Warm-Up

4-1: Antiderivatives & Indefinite Integrals

©2002 Roy L. Gover (www.mrgover.com)

Objectives:•Define the antiderivative (indefinite integral)•Learn basic antidifferentiation rules•Solve simple differential equations

Definition

If F(x) is an antiderivative of f(x) on an interval I, then F’ (x)=f(x) + c where c is a constant.

Example

Let be an antiderivative of f(x) then

3( ) 2F x x

2'( ) 6 ( )F x x f x c

The Problem...

Find the function given its derivative. The function is the antiderivative or indefinite integral.

Example

'( )f x xthe function whose derivative is . Confirm your answer.

x

If ,find f(x),

Try This2'( )f x x

the function whose derivative is . Confirm your answer.

2x

31( )

3f x x c

If ,find f(x),

Try This3'( )f x x

the function whose derivative is . Confirm your answer.

3x

41( )

4f x x c

If ,find f(x),

Try This5'( )f x x

the function whose derivative is . Confirm your answer.

5x

61( )

6f x x c

If ,find f(x),

Try This'( ) nf x x

the function whose derivative is .

nx

If ,find f(x),

11( )

1nf x x c

n

aka the power rule for integration.

Example

Find the antiderivative of

2x2.

Example

Find the antiderivative of cos x.

Try This

Find the antiderivative of sin x.

-cos x+c

Example

Find the general solution of the differential equation:

4dy

dx

The general solution to:

is: y=4x+c

because:

4dy

dx

[4 ]4

d x c

dx

Multiple solutions exist because:

Important Idea

[4 0]4

d x

dx

[4 6]4

d x

dx

[4 12]4

d x

dx

4 Solutions to :

0

10

20

30

40

1 2 3 4 5

c=0

c=6

c=12

c=18

4dy

dx y=4x+c

Definition

( )F f x dx c Integrand

Variable of Integration

Constant of Integration

Integral Sign

Example

Find the antiderivative:

2(2 3 4)x x dx

Power Rule for Integration1

1

nn x

x dx cn

1n

Special Case:

0 1 101

0 1 1

x xdx dx x dx x

Definition

Sum& Difference Rule for Integration

[ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx The integral of the sum is the sum of the integrals…

Definition

The integral of the difference is the difference of the integrals

ExampleEvaluate the indefinite

integral:

5

1dx

x

Try ThisFind the antiderivative:

3 2( 2)x dxHint: re-write and use the power rule.

5

332

5x x c

Assignment

page 255 9-14 all,15-27 odd

Assignment

page 255 9-14 all,15-27 odd

Warm-Up Find the antiderivative:

3

( )x x

dxx

Hint: Re-write as 2 fractions.

Solution

3

3

xx c

ExampleEvaluate the indefinite integral:

2

cos

1 cos

xdx

x

Lesson CloseSee page 250 of your text for basic integration rules.

Since integration is the inverse of differentiation, you already know most of the rules.

Memor

iz

e

Intro to Differential Equations.

We need to do some work on Differential equations…

ExampleSometimes we want a particular antiderivative...

Find the equation for y given that y’=2x+3 and y passes through the point (2,1)

ExampleFind the equation for y given that y’=2x+3 and y passes through the point (2,1)

You try

Find the general solution of F'(x) = x3 + 2x and the particular solution passing through point (2, 6) Note: (2, 6) means F(2) = 6

Particular SolutionsFind the general solution of F'(x) = x3 + 2x and the particular solution passing through point (2, 6)F(x) = ∫x3 + 2x = x4 + x2 + C 4F(2) = 6 = 24 + 22 + C 4

general solution

F(x) = x4 + x2 - 2 4

particular solution

-2 = c

Solving a Problem

A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

Assignment

page 255 #35-41odd, 55-61 odd, 63, 69.

QR code activity1) If you do not have a QR code scanner, partner with someone who does.2) Choose a random QR Code to scan, and solve the given problem.3) In the appropriate blank, record your answer and justification on the back of your worksheet.4) Use that answer to find the next problem. (e.g. The answer to #7 should be at the bottom of the QR code for #8)

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

s' = ∫s'' = ∫-32 feet/secposition = s velocity = s' acceleration = s''

s' = -32t + c

How can we find c?

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

position = s velocity = s' acceleration = s'' s' = -32t + c

initial velocity is 64 when time is zero 64 = -32(0) + c

64 = c

s' = -32t + 64

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

position = s velocity = s' acceleration = s'' s = ∫s' = ∫-32t + 64

s = -32t2 + 64t + c 2How do we find c this time?

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

position = s velocity = s' acceleration = s'' s = -32t2 + 64t + c

2initial height is 80 (time is 0)

80 = -32(0)2 + 64(0) + c 280 = c

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?

position = s velocity = s' acceleration = s''

80 = -32(0)

2 + 64(0) + c

280 = cs = -16t2 +64t + 80

So when does the ball hit the ground?

Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?s = -16t2 +64t + 80

The position will be zero.

0 = -16t2 +64t + 800 = -16(t - 5)(t + 1) t = 5 or -1