4.1 : anti-derivatives greg kelly, hanford high school, richland, washington

4.1 : Anti-derivatives

Greg Kelly, Hanford High School, Richland, Washington

First, a little review:

Consider:2 3y x

then: 2y x 2y x

2 5y x or

It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears.

However, when we try to reverse the operation:

Given: 2y x find y

2y x C

We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant.

If we have some more information we can find C.

Given: and when , find the equation for .2y x y4y 1x

2y x C 24 1 C

3 C2 3y x

This is called an initial value problem. We need the initial values to find the constant.

An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation.

4.1 Antiderivatives and Indefinite Integration

Exploration

Finding Antiderivatives For each of the following derivatives, describe the original

function .F

(a) '( ) 2F x x

(b) '( )F x x

2(c) '( )F x x

2

1(d) '( )F x

x

3

1(e) '( )F x

x

(f) '( ) cosF x x

2( )F x x

2( ) / 2F x x

3( ) / 3F x x

( ) 1/F x x

2( ) 1/ 2F x x

( ) sinF x x


2( ) 3f x x

Find the function that is the antiderivative of .F f

3( ) + C becauseF x x3

2[ ]3

d x Cx

dx

2

Here are some members of the

family of antiderivatives of ( ) 3 :f x x

31( ) 5F x x 3

2 ( ) 36F x x ...etc

33( )

3

xF x C


2[ ] 2xD x x

The family of all antiderivatives of ( ) 2 is

represented by

f x x

2( )F x x C


Find the general solution of the differential equation

' 2y 2y x C


:Notation

( )y f x dx ( )F x C

Integrand

Variable of

Integration

Constant of

Integration

is the antiderivative (indefinite integral) of ( )

with respect to .

y f x

x


3xdx 3 xdx 2

32

xC

23

2x C

RewriteIntegrate

Simplify


3

1dx

x 3x dx 2

1

2C

x

2

2

xC

xdx 1/ 2x dx 3 / 2

3/ 2

xC

3 / 22

3x C

2sin xdx 2 sin xdx 2 ( cos )x C

2cos x C


( 2)x dx 2

1 222

xC x C

2xdx dx

2

22

xx C


4 2(3 5 )x x x dx May we skip step 2?????

5 3 23 5 1

5 3 2x x x C

5 3 2

3 55 3 2

x x xC


1xdx

x

1/ 2 1/ 2( )x x dx 3 / 2 1/ 2

3/ 2 1/ 2

x xC 3 / 2 1/ 22

23x x C

Never integrate the numerator and denominator seperately!

Rewrite the Integrand


2 5 3 2x xdx

x

5/2 3/2 1/212 2220

5 3x x x C


2

sin

cos

xdxx

1 sin

cos cos

xdx

x x

sec tanx xdx sec x C

Front Cover#15?

????


Find the general solution of

2

1'( ) , 0f x F x x

x

2( )F x x dx1

1

xC

1C

x


Find the particular solution that

satisfies the initial condition

(1) 0F 1

(1)1

F C 0

1C

1( ) 1, 0F x x

x

1F x C

x


Solve the differential equation given the initial condition.

3 1, 2 3f x x f

231

2f x x x


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 as a function of the time .s t

0 initial timet

Given Initial Conditions:Acceleration due to gravity: 32 feet per second per second

''( ) 32s t

'(0) 64s (0) &80s

80 ft



0 initial timet Given Conditions:Acceleration due to gravity: 32 feet per second per second

''( ) 32s t

'(0) 64s (0) &80s

'( )s t ''( ) 32s t dt dt 132t C

164 32(0) C 1 64C

'( ) 32 64s t t



( ) '( ) ( 32 64)s t s t dt t dt 2

232 642

tt C

2216 64t t C Using (0) 80:s

2

280 16 0 64 0 C 2 80C 2( ) 16 64 80s t t t

(b) When does the ball hit the ground?

2( ) 64 816 0s t t t 0216( 4 5) 0t t

16( 5)( 1) 0t t

5 secondst

80 ft

5t

Basic Integration Rules

These two equations allow you to obtain integration formulas directly from differentiation formulas, as shown in the following summary.

Basic Integration Rulescont’d


Before you begin the exercise set, be sure you realize

that one of the most important steps in integration is

rewriting the integrand

in a form that fits the basic integration rules.

Practice Exercises

Original Rewrite Integrate Simplify

2dxx

22 1t dt

3

2

3xdx

x

3 4x x dx

4.1 Homework

HW 4.1 Wed: pg. 255, 5-14 all, 15-47 odd,55-63 odd 73,77, 79Thurs:MMM pg. 124-125Fri:More practice problems (4.1)

4.1 : anti-derivatives greg kelly, hanford high school, richland, washington

Documents

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