Consider: then: 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

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.

6.1 Antiderivatives with Slope Fields

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.


Initial value problems and differential equations can be illustrated with a slope field.


Definition Slope Field or Directional Field A slope field or a directional field for the first order differentiable equation

is a plot of sort line segments with slopes f(x,y) for a lattice of points (x,y) in the plane.

),( yxfdx

dy

2y x


'yyx0 0 0

0 1 0

0 2 0

0 3 0

1 0 2

1 1 2

2 0 4

-1 0 -2

-2 0 -4

2y x

If you know an initial condition, such as (1,-2), you can sketch the curve.

By following the slope field, you get a rough picture of what the curve looks like.

In this case, it is a parabola.


Integrals such as are called

definite integrals because we can find a

definite value for the answer.

4 2

1x dx

43

1

1

3x C

3 31 14 1

3 3C C

64 1

3 3C C

63

3 21

The constant always cancels when finding a definite integral, so we leave it out!


4 2

1x dx

Integrals such as are called indefinite integrals

because we can not find a definite value for the answer.

2x dx

2x dx31

3x C

When finding indefinite integrals, we always include the “plus C”.



Definition Indefinite Integral

The set of all antiderivatives of a function f(x) is theindefinite integral of f with respect to x and is denoted by

dxxf )(

CxFdxxf )()(


dxxn.1 1,1

1

nCn

xn

x

dx.2 Cx ln

dxex.3 Cex

dxxsin.4 Cx cos


dxxcos.5

dxx2sec.6

dxx2csc.7

dxxx tansec.8

Cx sin

Cx sec

Cx cot

Cx tan

dxxx cotcsc.9 Cx csc


dxxx )32( 3 Cxxx

34

24

dxxx3Cx 2

9

9

2

dttt

cos12 Ct

t sin

1


dxxxx )cos(tansec Cxx sec

dxxx )cscsin1( 2 Cxx cos

dtte t )( 22 Cte t

32

32


Find the position function for v(t) = t3 - 2t2 + t s(0) = 1

Cttt

ts 23

2

4)(

234

C2

)0(

3

)0(2

4

01

234C = 1

123

2

4)(

234

ttt

ts


cabin

cabind )(log cabin

log cabin + C houseboat

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

6.2 Integration by Substitution

52x dx Let 2u x

du dx5u du61

6u C

62

6

xC

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!


21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integrand.

The derivative of is .21 x 2 x dx1

2 u du3

22

3u C

3

2 22

13

x C

2Let 1u x 2 du x dx

Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.


4 1 x dx Let 4 1u x

4 du dx

1

4du dx

Solve for dx.1

21

4

u du3

22 1

3 4u C

3

21

6u C

3

21

4 16

x C


cos 7 5 x dx7 du dx

1

7du dx

1cos

7u du

1sin

7u C

1sin 7 5

7x C

Let 7 5u x


2 3sin x x dx 3Let u x23 du x dx

21

3du x dx

We solve for because we can find it in the integrand.

2 x dx

1sin

3u du

1cos

3u C

31cos

3x C


4sin cos x x dx Let sinu x

cos du x dx 4sin cos x x dx4 u du 51

5u C

51sin

5x C


24

0tan sec x x dx

The technique is a little different for definite integrals.

Let tanu x2sec du x dx

0 tan 0 0u

tan 14 4

u

1

0 u du We can find

new limits, and then we don’t have to substitute back.

new limit

new limit1

2

0

1

2u

1

2We could have substituted back and used the original limits.


24

0tan sec x x dx

Let tanu x2sec du x dx

4

0 u du

Wrong!The limits don’t match!

42

0

1tan

2x

2

21 1tan tan 0

2 4 2

2 21 11 0

2 2

u du21

2u 1

2

Using the original limits:

Leave the limits out until you substitute back. This is

usually more work than finding new limits


1 2 3

13 x 1 x dx

3Let 1u x

23 du x dx 1 0u

1 2u 12

2

0 u du23

2

0

2

3u

Don’t forget to use the new limits.

3

22

23

22 2

3 4 2

3



dxxx 212

dxx 43

dttt 8235

dtttan

d cossin3

d 2csccot


dxx

x

5

2 1

2

dxxx 2sincos4

2

xx

dxe

e ln

2

dxe

ex

x

5

5

3

1

0 xe

dx

dxxx 21

Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators.

In another generation or so, we might be able to use the calculator to find all integrals.

You must practice finding integrals by hand until you are good at it!


Start with the product rule:

d dv duuv u v

dx dx dx

d uv u dv v du

d uv v du u dv

u dv d uv v du

u dv d uv v du

u dv d uv v du

u dv uv v du This is the Integration by Parts formula.

6.3 Integrating by Parts

u dv uv v du

The Integration by Parts formula is a “product rule” for integration.

u differentiates to zero (usually).

dv is easy to integrate.

Choose u in this order: LIPET

Logs, Inverse trig, Polynomial, Exponential, Trig


cos x x dxpolynomial factor u x

du dx

cos dv x dx

sinv x

u dv uv v du LIPET

sin cosx x x C

u v v du

sin sin x x x dx


ln x dxlogarithmic factor lnu x

1du dx

x

dv dx

v x

u dv uv v du LIPET

lnx x x C 1ln x x x dx

x

u v v du


This is still a product, so we need to use integration by parts again.

2 xx e dx u dv uv v du LIPET

2u x xdv e dx

2 du x dx xv e u v v du

2 2 x xx e e x dx 2 2 x xx e xe dx

u x xdv e dx

du dx xv e

2 2x x xx e xe e dx 2 2 2x x xx e xe e C


A Shortcut: Tabular Integration

Tabular integration works for integrals of the form:

f x g x dx

where: Differentiates to zero in several steps.

Integrates repeatedly.


Also called tic-tac-toe method

2 xx e dx & deriv.f x & integralsg x

2x

2x

2

0

xexexexe

2 xx e dx 2 xx e 2 xxe 2 xe C

Compare this with the same problem done the other way:


3 sin x x dx 3x

23x

6x

6

sin x

cos x

sin xcos x

0

sin x

3 cosx x 2 3 sinx x 6 cosx x 6sin x + C


cos xe x dxLIPET

sinu x xdv e dx

cos du x dx xv e

u v v du cos sin x xx e e x dx cos sin x xx e e x dx

cosu x xdv e dx

sin du x dx xv e

cos sin cos x x xx e x e e x dx This is the expression we started with!


cos xe x dx

sinu x xdv e dxcos du x dx xv e

cos sin x xx e e x dx cos sin x xx e e x dx

cosu x xdv e dxsin du x dx xv e

cos sin cos x x xx e x e e x dx cos xe x dx 2 cos cos sinx x xe x dx x e x e

sin coscos

2

x xx x e x e

e x dx C



The goal of integrating by parts is to go from an integral that we don’t see how to integrate to an integral that we canevaluate.

dvu

duv

The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.)

So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by: dy

kydt

6.4 Exponential Growth and Decay

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.


Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

ln y kt Ce e

C kty e e


C kty e e

kty Ae Since is a constant, let .Ce Ce A


At , .0t 0y y00

ky Ae

0y A

1

0kty y e This is the solution to our

original initial value problem.

0kty y eExponential Change:

If the constant k is positive then the equation

represents growth. If k is negative then the equation represents decay.


Continuously Compounded Interest

If money is invested in a fixed-interest account where the interest is added to the account k times per year, the

amount present after t years is: 0 1kt

rA t A

k

If the money is added back more frequently, you will make a little more money.

The best you can do is if the interest is added continuously.


Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine.

We could calculate: 0lim 1kt

k

rA

k

Since the interest is proportional to the amount present, the equation becomes:

Continuously Compounded Interest:

0rtA A e

You may also use:

rtA Pe

which is the same thing.


Radioactive Decay

The equation for the amount of a radioactive element left after time t is:

The half-life is the time required for half the material to decay.


ktOeyy

Radioactive Decay

60 mg of radon, half-life of 1690 years. How much is left after 100 years?


ke16906030

ke1690

2

1

k16902

1ln

00041.k

)100)(00041.(60 ey mgy 58

ktOeyy

100 bacteria are present initially and double every 12 minutes. How long before there are 1,000,000


ke12100200

ke122

k122ln

0577.k

))(0577(.100000,000,1 te

te 0577.10000

ktOeyy

minutes159t

Half-life

0 0

1

2kty y e

1ln ln

2kte

ln1 ln 2 kt 0

ln 2 kt ln 2t

k

Half-life:

ln 2half-life

k


Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air.

If we solve the differential equation:

s

dTk T T

dt

we get: Newton’s Law of Cooling

0kt

s sT T T T e

where is the temperature of the surrounding medium, which is a constant.

sT


Years

Bears

6.5 Population Growth

We have used the exponential growth equationto represent population growth.

0kty y e

The exponential growth equation occurs when the rate of growth is proportional to the amount present.

If we use P to represent the population, the differential equation becomes: dP

kPdt

The constant k is called the relative growth rate.

/dP dtk

P


The population growth model becomes: 0ktP P e

However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal.

There is a maximum population, or carrying capacity, M.


A more realistic model is the logistic growth model where

growth rate is proportional to both the amount present (P)

and the fraction of the carrying capacity that remains: M P

M


The equation then becomes: dP M PkP

dt M

Our book writes it this way:Logistics Differential Equation

dP kP M P

dt M

We can solve this differential equation to find the logistics growth model.


PartialFractions

Logistics Differential Equation

dP kP M P

dt M

1 k

dP dtP M P M

1 A B

P M P P M P

1 A M P BP

1 AM AP BP

1 AM

1A

M

0 AP BP

AP BPA B1

BM

1 1 1 kdP dt

M P M P M

ln lnP M P kt C

lnP

kt CM P



dP kP M P

dt M

1 k

dP dtP M P M

1 1 1 kdP dt

M P M P M

ln lnP M P kt C

lnP

kt CM P

kt CPe

M P

kt CM Pe

P

1 kt CMe

P

1 kt CMe

P



kt CPe

M P

kt CM Pe

P

1 kt CMe

P

1 kt CMe

P

1 kt C

MP

e

1 C kt

MP

e e

CLet A e

1 kt

MP

Ae


Logistics Growth Model

1 kt

MP

Ae


Logistic Growth Model

Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.

Assuming a logistic growth model, when will the bear population reach 50? 75? 100?


Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.

Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

1 kt

MP

Ae 100M 0 10P 10 23P


1 kt

MP

Ae 100M 0 10P 10 23P

0

10010

1 Ae

10010

1 A

10 10 100A

10 90A

9A

At time zero, the population is 10.

100

1 9 ktP

e


1 kt

MP

Ae 100M 0 10P 10 23P

After 10 years, the population is 23.

100

1 9 ktP

e

10

10023

1 9 ke

10 1001 9

23ke

10 779

23ke

10 0.371981ke

10 0.988913k

0.098891k

0.1

100

1 9 tP

e


0.1

100

1 9 tP

e

Years

BearsWe can graph this equation and use “trace” to find the solutions.

y=50 at 22 years y=75 at 33 years y=100 at 75 years


Leonhard Euler 1707 - 1783

Leonhard Euler made a huge number of contributions to mathematics, almost half after he was totally blind.

(When this portrait was made he had already lost most of the sight in his right eye.)

6.6 Euler’s Method

Leonhard Euler 1707 - 1783

It was Euler who originated the following notations:

e (base of natural log)

f x (function notation)

(pi)

i 1

(summation)y (finite change)


There are many differential equations that can not be solved.We can still find an approximate solution.

We will practice with an easy one that can be solved.

2dy

xdx

Initial value: 0 1y


Recall the formula for local linearization


dxxyxyxL )(')()( 00

dxyxfyy ),( 0001 where ),( yxfdx

dy

2dy

xdx

0 1y 0.5dx


dxyxfyy ),( 0001

1)5)(.0(11 y

(0,1)

(.5,1)

dxyxfyy ),( 1112

5.1)5)(.1(12 y (1,1.5)


5.2)5)(.2(5.13 y

dxyxfyy ),( 2223

(1.5,2.5)

dxyxfyy ),( 3334

4)5)(.3(5.24 y (2,4)

2dy

xdx

0,1 0.5dx

2 dy x dx 2y x C

1 0 C 2 1y x

Exact Solution:


It is more accurate if a smaller value is used for dx.

This is called Euler’s Method.

It gets less accurate as you move away from the initial value.


The book refers to an “Improved Euler’s Method”. We will not be using it, and you do not need to know it.

The calculator also contains a similar but more complicated (and more accurate) formula called the Runge-Kutta method.

You don’t need to know anything about it other than the fact that it is used more often in real life.

This is the RK solution method on your calculator.


Documents

Consider: then: 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