7.1 antiderivatives objectives * find an antiderivative of a function. *evaluate indefinite...

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OBJECTIVES *Find an antiderivative of a function. *Evaluate indefinite integrals using the basic integration formulas. *Use initial conditions, or boundary conditions, to determine an antiderivative. Slide 4.2 - 1

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7.1 Antiderivatives

OBJECTIVES*Find an antiderivative of a function.

*Evaluate indefinite integrals using the basic integration formulas.

*Use initial conditions, or boundary conditions, to determine an antiderivative.

Slide 4.2 - 1

Who comes up with the ready-made functions we find derivatives for? Isn’t it hard sometimes to find a function for total cost, profit, etc.?

Sometimes it is easier to calculate the rate of change of something and get the function for the total from it.

This process, the reverse of finding a derivative, is antidifferentiation.

Slide 4.2 - 2Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 4.2 - 3Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 1: Can you think of a function that would have x2 as its derivative?

Antiderivatives

Slide 4.2 - 4Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

One antiderivative is x3/3. All other antiderivatives differ from this by a constant. So, we can represent any one of them as follows:

To check this, we differentiate .

x3

3C

d

dx

x3

3C

d

dx

x3

3

d

dxC 3

x2

3 0 x2

x3

3C

Slide 4.2 - 5Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

THEOREM

If two functions F and G have the same derivative over an interval, then

F(x) = G(x) + C, where C is a constant.

Antiderivatives

Slide 4.2 - 6Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Antiderivatives

Integrals and Integration

Antidifferentiating is often called integration.

To indicate the antiderivative of x2 is x3/3 +C, we

x2 dx x3

3C,

write

Slide 4.2 - 7Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Antiderivatives

The notationf x dx

is used to represent the antiderivative of f (x).

f x dx F x C,

More generally,

where

F(x) + C is the general form of the antiderivative of f (x).

Slide 4.2 - 8Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

1. k dx kx C (k is a constant)

2. xrdx xr1

r 1C, provided r 1

(To integrate a power of x other than 1, increase the

power by 1 and divide by the increased power.)

THEOREM : Basic Integration Formulas

1

1

13. ln , 0

ln , 0

(

4. or ax

ax ax ax

dxx dx dx x C x

x x

x dx x C x

b ebe dx e C e dx C

a a

We will generally assume that x > 0.)

Slide 4.2 - 9Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 2: Evaluate

4.2 Area, Antiderivatives, and Integrals

x9dx.

Slide 4.2 - 10Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

x9dx x10

10C

Check:

d

dx

x10

10C

10

x9

10 0 x9

Slide 4.2 - 11Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 3: Evaluate

4.2 Area, Antiderivatives, and Integrals

5e4 xdx.

Slide 4.2 - 12Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5e4 xdx 5

4e4 x C

d

dx

5

4e4 x C

5

4e4 x 4 0 5e4 x

Check:

Slide 4.2 - 13Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

THEOREM 4

(The integral of a constant times a function

is the constant times the integral of the function.)

(The integral of a sum or difference is the sum or difference of the integrals.)

4.2 Area, Antiderivatives, and Integrals

Rule A. kf (x)dx k f (x)dx

Rule B. f (x) g(x) dx f (x)dx g(x)dx

Slide 4.2 - 14Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 4: Evaluate

4.2 Area, Antiderivatives, and Integrals

(5x 4x3 )dx.

Slide 4.2 - 15Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

(5x 4x3 )dx 5x dx 4x3dx 5 x dx 4 x3dx 5

x2

2 4

x4

4C

5

2x2 x4 C

Slide 4.2 - 16Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 5: Evaluate and check by differentiation:

4.2 Area, Antiderivatives, and Integrals

a) 7e6 x x dx; b) 1 3

x 1

x4

dx;

Slide 4.2 - 17Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

a) 7e6 x x dx 7e6 xdx x dx

7e6 xdx x1 2

dx

7

6e6 x

x1 21

1

21

C

7

6e6 x

2

3x3 2 C

Antiderivatives

Example 5 (concluded):

Slide 4.2 - 18Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 5 (continued):Check:

Antiderivatives

d

dx

7

6e6 x

2

3x

3

2 C

7

6e6 x 6

3

22

3x1 2 0

7e6 x x

Slide 4.2 - 19Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 5 (continued):

Antiderivatives

44

4 1

3

3

3 1 1b) 1 1 3

3ln4 1

3ln31

3ln3

dx dx dx x dxx x x

xx x C

xx x C

x x Cx

Slide 4.2 - 20Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 5 (concluded):Check:

Antiderivatives

d

dxx 3ln x

x 3

3C

1 31

x 3 1

3x 4 0

1 3

x

1

x4

Slide 4.2 - 21Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 6: Find the function f such that

First find f (x) by integrating.

Antiderivatives

f (x) x2 and f ( 1) 2.

f (x) x2dxf (x)

x3

3C

Slide 4.2 - 22Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 6 (concluded): Then, the initial condition allows us to find C.

Thus,

Antiderivatives

f ( 1) ( 1)3

3C 2

1

3C 2

C 7

3

f (x) x3

3

7

3.

Slide 4.2 - 23Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2

7

Find a function f whose graph has slope '(x) 6 x 4

and goes through the point (1,1).

Example

f

3( ) 2 4 5f x x x

Slide 4.2 - 24Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

8

Suppose a publishing company has found that the marginal cost at a

level of production of x thousand books is given by

50'( ) and that the fixed cost

(cost before the first book can be prod

Example

C xx

uced) is $25,000.

Find the cost function C(x).

1/2C( ) 100 25,000x x

Slide 4.2 - 25Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

0.01

9

Find the cost function for the given marginal cost function.

C'(x)=0.03e

with a fixed cost of $8.

x

Example

0.01C( ) 3 5xx e