Math 1120 Class 2pi.math.cornell.edu/~web1120/slides/fall12/aug28.pdfMath 1120 Class 2 Dan Barbasch...

Math 1120 Class 2

Dan Barbasch

Aug. 28, 2012

Dan Barbasch () Math 1120 Class 2 Aug. 28, 2012 1 / 19

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) says the following:

Theorem: Let y = f (x) be continuous on the interval [a, b]. Then f (x) hasan antiderivative F (x), and∫ b

af (x) dx = F (b)− F (a).

What does this mean?!The LHS(left hand side) is the definite integral. The RHS(right hand side)involves evaluating the antiderivative F of f .



The definition of the definite integral (LHS)∫ ba f (x) dx is very deep. You

approximate y = f (x) by step functions over the interval, and take a limit.The definition is in section 5.3. You may need 5.2 for the notation andother technicalities.

The definite integral is what arises in applications. You can see them insection 5.1; area, distance travelled, displacement, average of a function.

When you compute a definite integral by finding an antiderivative andevaluating, you are invoking FTC (mostly without explicit mention).

Question: How many antiderivatives can a function y = f (x) have?The mean value theorem implies that if F ′(x) = G ′(x) over some interval,then there is a constant C such that F (x) = G (x) + C .

Question: How does this fit with FTC?



Example: Compute the area under the graph of y = 1− x2 and above thex−axis.

The notion of the area of a region is by definition a definite integral; thelimit of Riemann sums.

A =

∫ b

a|f (x)| dx

is the area bounded by x = a, x = b y = f (x) and the x−axis.



Solution to Exercise: The graph of y = 1− x2 in the region −1 ≤ x ≤ 1 is

FTC reduces the problem of computing the area to finding theantiderivative of f (x) = 1− x2. We know that F (x) = x − x3/3 satisfiesF ′(x) = 1− x2. So by FTC,

A =

∫ 1

−1

(1− x2

)dx =

(x − x3/3

)|1−1=

= (1− 1/3)− (−1 + 1/3) = 4/3.


FTC

The Fundamental Theorem of Calculus (FTC) also says the following:

Theorem: Assume that y = f (x) is continuous in the interval [a, b]. Then∫ xa f (t) dt (as a limit of Riemann sums) exists for any a ≤ x ≤ b. The

function F (x) :=∫ xa f (t) dt therefore exists for any a ≤ x ≤ b.

Furthermore F (x) is differentiable in the interval, and F ′(x) = f (x).

I say appropriate because the way one proves the theorem is by applyingproperties of the definite integral proved from the definition in terms ofRiemann sums and their limit.


Differentiating Integrals

Exercises: Compute the following.

1.d

dx

(∫ x

0

√1 + t2 dt

)

2.d

dx

(∫ x2

0

√1 + y2 dy

)

3. G (x) :=d

dx

(∫ tan x

sin xcsc2 t dt

)


Differentiating Integrals

The Key:∫ ba f (t) dt = F (b)− F (a) where F ′(x) = csc2(x). So∫ tan x

sin xcsc2 t dt = F (tan x)− F (sin x)

The chain rule implies

G (x) = F ′(tan x) (tan x)′ − F ′(sin x) (sin x)′ .

Exercise: Finish the calculation.


Answer

(tan x)′ =1

cos2 x, (sin x)′ = cos x ,

So

G (x) = csc2 (tan x) · 1

cos2 x− csc2 (sin x) · cos x


A Few Other Notions in 5.4

Area between the graph of a function y = f (x) and the x-axisSigned Area.

Exercise: Find the area of the region between the graph of y = (x − 2)2,the x-axis and the lines x = 1 and x = 3.

Exercise: Find the area of the region between the graph of ln xx , the x-axis,

and the lines x = 1/e and x = e.


Area, Marginal Cost, Distance, Displacement

If f (x) ≥ 0 for a ≤ x ≤ b, then the area of the region is A =∫ ba f (x) dx .

If f (x) 6≥ 0 throughout the interval, then the integral∫ ba f (x) dx

represents the signed area of the region.

The area is A =∫ ba |f (x)| dx .

Marginal cost is the derivative of total cost C (x) with respect to quantityx . Similar for Marginal Revenue.

Velocity is the derivative of position with respect to time v(t) = s ′(t).

So position (displacement) is the integral of velocity

s(b)− s(a) =∫ ba v(t) dt.

Total distance traveled is D =∫ ba |v(t)| dt.


(Hints to) Answers

The function y = (x − 2)2 is positive in the interval [1, 3], so

A =∫ 31 (x − 2)2 dx = (x−2)3

3 |31= (3−2)33 − (1−2)2

3

The function ln xx is nonpositive in the interval 1/e ≤ x ≤ 1 and

nonnegative in the interval 1 ≤ x ≤ e. Thus

A =

∫ e

1/e| ln x

x| dx = −

∫ 1

1/e

ln x

xdx +

∫ e

1

ln x

xdx .

The antiderivative of ln xx IS (ln x)2

2 because of the chain rule!ln xx = (ln x) (1/x) = (ln x) (ln x)′.


Examples

1. 5.4, problem 74. Suppose a company’s marginal revenue r(x) fromselling x eggbeaters is

r(x) = 2− 2

(x + 1)2

r(x) in thousands of $, x in thousands of units. How much money doesthe company earn from 3000 eggbeaters?

2. 5.4, problem 77. Suppose∫ x1 f (t) dt = x2 − 2x + 1. Find f (4) and

f (x).3. 5.4 problem 75b. The temperature T (t) (in ◦F ) of a room at time tminutes is given by T (t) = 85− 3

√25− t for 0 ≤ t ≤ 25.

Find the room’s average temperature for 0 ≤ t ≤ 25.


Hints to the answers

1. R(x) =∫ x0 r(u) du.

2. The function f (x) is the derivative of x2 − 2x + 1.

3. The average of a function y = f (x) over an interval a ≤ x ≤ b is

Avg(f ) := 1b−a

∫ ba f (x) dx .


Some Integrals

(1)

∫ 3

0(x − 2)2 dx (2)

∫ 3

1(x − 2)2 dx

(3)

∫13√xdx (4)

∫ 2

1

x +√x

x2dx

(5)

∫ e

1

ln x

xdx

(6)

∫cos2 x sin x dx (7)

∫csc x cot x dx


(Hints to) Answers

In the first three problems we use∫xn dx = xn+1

n+1 + C for n 6= −1. In

problem (4) we need to use the algrbraic identity x+√x

x2= x−1 + x−3/2.

For (6), the antiderivative is − cos3 x3 + C . Check the answer by

differentiating; the chain rule is crucial.

For (7), the expression is (sin x)−2 cos x = (sin x)−2(sin x)′.


SubstitutionIn (5)-(7) we had to rely on the chain rule to obtain the correct answer.The method of substitution makes this procedure systematic.

Substitute u = h(x) in∫f (x) dx .

The substitution u = h(x) must be 1-1; compute/solve x = h−1(u).

Compute dx =(h−1)′

(u) du.

Substitute

∫f (x) dx =

∫f(h−1(u)

) (h−1)′

(u) du.

If already x = r(u),

∫f (x) dx =

∫f (r(u))) r ′(u) du.

For definite integrals,∫ b

af (x) dx =

∫ h−1(b)

h−1(a)f(h−1(u)

) (h−1)′

(u) du,∫ b

af (x) dx =

∫ r(b)

r(a)f (r(u))) r ′(u) du.


Computing a Definite Integral

∫ b

af (x) dx

Check that the function is continuous in the interval [a, b], or at leastpiecewise continuous. NO ASYMPTOTES!

Find an antiderivative F (x) of f (x) F ′(x) = f (x).

Evaluate F (b)− F (a).


Examples of Substitutions

Exercise: Compute the area of the circle of radius r > 0.


We use r = 2, the general case is not that different.

A = 2

∫ 2

−2

√4− x2 dx .

Make the substitution x = 2 sin θ with −π/2 ≤ θ ≤ π/2. Thendx = 2 cos xdx . If x ranges between −2 ≤ x ≤ 2, then −π/2 ≤ θ ≤ π/2.

So we change the endpoints of the integral:

A = 2

∫ π/2

−π/2

√4− 4 sin2 θ · 2 cos θ dθ.

Because 4− 4 sin2 θ = 4(1− sin2 θ) = 4 cos2 θ,√

4− 4x2 = 2| cos θ|; thesquare root must be nonnegative, but cos θ may be negative.

!Again, note that −π/2 ≤ θ ≤ π/2, cos θ ≥ 0!


The area is

A = 8

∫ π/2

−π/2cos2 θ dθ.


The area is

A = 8

∫ π/2

−π/2cos2 θ dθ.

Important Formulas for Integration

sin(a± b) = sin a cos b ± sin b cos a cos(a± b) = cos a cos b ∓ sin a sin b

sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x

cos 2x = 2 cos2 x − 1 cos 2x = 1− 2 sin2 x .


Important Formulas for Integration

sin(a± b) = sin a cos b ± sin b cos a cos(a± b) = cos a cos b ∓ sin a sin b

sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x

cos 2x = 2 cos2 x − 1 cos 2x = 1− 2 sin2 x .

This gives

sin2 x =1− cos 2x

2cos2 x =

1 + cos 2x

2

These will be available on the formula sheets.No need to memorize, but make sure you practice with them!


So

A = 8

∫ π/2

−π/2

1 + cos 2θ

2dθ = 4

∫ π/2

−π/21 dθ + 4

∫ π/2

−π/2cos 2θ dθ = 4π.

In other words, the second integral is 0. Here’s a simple way to see this,using symmetry.Make the substitution u = 2θ; so du = 2dθ, and the limits of integrationchange from θ = −π/2, π/2 to u = 2θ = −π, π :∫ π/2

−π/2cos 2θ dθ =

∫ π

−πcos u du/2 =

1

2

∫ π

−πcos u du

By symmetry, this integral is 0.


Math 1120 Class 2pi.math.cornell.edu/~web1120/slides/fall12/aug28.pdfMath 1120 Class 2 Dan Barbasch...

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