Example We can also evaluate a definite integral by interpretation

of definite integral. Ex. Find by interpretation of definite integral. Sol. By the interpretation of definite integral, we know the

definite integral is the area of the region under the curve from 0 to a. From the graph, we see the region is a quarter disk with radius a and centered origin. Therefore,

2 2

0

aa x dx

2 2 2

0

1.

4

aa x dx a

Example Ex. By interpretation of definite integral, find

Sol. (1)

(2)

52

3

13

(1) sin (2) 2 . xdx xdx

5

3

3

sin 0

xdx

2

12 3 xdx

Properties of definite integral Theorem(linearity of integral) Suppose f and g are

integrable on [a,b] and are constants, then

is integrable on [a,b] and

Theorem(product integrability) Suppose f and g are integrable on [a,b], then is integrable on [a,b].

, f g

[ ( ) ( )] ( ) ( ) . b b b

a a af x g x dx f x dx g x dx

f g

Properties of definite integral Theorem(additivity with respect to intervals)

Remark In the above property, c can be any number, not necessarily between a and b.

When the upper limit is less than the lower limit in the definite integral, it is understood as

Especially, ( ) 0.a

af x dx

( ) ( ) . a b

b af x dx f x dx

( ) ( ) ( ) . b c b

a a cf x dx f x dx f x dx

Comparison properties of integral 1. If for then

2. If for then

3. If for then

4.

( ) 0f x , a x b ( ) 0.b

af x dx

( ) ( )f x g x , a x b ( ) ( ) . b b

a af x dx g x dx

, a x b( ) m f x M

( ) ( ) ( ). b

am b a f x dx M b a

( ) | ( ) | . b b

a af x dx f x dx

Estimation of definite integral Ex. Use the comparison properties to estimate the definite

integral

Sol. Denote Then when

Letting we get the only critical number

By the closed interval method, we find the range for f(x):

1 2 31

2

1 . x x dx

2 3( ) 1 . f x x x1

[ ,1],2

x2

2 3

2 3( ) .

2 1

x xf x

x x

( ) 0, f x2

.3

x

69 69 1( ) 1, and thus ( ) .

9 18 2

b

af x f x dx

Mean value theorems for integrals Second mean value theorem for integrals Let

g is integrable and on [a,b]. Then

there exists a number such that

Proof. Let Since

we have and

Hence

or By intermediate value theorem

[ , ],f C a b

( ) 0 (or ( ) 0) g x g x[ , ] a b

( ) ( ) ( ) ( ) . b b

a af x g x dx f g x dx

[ , ][ , ]max ( ), min ( ).

x a bx a bM f x m f x ( ) 0,g x( ) 0

b

ag x dx ( ) ( ) ( ) ( ). mg x f x g x Mg x

( ) ( ) ( ) ( ) , b b b

a a am g x dx f x g x dx M g x dx

( ) ( ).

( )

b

ab

a

f x g x dxm M

g x dx

Mean value theorems for integrals First mean value theorem for integrals Let

then there exists a number such that

Remark. We call the mean value of f on [a,b].

[ , ],f C a b

[ , ] a b( ) ( )( ).

b

af x dx f b a

1( ) ( )

b

af f x dx

b a

Example Ex. Suppose and

Prove that such that

Proof. By the first mean value theorem for integrals, there

exists such that Thus

By Rolle’s theorem, such that

[0,1] and (0,1), f C f D1

2

3

3 ( ) (0). f x dx f ( ) 0. f (0,1)

2[ ,1]3

1

2

3

1( ) ( ).

3 f x dx f

1

2

3

(0) 3 ( ) ( ). f f x dx f

(0,1) ( ) 0. f

Function defined by definite integrals with varying limit

Suppose f is integrable on [a,b]. For any given

the definite integral is a number. Letting x vary

between a and b, the definite integral defines a function:

Ex. Find a formula for the definite integral with varying limit

Sol. By interpretation of definite integral, we have

( )x

af t dt

[ , ],x a b

( ) ( ) .x

ag x f t dt

( ) .x

ag x tdt

2 21( ) ( )( ) .

2 2

x

a

x ag x tdt a x x a

Properties of definite integral with varying limit

Theorem(continuity) If f is integrable on [a,b], then the

definite integral with varying limit

is continuous on [a,b].

( ) ( )x

ag x f t dt

The fundamental theorem of calculus (I)

The Fundamental Theorem of Calculus, Part 1 If f is

continuous on [a,b], then the definite integral with varying

limit is differentiable on [a,b] and

Proof

is between x and as and

Therefore,

( ) ( )x

ag x f t dt

( ) ( ) ( ) ( )

x x x x x

a a xg f t dt f t dt f t dt f x

( ) ( ) ( ). x

a

dg x f t dt f x

dx

, x x 0, , x x ( ) ( ).f f x

0 0( ) lim lim ( ) ( ).

x x

gg x f f x

x

Definite integral with varying limits

The definite integral with varying lower limit is

Since we have

The most general form for a definite integral with varying

limits is To investigate its properties,

we can write it into the sum of two definite integrals with

varying upper limit

( ) ( ) ,b

xh x f t dt

( )

( )( ) ( ) .

b x

a xx f t dt

( ) ( ) ( ). b

x

dh x f t dt f x

dx

( ) ( ) , x

bh x f t dt

( ) ( ) ( )

( )( ) ( ) ( ) ( ) .

b x b x a x

a x a ax f t dt f t dt f t dt

Definite integral with varying limits

By the chain rule, we have the formula

( )( ) ( ( )) ( )

b x

af t dt f b x b x

( ) ( ) ( )

( )( ) ( ) ( )

( ( )) ( ) ( ( )) ( )

b x b x a x

a x a af t dt f t dt f t dt

f b x b x f a x a x

Example Ex. Find derivatives of the following functions

Sol.

(2) Let by chain rule,

2

0(1) sin ,

xt tdt 2

1 2(3) ln(1 ) ,x t dt2

2

sin(4) .

x t

xe dt2

0(2) cos ,

xt dt

2 2

0(1) sin sin .

xdt tdt x x

dx

2 2

0 0

1cos cos cos .

2

x ud d dut dt t dt x

dx du dx x

,u x

2

2

1 2 2 4

1(3) ln(1 ) ln(1 ) 2 ln(1 ).

x

x

d dt dt t dt x x

dx dx2 2

2 2 2 4 2sin sin

sin(4) 2 cos .

x x xt t t x x

x a a

d d de dt e dt e dt xe xe

dx dx dx

Example Ex. Find derivative Sol.

Ex. Find if Sol.

22

2(1) ,

x t

xt e dt 2

0(2) (5 ) .

xx t dt

0 0cos 1.

y xte dt tdtdy

dx

222 5 2 2

2(1) 2 8 .

x t x x

x

dt e dt x e x e

dx

2 2 2 2

0 0 0(2) (5 ) (5 ) (5 ) (5 ) .

x x xd dx t dt x t dt t dt x x

dx dx

0 0

coscos 0 cos 0 .

y xt yy

d dy dy xe dt tdt e x

dx dx dx e

Example Ex. Find the limit

Sol. By L’Hospital’s Rule and equivalent substitution,

Question:

2

030

arctanlim .

ln(1 ) x

x

tdt

x

2 2

0 03 3 20 0 0

arctan arctan 2 arctan 2lim lim lim .

ln(1 ) 3 3

x x

x x x

tdt tdt x x

x x x

2

030

arctanlim ?

ln(1 )

x

x

tdt

x

Example Ex. Find the limit

Sol.

2

2

2

0

0 2

0

(1) lim .

x t

xx t

e dt

e dt

2 2 2

2 2

0 0

20 0

2 2 0(1) lim lim 0.

1

x xt x t

x xx x

e dt e e dt

e e

2 3

2

0

0

0

(2) lim .( sin )

x

xx

t dt

t t t dt

3 2

0 0

2 6(2) lim lim 12.

( sin ) 1 cosx x

x x x

x x x x

Example Ex. Suppose b>0, f continuous and increasing on [0,b].

Prove the inequality

Sol. Let

Then F(0)=0 and when

This implies F(t) is increasing, thus

0 02 ( ) ( ) .

b bxf x dx b f x dx

0 0( ) 2 ( ) ( ) .

t tF t xf x dx t f x dx

0 0( ) 2 ( ) ( ) ( ) ( ) ( ) .

t tF t tf t f x dx tf t tf t f x dx

[0, ],t b

( ) ( ) [ ( ) ( )] 0. tf t tf t f t f

( ) (0).F b F

Example Ex. Suppose f is continuous and positive on [a,b]. Let

Prove that there is a unique solution in (a,b) to F(x)=0. Sol.

1( ) ( ) .

( )

x x

a bF x f t dt dt

f t

( ) 0, ( ) 0. ( ) 0 F a F b F

1( ) ( ) 0.

( ) F x f x

f x

Fundamental theorem of calculus (II) The Fundamental Theorem of Calculus, Part 2 If f is

continuous on [a,b] and F is any antiderivative of f, then

Proof Let then g is an antiderivative of f.

So F(x)=g(x)+C. Therefore,

Remark The formula is called Newton-Leibnitz formula

and often written in the form

( ) ( ) ( ). b

af t dt F b F a

( ) ( ) ,x

ag x f t dt

( ) ( ) ( ( ) ) ( ( ) ) ( ) .b a b

a a aF b F a f t dt C f t dt C f t dt

( ) ( ) ( ) . b bb

a aaf t dt F x F x

Example Ex. Evaluate

Sol.

Ex. Find the area under the parabola from 0 to 1. Sol.

24

0tan . xdx

2 24 4 400 0

tan (sec 1) tan 1 .4

xdx x dx x x

2y x13

1 2

00

1.

3 3

xA x dx

Example Ex. Evaluate

Sol.

Ex. Evaluate Sol.

2

0|1 | . x dx

2 1 2

0 0 1|1 | |1 | |1 | x dx x dx x dx

1 22 21 2

0 10 1

(1 ) ( 1) 1.2 2

x xx dx x dx x x

2

0| sin cos | . x x dx

240

4

sin cos cos sin 2( 2 1). x x x x

Example Anything wrong in the following calculation?

33

211

1 1 4.

3

dxx x

Differentiation and integration are inverse

The fundamental theorem of calculus is summarized into

The first formula says, when differentiation sign meets integral sign, they cancel out.

The second formula says, first differentiate F, and then integrate the result, we arrive back to F.

( ) ( ).x

a

df t dt f x

dx

( ) ( ) ( ). b

aF x dx F b F a

Homework 12 Section 5.1: 21

Section 5.2: 22, 37, 53, 59, 67

Section 5.3: 18, 50, 54, 62