exercise 7.5 Show that for 0 < x < π,

sinx+sin 3x

3+

sin 5x5

+ . . . =π

4.

Exercise 7.6 Show that for −π < x < π,

x sinx = 1− 12

cosx− 21.3

cos 2x+2

2.4cos 3x− 2

3.5cos 4x+ . . . ,

and find the sum of the series

11.3

− 13.5

cos 4x+1

5.7− 1

7.9+ . . . .

Exercise 7.7 Show that for 0 ≤ x ≤ π,

x(π − x) =π2

6−[cos 2x

12+

cos 4x22

+cos 6x

32+ . . .

],

x(π − x) =8π

[sinx13

+sin 3x

33+

sin 5x53

+ . . .

].

Exercise 7.8 Assuming that the Fourier series of f converges uni-formly on [−π, π), show that

1π

∫ π

−π[f(x)]2dx =

a20

2+

∞∑n=1

(a2n + b2n).

Exercise 7.9 Using Exercises 7.7 and 7.8 show that

(a)∞∑

n=1

1n4

=π4

90, (b)

∞∑n=1

(−1)n−1

n2=π2

12

(c)∞∑

n=1

1n6

=π6

945(d)

∞∑n=1

(−1)n−1

(2n− 1)3=π3

32

Exercise 7.10 Write down the Fourier series of f(x) = x for x ∈[1, 2) so that it converges to 1/2 at x = 1.

8

Functions of Several Variables

8.1 Introduction

Functions of more than one variables come naturally in applications.For example, in physics one come across the relation

PV

T= c, constant,

where P, V, T represents the pressure, volume and temperature of anideal gas. Since

P =cT

V= c, V =

cT

P= c, T =

PV

c

each of P, V, T can be thought of as a function of the remaining twovariables.

In the following we shall use some standard notations:

For k ∈ N, we denote by Rk the set of all k-tuples (x1, . . . , xk)with xi ∈ R for i ∈ 1, . . . , k. Also, for u = (x1, . . . , xk) ∈ Rk, wedenote by ‖u‖ the positive square root of x2

1 + x22 + . . . , x2

k, i.e.,

‖u‖ =√x2

1 + x22 + . . . , x2

k.

Definition 8.1 A subset D of R2 is said to be a bounded set ifthere exits M > 0 such that ‖u‖ ≤M for every u ∈ D.

If un = (xn, yn) for n ∈ N, for sequences (xn) and (yn) in R, thenwe say that (un) is a sequence in R2.

101

102 Functions of Several Variables

Definition 8.2 A sequence (un) in R2 is said to converge to a pointu ∈ R2, and we write un → u, if ‖un − u‖ → 0 as n→∞.

Definition 8.3 By a function of several variables we mean a func-tion f : D → R, where D is a subset of Rk for some k ∈ 2, 3, . . ..We write this fact by

z = f(x1, . . . , xk), (x1, . . . , xk) ∈ D,

and say that z is a function of (x1, . . . , xk).

The set D is called the domain of the function f , or we saythat f is defined on D.

In this course, for the sake of simplicity of presentation, we shallconsider functions of two variables, i.e., D ⊆ R2 and f : D → R, andwe write this by

z = f(x, y), (x, y) ∈ D.

Whatever we do with two variables can also be extended to more thantwo variables. Geometrically a function of two variables represent asurface S in the 3-dimensional space, i.e., the surface is the set of allpoints (x, y, z) ∈ R3 such that z = f(x, y) with (x, y) ∈ D, i.e.,

S = (x, y, z) ∈ R3 : z = f(x, y), (x, y) ∈ D.

EXAMPLE 8.1 Here are two examples functions of of two vari-ables:

(a) z = f(x, y) :=√

1− x2 − y2, D := (x, y) : x2 + y2 ≤ 1.

(b) z = f(x, y) :=xy

x2 + y2, D := (x, y) : x2 + y2 6= 0.

8.2 Limit and Continuity

We shall discuss limit, continuity and differentiability of functions ofseveral variables. One of the primary concept required to do these isthat of a neighbourhood of a point in R2.

Definition 8.4 By a neighbourhood of a point u0 = (x0, y0) ∈ R2

we mean the set of all points u = (x, y) ∈ R2 such that ‖u− u0‖ < δfor some δ > 0. Such a set, u ∈ R2 : ‖u − u0‖ < δ is called a

Limit and Continuity 103

δ-neighbourhood of u0 = (x0, y0), or an open disc with centre u0

and radius δ.

A set of the form u ∈ R2 : 0 < ‖u − u0‖ < δ is called a adeleted δ-neighbourhood of u0 = (x0, y0).

Definition 8.5 Suppose f is defined on a set D ⊆ R2, and u0 =(x0, y0) ∈ R2. We say that f has the limit α as u = (x, y) ∈ Dapproaches u0 if for every e > 0 there exists a δ > 0 such that

|f(u)− α| < ε

whenever u ∈ D and 0 < ‖u− u0‖ < δ. We write the above fact by

limu→u0

f(u) = α or limx→0,x0→y→y0

f(x, y) = α.

The following results can be observed (Exercise)

(a) Suppose limit exists for a function f : D → R2 at a pointu0 ∈ R2, then the limit is unique.

(b) If limu→u0 f(u) = α exists, then for every sequence (un) inD, if un → u0 then f(un) → α.

By (b) above it follows that if (un) and vn) are sequences in Dwith un → u0 and vn → u0 but sequences (f(un)) and (f(vn) havedifferent limits, then limu→u0 f(u) does not exist.

Definition 8.6 A function f is defined on a set D ⊆ R2 is said tobe continuous at a point u0 ∈ D if limu→u0 f(u) exists and is equalto f(u0).

Thus, if f : D → R is continuous at u0 ∈ D, then fore verysequence (un) which converges to u0, we have f(un) → f(u0).

Remark 8.1 We note that in order to define limit of a function fat a point u0(x0, y0), it is not necessary that the function is definedat u0, whereas to define continuity of f at u0 it is necessary that u0

belongs to the domain of f .

EXAMPLE 8.2 Let f(x, y) =√

1− x2 − y2, D := (x, y) : x2 +y2 ≤ 1. Then lim

(x,y)→(0,0)f(x, y) = 1. In fact, f is continuous at all

points in D.


EXAMPLE 8.3 Let f(x, y) := xyx2 − y2

x2 + y2, D := (x, y) : x2 +y2 6=

0. Then taking x = r cos θ and y = r sin θ we have r2 = x2 + y2,and

|f(x, y)| =∣∣∣∣r24 sin 4θ

∣∣∣∣ ≤ r2

4→ 0 as r2 → 0.

Hence lim(x,y)→(0,0)

f(x, y) = 0. Note that f is not defined at (0, 0).

EXAMPLE 8.4 Let f(x, y) :=xy

x2 + y2, D := (x, y) : x2+y2 6= 0.

Then lim(x,y)→(0,0)

f(x, y) does not exist. To see this, for each m ∈ R,

consider the straight line Lm := (x, y) : y = mx, i.e., the straightline passing through the origin with slope m. Then we see that for(x, y) ∈ Lm, f(x, y) =

m

1 +m2. Thus, for ε > 0 it is not possible to

find a δ-neighbour hood of (0, 0) such that |f(x, y) − α| < ε for allpoints in that neighbourhood.

For instance, we can identify sequences (un) and (vn) in D havingthe same limit (0, 0), but (f(un)) and (f(vn) have different limits.

EXAMPLE 8.5 Let z = f(x, y) :=x2y

x4 + y2, D := (x, y) : x2 +

y2 6= 0. Then lim(x,y)→(0,0)

f(x, y) does not exist. To see this, for each

m ∈ R, consider the set Am := (x, y) : y = mx2. Then we see thatfor (x, y) ∈ Am, f(x, y) =

m

1 +m2. Again, by the same argument as

in last example, the function does not have limit at (0, 0).

Remark 8.2 Suppose a function f does not have a limit at a point(x0, y0), and suppose that (x0, y0) is not in the domain of definitionof f . Then, no matter whatever value we assign to f at (x0, y0), theextended function cannot be continuous.

It is possible that

• lim(x,y)→(x0,y0)

f(x, y) does not exist but one or both of the limits

limx→x0

limy→y0

f(x, y), limy→y0

limx→x0

f(x, y)

exist.

• Any one or both of limx→x0

limy→y0

f(x, y) and limy→y0

limx→x0

f(x, y) may

Limit and Continuity 105

not exist, but lim(x,y)→(x0,y0)

f(x, y) exists.

To illustrate the above we consider a few examples.

EXAMPLE 8.6 (a) Let

f(x, y) :=(y − x)(1 + x)(y + x)(1 + y)

, D := (x, y) : x+ y 6= 0.

Then we see that

limy→0

limx→0

f(x, y) = limy→0

y

y= 1, lim

x→0limy→0

f(x, y) = limx→0

−(1 + x) = −1,

andlim

y=mx,x→0f(x, y) =

m− 1m+ 1

.

Thus, separate limit exit, but the limit does not exists at (0, 0).

EXAMPLE 8.7 Let

f(x, y) := x sin1y

+ y sin1x

D := (x, y) : xy 6= 0.

Then we see that separate limit do not exist at (0, 0), but

|f(x, y)| ≤ |x|+ |y| so that lim(x,y)→(0,0)

f(x, y) = 0.

8.2.1 Some Topological Notions

For stating two important theorems concerning continuous functions,we need to use a few more definitions.

Definition 8.7 A point (x0, y0) ∈ R2 is said to be an interiorpoint of a set D ⊆ R2 if D contains a neighbourhood of (x0, y0).

Definition 8.8 A point (x0, y0) is said to be a boundary point ofa set D ⊆ R2 if every neighbourhood of (x0, y0) contains some pointof D and some point of Dc.

Definition 8.9 A subset G of R2 is said to be an open set if everypoint in G is an interior point of G.

Definition 8.10 A subset F f R2 is said to be a closed set if Fcontains all its boundary points.


It can be shown that a set D is closed iff it contains all its bound-ary points, iff its compliment Dc is open in R2.

Definition 8.11 Let D ⊆ R2 and ϕ : [a, b] → D be a function.For t ∈ [a, b], let ϕ(t) = (ϕ1(t), ϕ2(t)), where ϕ1 and ϕ2 are realvalued functions defined on [a, b]. Then ϕ : [a, b] → D is said to becontinuous at t0 ∈ [a, b] if both ϕ1 and ϕ2 are continuous at t0.

Definition 8.12 Let D ⊆ R2. By a curve in D in D we mean acontinuous function γ : [0, 1] → D. The point u0 := γ(0) is calledthe initial point of the curve γ and u1 := γ(1) is called the final orterminal point of γ, and we say that γ is a curve joining u0 to u1.

Definition 8.13 A subset D of R2 is said to be a connected setif any two points in D can be joined by a curve in D, that is, for anyu0 and u1 in D, there exists a curve γ : [0, 1] → D in D such thatu0 = γ(0) and u1 := γ(1).

We observe that if any two points in D ⊆ R2 can be joined by apolygonal line, then D is connected.

Definition 8.14 A subset D of R2 is said to be a domain in R2 if itconsists of an open connected set and possibly some of its boundarypoints.

Definition 8.15 A domain which is also an open set is called anopen domain, and a domain which is also a closed set is called aclosed domain.

8.2.2 Two Theorems

Theorem 8.1 Suppose f is a continuous function defined on a closedand bounded domain D ⊆ R2. Then we have following:

(a) (On attaining maximum and minimum) There exist points(x1, y1) and (x2, y2) in D such that

f(x1, y1) ≤ f(x, y) ≤ f(x2, y2) ∀ (x, y) ∈ D.

(b) (Intermediate value theorem) If (x1, y1) and (x2, y2) are inD and c ∈ R is such that

f(x1, y1) < c < f(x2, y2)

then there exists (x0, y0) ∈ D such that f(x0, y0) = c.

Partial Derivatives 107

8.3 Partial Derivatives

Definition 8.16 Suppose f is a (real valued) function defined in aneigbourhood of a point (x0, y0). Then f is said to have the partial

derivative with respect to x at (x0, y0) ifd

dxf(x, y0) exists at x0,

and it is denoted by ∂f∂x (x0, y0). Thus, if f has partial derivative with

respect to x at (x0, y0), then

∂f

∂x(x0, y0) =

d

dxf(x, y0)∣∣

x=x0

= lim∆x→0→0

f(x0 + ∆x, y0)− f(x0, y0)∆x

,

and it is called the partial derivative of f with respect to x at(x0, y0).

Similarly, we can define partial derivative of f with respectto y at (x0, y0).

Thus, if f has partial derivative at (x0, y0) if f(x0, y), as a function

of y, is differentiable at y0, and in that case the quantityd

dyf(x0, y)∣∣

y=y0

,

denoted by∂f

∂y(x0, y0) is called the partial derivative of f with

respect to y at (x0, y0), i.e.,

∂f

∂y(x0, y0) :=

d

dyf(x0, y)∣∣

y=y0

.

Partial derivatives∂f

∂x(x0, y0) and

∂f

∂y(x0, y0) are also denoted by

fx(x0, y0) and fy(x0, y0) respectively.

We denote

fxx(x0, y0) := (fx)x (x0, y0), fxy(x0, y0) := (fx)y (x0, y0),

fyx(x0, y0) := (fy)x (x0, y0), fyy(x0, y0) := (fy)y (x0, y0).

Thus,

fxx(x0, y0) :=d

dxfx(x, y0)∣∣x=x0

, fxy(x0, y0) :=d

dyfx(x0, y)∣∣y=y0

,

fyx(x0, y0) :=d

dxfx(x, y0)∣∣x=x0

, fyy(x0, y0) :=d

dyfy(x0, y)∣∣y=y0

.


EXAMPLE 8.8 Let z = f(x, y) := xy

x2+y2 , (x, y) 6= 0,0, (x, y) = (0, 0).

We have already observed that this function is not continuous at(0, 0). However,

fx(0, 0) = 0 = fy(0, 0).

EXAMPLE 8.9 Let z = f(x, y) :=

xy(x2−y2)

x2+y2 , (x, y) 6= 0,0, (x, y) = (0, 0).

In this case it is seen that fx(0, 0) = 0 = fy(0, 0). Note that

(fx)y(0, 0) = lim∆y→0

fx(0,∆y)− fx(0, 0)∆y

,

where

fx(0,∆y) = lim∆x→0

f(∆x,∆y)− f(0,∆y)∆x

= −∆y.

Hence(fx)y(0, 0) = −1.

Also,

(fy)x(0, 0) = lim∆x→0

fy(∆x, 0)− fy(0, 0)∆x

,

wherefy(∆x, 0) = lim

∆y→0

f(∆x,∆y)− f(∆x, 0)∆y

= ∆x.

Hence(fy)x(0, 0) = 1.

The above example shows that, in general, fxy need not be equalto fyx. However,

Theorem 8.2 If fxy and fyx exist and are continuous in a neigh-bourhood D0 of (x0, y0), then fxy = fyx on D0.

8.3.1 Partial Increments and Total Increment

Definition 8.17 Suppose f is a (real valued) function defined in aneigbourhood of a point (x, y), and let z = f(x, y). Then


• Partial increment of z with respect to x is

∆xz := f(x+ ∆x, y)− f(x, y).

• Partial increment of z with respect to y is

∆yz := f(x, y + ∆y)− f(x, y).

• Total increment of z is

∆z := f(x+ ∆x, y + ∆y)− f(x, y).

Theorem 8.3 Suppose fx and fy exist and are continuous in aneighbourhood D1 of a point (x, y). Then there exist functions ϕand ψ defined in the neighbourhood D2 of (0, 0) such that

∆z = fx(x, y)∆x+ fy(x, y)∆y + φ(∆x,∆y)∆x+ ψ(∆x,∆y)∆y

for all (x, y) ∈ D1 and (∆x,∆y) ∈ D2, where

φ(∆x,∆y) → 0, ψ(∆x,∆y) → 0 as (∆x,∆y) → (0, 0).

Proof. We may observe that

∆z := f(x+ ∆x, y + ∆y)− f(x, y)= [f(x+ ∆x, y + ∆y)− f(x, y + ∆y)] + [f(x, y + ∆y)− f(x, y)].

Since f has partial derivatives in a neighbourhood of (x, y), by meanvalue theorem, there exists ξ between x and x+ ∆x such that

f(x+ ∆x, y + ∆y)− f(x, y + ∆y) = fx(ξ, y + ∆y),

there exists η between y and y + ∆y such that

f(x, y + ∆y)− f(x, y) = fy(x, η).

Thus,∆z = fx(ξ, y + ∆y)∆x+ fy(x, η)∆y.

Further, since fx and fy are continuous at (x, y),

fx(ξ, y + ∆y) → fx(x, y), fy(x, η) → fy(x, y)

as (∆x,∆y) → (0, 0), i.e., as ∆ρ :=√

(∆x)2 + (∆y)2 → 0. Thus,

∆z = fx(x, y)∆x+ fy(x, y)∆y + φ(∆x,∆y)∆x+ ψ(∆x,∆y)∆y,

where φ(∆x,∆y) → 0 and ψ(∆x,∆y) → 0 as ∆ρ→ 0.


8.3.2 Total Differential, Gradient, Differentiability

Suppose f is a function of two variables defined in a neighborhoodof a point u0 := (x0, y0).

Definition 8.18 Suppose f has partial derivatives at u0 := (x0, y0).Then the expression

fx∆x+ fy∆y

evaluated at u0 is called the total differential of f at u0, and thepair (fx(u0), fy(u0)) is called the gradient of f at u0, denoted by

(grad f)(u0) or (∇f)(u0).

Definition 8.19 The function f is said to be differentiable atu0 := (x0, y0) if there exists a pair (α, β) of real numbers such that

lim∆ρ→0

1∆ρ

[f(x0 + ∆x, y0 + ∆y)− f(x0, y0)]− (α∆x+ β∆y) = 0,

where ∆ρ :=√

(∆x)2 + (∆y)2. The pair (α, β) is called the deriva-tive of f and is denoted by f ′(u0).

From Theorem 8.3, the following result is obvious, thus providinga sufficient condition for differentiability at a point.

Theorem 8.4 If fx and fy exist and are continuous in a neighbour-hood of u0 := (x0, y0), then f is differentiable at (x0, y0), and

f ′(u0) = (∇f)(u0).

Here is a necessary condition for differentiability.

Theorem 8.5 Suppose f is differentiable at u0 := (x0, y0). Then,then f is continuous, partial derivatives fx and fy exist at u0, and

f ′(u0) = (∇f)(u0).

Remark 8.3 If a function f is either not continuous or if fx andfy does not exist at (x0, y0), then By Theorem 8.5 we can concludethat f is not differentiable at (x0, y0). Suppose fx and fy exist at


u0 = (x0, y0). Then it follows that, f is differentiable at (x0, y0) ifand only if

lim∆ρ→0

1∆ρ

[f(x0+∆x, y0+∆y)−f(x0, y0)]−[fx(u0)∆x+fy(u0)∆y)] = 0.

and in that case the derivative is (∇f)(u0) := (fx(u0), fy(u0)).


8.3.3 Derivatives of Composition of Functions

Suppose z = F (u, v) where u and v are functions of (x, y), i.e., u =ϕ(x, y) and v = ψ(x, y) for some functions ϕ and ψ. Then z itself isa function of (x, y). Thus,

z = F (ϕ(x, y), ψ(x, y)).

Then we have

∆xz = F (ϕ(x+ ∆x, y), ψ(x+ ∆x, y))− F (ϕ(x, y), ψ(x, y)).

But,

∆xu = ϕ(x+ ∆x, y)− ϕ(x, y), ∆xv = ψ(x+ ∆x, y)− ψ(x, y).

Hence, assuming all necessary conditions, we have

∆xz = F (u+ ∆xu, v + ∆xv)− F (u, v)= Fu∆xu+ Fv∆xv + ϕ1∆xu+ ϕ2∆xv,

where

ϕ1(∆xu,∆xv) → 0, ϕ2(∆xu,∆xv) → 0 as (∆x,∆y) → (0, 0).

Thus,∆xz

∆x= Fu

∆xu

∆x+ Fv

∆xv

∆x+ ϕ1

∆xu

∆x+ ϕ2

∆xv

∆x.

Now taking limit as (∆x,∆y) → (0, 0), we have

∂z

∂x= Fu

∂u

∂x+ Fv

∂v

∂x.

Thus we have proved the following theorem.


Theorem 8.6 Suppose ϕ and ψ are defined in a neighbourhood D ofa point (x0, y0) and z = F (u, v) where u = ϕ(x, y), v = ψ(x, y) for(x, y) ∈ D. Assume that Fu, Fv exist and are continuous in a neigh-bourhood of (u0, v0), where u0 = ϕ(x0, y0), v0 = ψ(x0, y0). Then forall (x, y) in a in a neighbourhood of (x0, y0), we have

∂z

∂x= Fu

∂u

∂x+ Fv

∂v

∂x.

A special case:

Suppose f is a function of (x, y) in a nbd of a point (x0, y0), andx and y are functions of another variable t with t ∈ [a, b]. Thenz = f(x, y) is a function of t and we have

dz

dt=∂z

∂t= fx

dx

dt+ fy

dy

dt.

EXAMPLE 8.10 Consider z = x2+√y where y = sinx, 0 < x < π.

Thendz

dx=∂z

∂x+∂z

∂y

dy

dx= 2x+

cosx2√y.

Euler’s Theorem:

Theorem 8.7 Suppose f is a homogeneous function of degree n ina domain D, i.e., f(λx, λy) = λnf(x, y) for all λ ∈ R, (x, y) ∈ D.Then

x∂f

∂x+ y

∂f

∂y= nf(x, y).

Proof. We may write

f(x, y) = f

(x, x

y

y

)= xnf

(1,y

y

)= xng(u),

where u = y/x and g(u) = f(1, u). Then by Theorem 8.6,

∂f

∂x= nxn−1g(u) + xng′(u)(−y/x2),

∂f

∂y= xng′(u)(1/x) = xn−1g′(u).

From these expressions the conclusion of the theorem follows.

Derivatives of Implicitly Defined Functions 113

8.4 Derivatives of Implicitly Defined Func-tions

Definition 8.20 A function f is said to be implicitly defined onan interval J if there exists a function F defined on a domain D if

(x, f(x)) ∈ D and F (x, f(x)) = 0 ∀x ∈ J.

Theorem 8.8 Suppose f is implicitly defined on an interval J byF (x, y) = 0, y = f(x) for x ∈ J . Assume further that Fx and Fy aredefined and are continuous in a nbd of a point (x0, y0) where x0 ∈ Jand y0 = f(x0). If Fy 6= 0 at (x0, y0), then f ′(x) exists in a nbd ofx0 and

f ′(x0) = −Fx

Fy

∣∣(x0,y0)

.

Proof. Let x0 ∈ J and y0 = f(x0), ∆y = f(x0 + ∆x) − f(x0).Then, since F (x0, y0) = 0 = F (x0 + ∆x, y0 + ∆y), we have

0 = ∆F = Fx∆x+ Fy∆y + ϕ1∆x+ ϕ2∆y,

where ϕ1(∆x,∆y), ϕ1(∆x,∆y) → 0 as (∆x,∆y) → 0. Hence,

0 = Fx + Fy∆y∆x

+ ϕ1 + ϕ2∆y∆x

= (Fx + ϕ1) + (Fy + ϕ2)∆y∆x

.

so that∆y∆x

= −Fx + ϕ1

Fy + ϕ2→ −Fx

Fyas (∆x,∆y) → 0.

Thus f ′(x) exists in a nbd of x0 and f ′(x0) = −Fx/Fy at (x0, y0).

The proof of the following theorem is omitted.

Theorem 8.9 (Implicit function Theorem Suppose F is defined ina nbd D of a point (x0, y0), and Fx and Fy exist and are continuousin D. If F (x0, y0) = 0 and Fy 6= 0 at (x0, y0), then we have following:

(i) There exists an open interval J containing x0 and a functionf : J → R such that (x, f(x) ∈ D for all x ∈ J , and

F (x, f(x)) = 0 ∀x ∈ J.

(ii) f ′(x) exists in a an open subinterval J0 of J with x0 ∈ J0

f ′(x) = −Fx

Fy

∣∣(x,f(x))

∀x ∈ J0.


8.4.1 Level Curve and Level Surface

Suppose u is a function defined in a domain D ⊆ R2. Then, for eachc ∈ R, the set of all (x, y) ∈ D such that f(x, y) = c is called a levelcurve of u. If v is a function defined in a domain D′ ⊆ R3, then,for each c ∈ R, the set of all (x, y, z) ∈ D′ such that f(x, y, z) = c iscalled a level surface of v.

If Γ is a level curve of a function u, then it can be easily seenthat ~v := (uy,−ux) is a vector in the direction of the tangent at thepoint (x, y). Indeed, if y = f(x) is implicitly defined by F (x, y) :=u(x, y) − c = 0, then the level curve of u is the graph of f , and itstangent is along the direction of (1,−ux/uy), i.e., along (uy,−ux)whenever uy 6= 0, and if uy = 0 then its direction is obviously along(0,−ux). Thus, we see that

5u · ~v = (ux, uy) · (uy,−ux) = uxuy − uyux = 0,

so that the gradient 5u of u is perpendicular to the tangent line,and hence it is along the normal to the level curve.

8.5 Directional Derivatives

Suppose u is a function defined in a domain D ⊆ R2. We would liketo know the change in the values of u as (x, y) varies from (x0, y0)along a particular direction ~s. Thus, taking ~s = (∆x,∆y) and writing∆s =

√(∆x)2 + (∆yx)2, we have

∆u = u(x0+∆x, y+∆y)−u(x0, y0) = ux∆x+uy∆y+ϕ1∆x+ϕ2∆y.

Hence,∆u∆s

= ux∆x∆s

+ uy∆y∆s

+ ϕ1∆x∆s

+ ϕ2∆y∆s

and consequently,

lim∆s→0

∆u∆s

= ux cosα+ uy cosβ,

where α and β are the angles that ~s makes with the x-axis and y-axisrespectively. Thus

∂u

∂s:= lim

∆s→0

∆u∆s

= 5u · ~ns,

where ~ns is the unit vector along ~s.

Directional Derivatives 115

Definition 8.21 The quantity ∂u∂s is called the directional deriva-

tive of u along ~s.

Remark 8.4 We may observe taking ~e1 = (1, 0) and ~e2 = (0, 1), weget

∂u

∂e1=∂u

∂x,

∂u

∂e2=∂u

∂y.

EXAMPLE 8.11 The directional derivative of u in the direction of~s := Ou at a (x, y) is given by

∂u

∂s:= 5u · ~ns, where 5u = (ux, uy)

and ~ns = ~s/|~s|. Hence∂u

∂s= |Ou|.

EXAMPLE 8.12 Let u(x, y) = xy. Then the directional derivativeof u in the direction of ~s := (3, 4) at the point (1, 2) is given by∂u

∂s:= 5u · ~ns at (1, 2), where 5u = (y, x) and ~ns = ~s/|~s| = 1

5(3, 4).Hence

∂u

∂s=

15(3y + 4x)

∣∣(1,2)

= 2.

EXAMPLE 8.13 Let u(x, y) = x2 + y2. Then the directionalderivative of u in the direction of ~s := (s1, s2) at the point (a, b)

is given by∂u

∂s:= 5u · ~ns at (a, b), where 5u = (2x, 2y) and ~ns =

~s/|~s| = (s1, s2)/√s21 + s22. Hence

∂u

∂s=√s21 + s22(2xs1 + 2ys2)

∣∣(a,b)

=2√

s21 + s22(as1 + bs2).

Exercise 8.2 Suppose f is differentiable at u0 := (x0, y0). Then,prove that directional derivatives of f in any direction ~s exist at u0,and

f ′(u0) =∂u

∂sat u0.


8.6 Taylor’s Formula

Recall that if ϕ is a function of one variable having continuous deriva-tives up to the order n+ 1 in a nbd of a point x0, then the Taylor’sformula for any x in that nbd we have

ϕ(x) = ϕ(x0) +n∑

k=1

f (k)(x0)k!

(x− x0) +f (k+1)(ξ)

k!(x− x0)

for some ξ lying between x0 and x. Writing ∆x = x− x0, the aboveformula can written as

ϕ(x) = ϕ(x0)+n∑

k=1

1k!

(∆x

d

dx

)k

f∣∣(x0)

+1

(n+ 1)!

(∆x

d

dx

)n+1

f∣∣(ξ).

Here, we used the notation(∆x

d

dx

)f :=

(∆x

df

dx

),

(∆x

d

dx

)k

f :=(

∆xd

dx

)(∆x

df

dx

)k−1

.

We obtain similar formula for functions of two variables as well.For this purpose let us define(

∆x∂

∂x+ ∆y

∂

∂y

)f :=

(∆x

∂f

∂x+ ∆y

∂f

∂y

),

and for k = 2, 3, . . .,(∆x

∂

∂x+ ∆y

∂

∂y

)k

f :=(

∆x∂

∂x+ ∆y

∂

∂y

)(∆x

∂

∂x+ ∆y

∂

∂y

)k−1

f.

It can be seen that(∆x

∂

∂x+ ∆y

∂

∂y

)k

f =k∑

r=0

(kCr

)(∆x)r(∆y)k−r

(∂

∂x

)r ( ∂

∂y

)k−r

f.

Theorem 8.10 Suppose f is a function of two variables having con-tinuous partial derivatives up to the order n+1 in a nbd D of a point(x0, y0). Then the Taylor’s formula for any x in D is given by

f(x, y) = ϕ(x0, y0) +n∑

k=1

1k!

(∆x

∂

∂x+ ∆y

∂

∂y

)k

f∣∣(x0,y0)

+1

(n+ 1)!

(∆x

∂

∂x+ ∆y

∂

∂y

)n+1

f∣∣(x0+ξ∆x,y0+ξ∆y)

.

for some 0 < ξ < 1.

Maxima and Minima 117

Proof. Let ϕ(t) = f(x0 + t∆x, y0 + t∆y). Then we have

ϕ(k)(t) =(

∆x∂

∂x+ ∆y

∂

∂y

)k

f∣∣t. (∗)

Hence, using the Taylor’s formula for functions of one variable wehave

ϕ(1) = ϕ(0) +n∑

k=1

ϕ(k)(0)k!

+ϕ(k+1)(ξ)

k!

for some ξ lying between 0 and 1. In view of (∗), this is exactly therequired formula.

8.7 Maxima and Minima

Definition 8.22 Suppose f : D → R and u0 ∈ D. Then

(i) f is said to have maximum at u0 (or f attains local max-imum at u0) if there exists a nbd U0 ⊆ D0 of u0 such that f(u) <f(u0) for all u ∈ U0, u 6= u0, and

(ii) f is said to have minimum at u0 (or f attains local minimumat u0) if there exits a nbd V0 ⊆ D0 of u0 such that f(u) > f(u0) forall u ∈ V0, u 6= u0.

(iii) f is said to have extremum at u0 if f attains either maxi-mum or minimum at u0.

Remark 8.5 (a) By the above definition, if a function f has max-imum (resp. minimum) at a point u0, then there exists a nbd of u0

such that f can not attain maximum (resp. minimum) at any otherpoint. For example, the function f(x) = 1 for 0 ≤ x ≤ 1 does notattain maximum or minimum at any point in [0, 1].

(b) In the above we have defined maximum and minimum onlylocally, i.e., in a nbd of a point. In case there exists u0 ∈ D such thatf(x) < f(u0) (resp. f(x) > f(u0)) for all u ∈ D, then we say that fattains global maximum at u0 (global minimum at u0).

(c) Some authors write ‘≤’ and ‘≥’ in place of ‘<’ and ‘>’ in theabove definitions. In this course we do not follow that convention.

Following theorem prescribes a necessary condition for a functionto have a maximum or minimum at a point.


Theorem 8.11 Suppose f : D → R has maximum or minimum atu0 ∈ D, and suppose fx and fy exist at u0. Then fx = 0 = fy at u0.

Proof. Suppose f : D → R has maximum at u0 = (x0, y0) ∈ D.Then

f(x0 + h, y0) < f(x0, y0)

for all h in a nbd J0 of 0. Hence, the function ϕ : J0 → R defined byϕ(h) = f(x0 +h, y0) has a maximum at x0. Therefore, ϕ′(0) = 0,i.e.,fx(u0) = 0. Similarly, we can discuss the case of attaining minimumat u0.

Definition 8.23 Let f : D → R, and u0 ∈ D. If fx and fy exist atu0 and fx = 0 = fy at u0, then u0 is called a critical point of f . Acritical point of f at which f is neither a maximum nor a minimumis called a saddle point of f .

EXAMPLE 8.14 Let f(x, y) = (x− 1)2 + (y− 2)2− 1, (x, y) ∈ R2.Then we see that f(x, y) > f(1, 2) = −1 for all (x, y) 6= (1, 2). Thus,f attains minimum at (1, 2) ∈ R2.

EXAMPLE 8.15 Let f(x, y) = 12 − sin(x2 + y2), (x, y) ∈ R2. Then

we see that f(x, y) < f(0, 0) = 12 for all (x, y) 6= (0, 0). Thus, f

attains maximum at (1, 2) ∈ R2.

EXAMPLE 8.16 Let f(x, y) = x2 + y2. Then fx = 0 = fy at(0, 0) ∈ R2, but, f attains neither maximum nor minimum at (0, 0).Thus, (0, 0) is a critical point which is a saddle point of f .

Now we prescribe a sufficient condition for a function to have amaximum or minimum at a point.

Theorem 8.12 Suppose u0 ∈ D is a critical point of f : D → R,and f has continuous partial derivatives up to the order three in anbd of u0. Then we have the following:

(i) f has a maximum at u0, if

fxxfyy − f2xy > 0, fxx < 0 at u0.

(ii) f has a minimum at u0 if

fxxfyy − f2xy > 0, fxx > 0 at u0.

(iii) u0 is a saddle point of f if fxxfyy − f2xy < 0 at u0.

Maxima and Minima 119

EXAMPLE 8.17 Let f(x, y) = x2 + y2 − xy + 3x− 2y + 1. Thenfx = 0 = fy at (4/3, 1/3) ∈ R2, fxxfyy − f2

xy = 3 and fxx = 2 > 0at (4/3, 1/3). Hence, by the above theorem, f has a minimum at(4/3, 1/3).

8.7.1 Method of Lagrange Multipliers

Suppose we want to know the lengths of sides of a rectangle of max-imum area for a fixed perimeter. Thus, the problem is to find themaximum of the function f(x, y) := xy subjected that 2(x+y) = ` isfixed. To solve this problem what one may do is to find y in terms ofx, from the equation 2(x+ y) = `, and obtain a function substitutethe value in the expression for f(x, y), and maximize the resultingfunction. Thus, from 2(x + y) = `, we have y = `

2 − x, so that theproblems is to maximize ψ(x) := xy = x

(`2 − x

). It is easily seen

that the maximum is attained at x = y = `/2.

To summarize, we had to maximize the function f(x, y) subjectedto the constraint ϕ(x, y) := 2(x+ y)− ` = 0. For that, the variable yappearing in ϕ(x, y) = 0 is expressed as a function of x as y = g(x),and them maximized the function ψ(x) := f(x, g(x)).

In more complicated problems it may not be easy to obtain ex-plicitly the function y = g(x) such that f(x, g(x)) = 0. The methodwe propose to discuss in this section is to find some set of pointswhich would contain the point at which the function f attains itsmaximum or minimum.

Suppose we would like to find maximum or minimum of a func-tion f which is defined in some open set D ⊆ R2, subjected tothe constrain ϕ(x, y) = 0, where ϕ is also defined in D. Supposethere a point u0 = (x0, y0) at which f attains maximum such thatϕ(x0, y0) = 0. Let us also assume that fx, fy, ϕx, ϕy exist in a nbd ofu0, and there exists a function y = g(x) defined in a nbd of x0 suchthat u(x) := f(x, g(x)) = 0. Hence, we must have du/dx = 0 at x0.Thus, we have the following necessary conditions:

du

dx:= fx + fyy

′ = 0, ϕx + ϕyy′ = 0 at u0.

Hence, we must have

(fx + fyy′) + λ(ϕx + ϕyy

′) = 0 at u0 ∀λ ∈ R,


i.e,(fx + λϕx) + (fy + λϕy)y′ = 0 at u0 ∀λ ∈ R,

Thus, if we can find λ, x0, y0 such that

ϕ = 0, fx + λϕx = 0, fy + λϕy = 0 at u0 ∀λ ∈ R,

then u0 := (x0, y0) is a possible point at which f takes maximum orminimum value such that ϕ(u0) = 0. Note that writing

F (x, y, λ) := f(x, y) + λϕ(x, y)

the above condition is same as solving for (λ, x, y) such that

ϕ = 0, Fx = 0 = Fy.

The λ above is called the lagrange multiplier, and the methodusing lagrange multiplier is the above procedure of finding λ, x, y)such that

ϕ = 0, Fx = 0 = Fy,

so that the required point at which maximum or minimum the func-tion f can be one among these points.

Let us look the example that we have already discussed:

EXAMPLE 8.18 Suppose we want to find the point at which thefunction f(x, y) := xy attains subject to the constraint ϕ(x, y) :=2(x+ y)− ` = 0. We consider the equation

ϕ = 2(x+y)−` = 0, fx+λϕx := y+2λ = 0, fy+λϕy = x+2λ = 0.

Thus,x = −2λ = y, ` = 2(x+ y) = 4x

so that x = y = `/4.

Exercise 8.3 Find the maximum and minimum of f(x, y, z) :=x2y2z2 subject to the constraint that x2 + y2 + y2 = 1.

Exercise 8.4 Among all parallelopipeds of a given volume V , findthe one which has minimum surface area.

Exercise 8.5 Find points on the surface given by z2 = xy+4 subjectclosest to the origin.

Exercise 8.6 Show that, among all parallelopipeds with the givensurface area A, the cube has the maximum volume.

Exercise 8.7 Show that, of all triangles inscribed in a circle, theequilateral triangle has the greatest area.

9

Appendix

Theorem 9.1 Let f : [a, b] → R be a bounded function. Then f isintegrable over [a, b], if and only if there exists γ ∈ R, and for everyε > 0, there exists a partition P of [a, b] such that∣∣∣∣S(P, f, T )−

∫ b

af(x) dx

∣∣∣∣ < ε

for all set T of tags on P , and in that case γ =∫ ba f(x) dx.

Proof. Suppose f is integrable and ε > 0 be given. Then, byTheorem 2.4, there exists a partition P of [a, b] such that

U(P, f)− L(P, f) < ε.

Hence, from the relations

L(P, f) ≤ S(P, f, T ) ≤ U(P, f),

L(P, f) ≤∫ b

af(x) dx ≤ U(P, f),

it follows that∣∣∣∣S(P, f, T )−∫ b

af(x) dx

∣∣∣∣ ≤ U(P, f)− L(P, f) < ε

for every set T of tags on P .

Conversely, let γ ∈ R and for ε > 0, let P = xi : i = 1, . . . , kbe a partition of [a, b] such that

|S(P, f, T )− γ| < ε

121

122 Appendix

for all set T of tags on P . For each i ∈ 1, . . . , k, let t′i, t′′i ∈ [xi−1, xi]

be such that

f(t′i)−mi < ε, Mi − f(t′′i ) < ε.

LetT ′ = t′i : i = 1, . . . , k, T ′′ = t′′i : i = 1, . . . , k.

Then we have

S(P, f, T ′)−L(P, f) < ε(b−a), U(P, f)−S(P, f, T ′′) < ε(b−a),

so that

U(P, f)− L(P, f) = [U(P, f)− S(P, f, T ′′)] + [S(P, f, T ′′)− S(P, f, T ′)]+[S(P, f, T ′)− L(P, f)]

< 2ε(b− a) + [S(P, f, T ′′)− S(P, f, T ′)].

But, by hypothesis,

|S(P, f, T ′′)−S(P, f, T ′)| ≤ |S(P, f, T ′′)− γ|+ |γ−S(P, f, T ′)| < 2ε.

Thus, we have

U(P, f)−L(P, f) < 2ε(b−a)+2ε = 2(b−a+1)ε = Mε, M = 2(b−a+1).

Had we started with ε/[2(b− a+1)] in place of ε, then we would getU(P, f)− L(P, f) < ε. Thus, f is integrable. Since

L(P, f) ≤∫ b

af(x) dx ≤ U(P, f), L(P, f) ≤ S(P, f, T ) ≤ U(P, f),

it also follows that∣∣∣∣S(P, f, T )−∫ b

af(x) dx

∣∣∣∣ < Mε

for all tags T of P . Hence,∣∣∣∣γ − ∫ b

af(x) dx

∣∣∣∣ ≤ |γ−S(P, f, T )|+∣∣∣∣S(P, f, T )−

∫ b

af(x) dx

∣∣∣∣ < (1+M)ε

for all ε > 0. From this it follows that γ =∫ ba f(x) dx.

123

Theorem 9.2 Suppose f and g are integrable on [a, b] and c ∈ R.Then f + g and c f are integrable, and∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx∫ b

ac f(x) dx = c

∫ b

af(x) dx.

Proof. Let ε > 0 be given. Let P1 and P2 be partitions of [a, b] suchthat

U(P1, f)− L(P1, f) < ε, U(P2, g)− L(P2, g) < ε.

Let P = P1 ∪ P2. Since

L(P1, f) + L(P2, g) ≤ L(P, f) + L(P, g) ≤ L(P, f + g), (∗)

U(P, f + g) ≤ U(P, f) + U(P, g) ≤ U(P1, f) + U(P2, g), (#)

it follows that

U(P, f + g)− L(P, f + g) ≤ 2 ε.

Thus, f + g is integrable. Also, we have

L(P1, f)+L(P2, g) ≤∫ b

af(x) dx+

∫ b

ag(x) dx ≤ U(P1, f)+U(P2, g),

L(P, f)+L(P, g) ≤ L(P, f+g) ≤∫ b

a[f(x)+g(x)] dx ≤ U(P, f+g) ≤ U(P, f)+U(P, g).

Hence, from (∗), (#), we have∣∣∣∣∫ b

af(x) dx+

∫ b

ag(x) dx−

∫ b

a[f(x) + g(x)] dx

∣∣∣∣ < 2 ε.

This is true for all ε > 0. Hence∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx.

Next, suppose that c > 0. Then we note that

L(P, cf) = cL(P, f), U(P, cf) = cU(P, f),

124 Appendix

for all partitions P of [a, b]. Hence, we have∫ b

ac f(x) dx = c

∫ b

af(x) dx, ∀ c > 0.

To show obtain the case for c < 0, we first prove∫ b

a[−f(x)] dx = −

∫ b

af(x) dx.

Let P = xi : i = 1, . . . , k be a partition of [a, b]. Then we have

inff(x) : xi−1 ≤ x ≤ xi = − sup[−f(x)] : xi−1 ≤ x ≤ xi,supf(x) : xi−1 ≤ x ≤ xi = − inf[−f(x)] : xi−1 ≤ x ≤ xi.

Hence,

L(P,−f) = −U(P, f), U(P,−f) = −L(P, f),

and hence,

U(P,−f)− L(P,−f) = U(P, f)− L(P, f).

From this, it follows that −f is integrable, and∫ ba [−f(x)] dx =

−∫ ba f(x) dx. Now, suppose, c < 0. Then, −c > 0, so that from the

above result, we have∫ b

a[c f(x)] dx =

∫ b

a[−c (−f(x))] dx = (−c)

∫ b

a[−f(x)] dx = c

∫ b

af(x) dx.

Index

Absolute convergence, 23Alternating series, 22

boundedabove, 4below, 4

Cauchy sequence, 12Cauchy’s test, 20Comparison test, 18convergence, 1

de’Alembert’s test, 19divergence, 3

geometric series, 16greatest lower bound, 6

property, 7

infimum, 6Integrability, 30

least upper bound, 6property, 6

lower bound, 6lower sum, 28

ratio test, 19root test, 20

sequence, 1alternating, 4bounded, 4monotonically decreas-ing, 6

monotonically increas-ing, 6strictly decreasing, 6strictly increasing, 6

series, 15convergent, 15divergent, 15

subsequence, 7supremum, 6

upper bound, 6upper sum, 28

125

Ma101 Notes by Thamban Nair

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