Section 17.4Integration

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A physicist who knows the velocity of a particle might wish to know its position at a given time.

A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time.

Introduction

In each case, the problem is to find a function F whose derivative is a known function f.

If such a function F exists, it is called an antiderivative of f.

Antiderivatives

Definition

A function F is called an antiderivative of f on

an interval I if F’(x) = f (x) for all x in I.

Find the integral. (Find the antiderivative.)

= ? dxx n

€

1n +1 x n +1 + C

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is

F(x) + C

where C is an arbitrary constant.

Theorem

Antiderivatives

Going back to the function f (x) = x2, we see that the general antiderivative of f is ⅓ x3 + C.

Notation for Antiderivatives

The symbol is traditionally used to represent the most general an antiderivative of f on an open interval and is called the indefinite integral of f .

Thus, means F’(x) = f (x)

( )f x dx

( ) ( )F x f x dx

€

x 2 dx =x 3

3+ C∫

€

x 3

3+ C

€

x 2because the derivative of is

Every antiderivative F of f must be of the form F(x) = G(x) + C, where C is a constant.

Example:

Represents every possible antiderivative of 6x.

Constant of Integration

€

6x dx = 3x 2 + C∫

1

if 11

nn x

x dx C nn

Example:4

3

4

xx dx C

Power Rule for the Indefinite Integral

1 1lnx dx dx x C

x

x xe dx e C

Indefinite Integral of ex and bx

ln

xx b

b dx Cb

Power Rule for the Indefinite Integral

Sum and Difference Rules

f g dx fdx gdx

Example:

2 2x x dx x dx xdx 3 2

3 2

x xC

( ) ( )kf x dx k f x dx ( constant)k

4 43 32 2 2

4 2

x xx dx x dx C C

Constant Multiple Rule

Example:

Integration by Substitution

Method of integration related to chain rule. If u is a function of x, then we can use the formula

/

ff dx du

du dx

Example: Consider the integral:

92 33 5x x dx3 2pick +5, then 3 u x du x dx

10

10

uC

9u du 103 5

10

xC

Sub to get Integrate Back Substitute

Integration by Substitution

Example: Evaluate

3/ 21

10 3/ 2

uC

3/ 225 7

15

xC

25 7x x dxPick u, compute du

Sub in

Sub in

Integrate€

Let u = 5x 2 − 7, du =10x dx

€

1

10⋅10x 5x 2 − 7 dx =

1

10u

12 du∫∫

3ln

dx

x xLet ln then u x xdu dx

3

3ln

dxu du

x x

2

2

uC

2ln

2

xC

Example: Evaluate

Examples on your own:

Find the integral of each:1.) 2.)

3.) 4.)

€

8 dx∫

€

(2x + 6) dx∫

€

(6x 2 −12x + 8) dx∫

€

(9x 2 +12x − 9) dx∫€

F(x) = 8x + c

€

F(x) =2x 2

2+ 6x + c

€

F(x) = x 2 + 6x + c

€

F(x) =6x 3

3−

12x 2

2+ 8x + c

€

F(x) = 2x 3 − 6x 2 + 8x + c

€

F(x) =9x 3

3+

12x 2

2− 9x + c

€

F(x) = 3x 3 + 6x 2 − 9x + c

Find the integral of each:5.) 6.)

7.) 8.)

€

(x − 2)10 dx∫

€

2(2x − 3)4 dx∫

€

(5 − x)6 dx∫

€

(3x −1)4 dx∫€

u = x − 2

du =1 dx

€

u10 du∫

€

F(x) =u11

11+ C

€

F(x) =(x − 2)11

11+ C€

u = 2x − 3

du = 2 dx

€

u4 du∫

€

F(x) =u5

5+ C

€

F(x) =(2x − 3)5

5+ C

€

u = 5 − x

du = −1 dx

€

−u6 du∫

€

−1 u6 du∫

€

F(x) = −u7

7+ C

€

F(x) = −(5 − x)7

7+ C€

u = 3x −1

du = 3 dx

€

1

3u4 du∫

€

1

3u4 du∫

€

F(x) =1

3⋅

u5

5+ C

€

F(x) =(3x −1)5

15+ C

Find the integral of each:9.) 10.)

11.) 12.)

€

1

x + 2dx∫

€

3

x − 4dx∫

€

3x 2 x 3 + 5 dx∫

€

x x 2 − 53 dx∫€

u = x + 2

du =1 dx

€

1

udu∫

€

F(x) = ln u + C

€

F(x) = ln x + 2 + C€

u = x − 4

du =1 dx

€

31

udu∫

€

F(x) = 3ln u + C

€

F(x) = 3ln x − 4 + C

€

u = x 3 + 5

du = 3x 2 dx

€

u du∫

€

u1

2 du∫

€

F(x) =u

32

32

+ C

€

F(x) =2

3x 3 + 5( )

32 + C€

u = x 2 − 5

du = 2x dx

€

u1

3

2du∫

€

F(x) =1

2⋅

u4

3

43

+ C

€

F(x) =3

8x 2 − 5( )

43 + C

Find the integral of each:13.) 14.)

€

10

x 2 dx∫

€

4

x 3 dx∫

€

10x −2 du∫

€

10 x −2 du∫

€

F(x) =10x −1

−1+ C

€

F(x) =−10

x+ C

€

4x −3 du∫

€

4 x −3 du∫

€

F(x) = 4x −2

−2+ C

€

F(x) = −2

x 2 + C