6.1 Antiderivatives with Slope
Fields
Consider:
then:
y  x2  3
y  x2  5
or
y  2 x
y  2 x
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: y   2 x
y  x2  C
find
y
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: y   2 x and y  4 when x  1 , find the equation for y .
y  x2  C
4  12  C
3C
y  x2  3
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.
6.1 Antiderivatives with Slope
Fields
Definition Slope Field or Directional Field
A slope field or a directional field for the first order
differentiable equation
dy
 f ( x, y )
dx
is a plot of sort line segments with slopes f(x,y)
for a lattice of points (x,y) in the plane.
Initial value problems and differential equations can be
illustrated with a slope field.
6.1 Antiderivatives with Slope
Fields
y  2 x
x y y'
0 0
0 1
0 2
0 3
1 0
1 1
2 0
-1 0
-2 0
0
0
0
0
2
2
4
-2
-4
6.1 Antiderivatives with Slope
Fields
y  2 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.
6.1 Antiderivatives with Slope
Fields

4
1
2
Integrals such as
x dx

4
1
4
1 3
x C
3
1
x dx
are called
definite integrals because we can find a
definite value for the answer.
1 3
 1 3

  4  C    1  C 
3
 3

64
1
C  C
3
3
2
63

3
The constant always cancels
when finding a definite
integral, so we leave it out!
 21
6.1 Antiderivatives with Slope
Fields
Integrals such as
 x dx
2
are called indefinite integrals
because we can not find a definite value for the answer.
2
x
 dx
1 3
x C
3
When finding indefinite
integrals, we always
include the “plus C”.
6.1 Antiderivatives with Slope
Fields
Definition Indefinite Integral
The set of all antiderivatives of a function f(x) is the
indefinite integral of f with respect to x and is denoted by
 f ( x)dx
 f ( x)dx  F ( x)  C
6.1 Antiderivatives with Slope
Fields
1.  x dx 
n
dx
2. 

x
x n1
 C , n  1
n 1
ln x  C
3.  e dx  e  C
x
x
4.  sin x dx   cos x  C
6.1 Antiderivatives with Slope
Fields
5.  cos x dx 
sin x  C
6.  sec x dx 
tan x  C
7.  csc x dx 
 cot x  C
8.  sec x tan x dx 
sec x  C
2
2
9.  csc x cot x dx   csc x  C
6.1 Antiderivatives with Slope
Fields
 (x
x

3
3
 2 x  3)dx
x dx
1

  t 2  cos t  dt
4
x
2
 x  3x  C
4
9
2
2
x C
9
1
  sin t  C
t
6.1 Antiderivatives with Slope
Fields
 sec x(tan x  cos x)dx
sec x  x  C
 (1  sin
x  cos x  C
 (e
2t
2
x csc x) dx
 t ) dt
2
e 2t t 3
 C
2 3
6.1 Antiderivatives with Slope
Fields
Find the position function for v(t) = t3 - 2t2 + t s(0) = 1
4
3
2
t
2t t
s(t )  
 C
4 3
2
4
3
2
0 2(0) (0)
1 

C
4
3
2
4
3
2
t
2t t
s(t )  
 1
4
3
2
C=1
6.1 Antiderivatives with Slope
Fields

d (cabin )

cabin
log cabin + C
log cabin
houseboat
6.2 Integration by Substitution
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
  x  2
5
dx
 u du
Let u  x  2
du  dx
5
1 6
u C
6
 x  2
6
6
C
The variable of integration
must match the variable in
the expression.
Don’t forget to substitute the
value for u back into the
problem!
6.2 Integration by Substitution

1  x  2 x dx
2
u
1
2
1 x
2
Let u  1  x
du  2 x dx
2
The derivative of
du
3
2
2
u C
3

One of the clues that we look for is
if we can find a function and its
derivative in the integrand.
2
1 x
3
3
2 2

C
is
2x dx .
Note that this only worked because
of the 2x in the original.
Many integrals can not be done by
substitution.
6.2 Integration by Substitution

4 x  1 dx
Let u  4x 1
du  4 dx
1
2
1
 u  4 du
3
2
1
du  dx
4
2
1
u  C
3
4
3
2
1
u C
6
3
1
 4 x  1 2  C
6
Solve for dx.
6.2 Integration by Substitution
 cos  7 x  5 dx
1
 cos u  7 du
1
sin u  C
7
1
sin  7 x  5   C
7
Let u  7 x  5
du  7 dx
1
du  dx
7
6.2 Integration by Substitution
 
2
3
x
sin
x
dx

1
sin u du

3
1
 cos u  C
3
1
3
 cos x  C
3
Let u  x3
du  3x 2 dx
1
du  x 2 dx
3
2
We solve for x dx
because we can find it
in the integrand.
6.2 Integration by Substitution
 sin
4
x  cos x dx
 sin x 
4
u
 du
4
cos x dx
1 5
u C
5
1 5
sin x  C
5
Let u  sin x
du  cos x dx
6.2 Integration by Substitution


4
0
2
tan x sec x dx
new limit
1
 u du
0
new limit
1
1 2
u
2 0
1
2
The technique is a little different
for definite integrals.
Let u  tan x
2
du  sec x dx
We can find
new limits,
u 0  tan 0  0
and then we

 
don’t have
u    tan  1
to substitute
4
4
 
back.
We could have substituted back and
used the original limits.
 
6.2 Integration by Substitution

4
0

Using the original limits:

tan x sec2 x dx
du  sec x dx

4
0
Let u  tan x
2
u du
 u du
1
 1
2
  tan    tan 0 
2
4
2


Wrong!
2
Leave the
limits out until
you substitute
back.
1 2
 u

2
1
2 4
  tan x 
2
0
This is
The
match!
1 limits
1 don’t
usually
2
2
1
 1   0 
more work
2
2
2 than finding
new limits
6.2 Integration by Substitution

1
1

2
0
3x
2
x  1 dx
3
Let u  x3  1
du  3x dx
2
1
2
u du
2
u
3
3 2
2
2
2
3
u 1  2
Don’t forget to use the new limits.
0
3
2
u  1  0
2
 2 2
3
4 2

3
6.2 Integration by Substitution
 2 x 1  x dx
2

3x  4 dx
 t 5  3t  dt
2 8
 tan t dt
3
sin
  cos d

cot  csc 2  d
6.2 Integration by Substitution

x2
dx
x 1
5
2




4
5x
e
 3  e5 x dx
cos x sin 2 x dx

dx
0
e
dx
x ln x

1  x x 2 dx
1
x
2

e2
e
6.2 Integration by Substitution
In another generation or so, we might be able
to use the calculator to find all integrals.
Until then, remember that half the AP exam and
half the nation’s college professors do not allow
calculators.
You must practice finding integrals by hand until
you are good at it!
6.3 Integrating by Parts
Start with the product rule:
d
dv
du
 uv   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
u differentiates to
dv is easy to
zero (usually).
integrate.
The Integration by Parts formula is a “product rule” for
integration.
Choose u in this order:
LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
6.3 Integrating by Parts
 x  cos x dx
polynomial factor
u v   v du
x  sin x   sin x dx
 u dv  uv   v du
LIPET
ux
dv  cos x dx
du  dx
v  sin x
x  sin x  cos x  C
6.3 Integrating by Parts
 ln x dx
logarithmic factor
u v   v du
1
ln x  x   x  dx
x
 u dv  uv   v du
LIPET
u  ln x
dv  dx
1
du  dx
x
vx
x ln x  x  C
6.3 Integrating by Parts
 x e dx
u v   v du
x e   e  2 x dx
2 x
2 x
 u dv  uv   v du
x
x

x e  2 xe   e dx
2 x
x
u  x2
dv  e x dx
du  2 x dx
v  ex
ux
x e  2  xe dx
2 x
LIPET
x

dv  e x dx
This is still a product, so
x we
v

e
du  dx
need to use integration by
parts again.
x e  2 xe  2e  C
2 x
x
x
6.3 Integrating by Parts
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
6.3 Integrating by Parts
2 x
x
 e dx
f  x  & deriv. g  x  & integrals
2
x

ex
 2x
e
x
 2
ex
0
ex
Compare this with
the same problem
done the other way:
x
2 x
x

2
e
x
e

2
xe
C
 x e dx 
2 x
6.3 Integrating by Parts
x
3
sin x dx
3
x

 3x 2
sin x
 cos x
 6x
 6
 sin x
cos x
0
sin x
 x 3 cos x  3x 2 sin x  6x cos x  6sin x + C
6.3 Integrating by Parts
LIPET
e
x
u  cos x
cos x dx
u v   v du
cos x  e x   e x   sin x dx
cos x  e x   e x sin x dx
du   sin x dx
v  ex
u  sin x
dv  e x dx
du  cos x dx
cos x  e  sin x  e   e cos x dx
x
x
x
dv  e dx
x
ve
x
This is the
expression we
started with!
6.3 Integrating by Parts
e
x
u  cos x dv  e dx
x
du   sin x dx v  e
x
cos x dx
cos x  e x   e x   sin x dx
cos x  e x   e x sin x dx
x
u  sin x
dv  e dx
du  cos x dx v  e x
 e cos x dx  cos x  e  sin x  e   e
2  e cos x dx  cos x  e  sin x  e
sin x  e  cos x  e
C
 e cos x dx 
2
x
x
x
x
x
x
x
x
x
x
cos x dx
6.3 Integrating by Parts
The goal of integrating by parts is to go from
an integral  u dv that we don’t see how to
integrate to an integral  v du that we can
evaluate.
6.4 Exponential Growth and
Decay
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,
dy
the change can be modeled by:
dt
 ky
6.4 Exponential Growth and
Decay
dy
 ky
dt
1
dy  k dt
y
1
 y dy   k dt
ln y  kt  C
Rate of change is proportional
to the amount present.
Divide both sides by y.
Integrate both sides.
6.4 Exponential Growth and
Decay
ln y
e
e
kt C
y  e e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.
6.4 Exponential Growth and
Decay
y  e e
C kt
y  Ae
kt
y0  Ae
y0  A
Since
 eC
is a constant, let  e
C
A.
1
k 0
At
t  0 , y  y0
y  y0 e
kt
.
This is the solution to our
original initial value
problem.
6.4 Exponential Growth and
Decay
Exponential Change:
y  y0 e
kt
If the constant k is positive then the equation
represents growth. If k is negative then the equation
represents decay.
6.4 Exponential Growth and
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
kt
 r
amount present after t years is:
A  t   A0  1  
 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.
6.4 Exponential Growth and
Decay
Of course, the bank does not employ some clerk to
continuously calculate your interest with an adding machine.
We could calculate:
 r
lim A0 1  
k 
 k
kt
Since the interest is proportional to the amount present,
the equation becomes:
Continuously Compounded
Interest:
A  A0e
rt
You may also use:
A  Pe
rt
which is the same thing.
6.4 Exponential Growth and
Decay
Radioactive Decay
The equation for the amount of
a radioactive element left after
time t is:
y  yO e
kt
The half-life is the time required for half the material to decay.
6.4 Exponential Growth and
Decay
Radioactive Decay
60 mg of radon, half-life of
1690 years. How much is left
after 100 years?
y  yO e
kt
30  60e1690k
1
 e1690k
2
1
ln  1690k
2
k  .00041
y  60e
( .00041)(100)
y  58 mg
6.4 Exponential Growth and
Decay
100 bacteria are present initially and double
every 12 minutes. How long before there are
1,000,000
y  yO e
kt
200  100e
12 k
k  .0577
1,000,000  100e(.0577)(t )
2e
10000  e.0577t
ln 2  12k
t  159 minutes
12 k
6.4 Exponential Growth and
Decay
1
 kt
y0  y0 e
2
1
ln    ln e  kt
2
0
ln1  ln 2  kt
Half-life
 
Half-life:
ln 2  kt
ln 2
t
k
ln 2
half-life 
k
6.4 Exponential Growth and
Decay
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.
dT
If we solve the
 k T  Ts 
differential equation: dt
Newton’s Law of Cooling
we get:
T  Ts  T0  Ts  e kt
where Ts is the temperature
of the surrounding medium,
which is a constant.
6.5 Population Growth
100
80
60
Bears
40
20
0
20
40
Years
60
80
100
6.5 Population Growth
kt
y

y
e
We have used the exponential growth equation
0
to represent population growth.
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 / dt
dP
 kP
dt
The constant k is called the relative growth rate.
P
k
6.5 Population Growth
kt
P

P
e
The population growth model becomes:
0
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.
6.5 Population Growth
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
6.5 Population Growth
The equation then becomes:
dP
M P
 kP 

dt
M


Logistics Differential Equation
Our book writes it this way:
dP k

P M  P
dt M
We can solve this differential equation to find the logistics
growth model.
6.5 Population Growth
Logistics Differential Equation
dP k

P M  P
dt M
1
k
dP 
dt
P  M  P
M
1 1
1 
k
dt
 
 dP 
M P M P
M
ln P  ln  M  P   kt  C
P
ln
 kt  C
M P
1
A
B
 
P  M  P P M  P
1  A  M  P   BP
1  AM  AP  BP
1  AM
1
A
M
Partial
Fractions
0   AP  BP
AP  BP
AB
1
B
M
6.5 Population Growth
Logistics Differential Equation
dP k

P M  P
dt M
1
k
dP 
dt
P  M  P
M
1 1
1 
k
dt
 
 dP 
M P M P
M
ln P  ln  M  P   kt  C
P
ln
 kt  C
M P
P
 e kt C
M P
M P
 e  kt C
P
M
 1  e  kt C
P
M
 1  e  kt C
P
6.5 Population Growth
Logistics Differential Equation
P
 e kt C
M P
M P
 e  kt C
P
M
 1  e  kt C
P
M
 1  e  kt C
P
M
P
1  e  kt C
P
M
1  e C  e  kt
Let A  e
C
M
P
1  Ae  kt
6.5 Population Growth
Logistics Growth Model
M
P
1  Ae  kt
6.5 Population Growth
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?
6.5 Population Growth
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?
M
P
1  Ae  kt
M  100
P0  10
P10  23
6.5 Population Growth
M
P
1  Ae  kt
M  100
100
10 
1  Ae0
At time zero, the population is 10.
P0  10
P10  23
10 A  90
100
10 
1 A
10 10 A  100
A9
100
P
1  9e  kt
6.5 Population Growth
M
P
1  Ae  kt
M  100
100
23 
1  9e  k 10
100
1  9e

23
77
10 k
9e

23
e10 k  0.371981
10 k
100
P0  10 P10  23 P 
1  9e  kt
After 10 years, the population is 23.
10k  0.988913
k  0.098891
100
P
1  9e 0.1t
6.5 Population Growth
100
100
P
1  9e 0.1t
We can graph
this equation
and use “trace”
to find the
solutions.
80
60
Bears
40
20
0
y=50 at 22 years
20
40
Years
y=75 at 33 years
60
80
100
y=100 at 75 years
6.6 Euler’s Method
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.)
Leonhard Euler 1707 - 1783
6.6 Euler’s Method
It was Euler who originated
the following notations:
f  x  (function notation)
e (base of natural log)

(pi)
i


(summation)
y
1

(finite change)
Leonhard Euler 1707 - 1783
6.6 Euler’s Method
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.
dy
 2x
dx
Initial value:
y0  1
6.6 Euler’s Method
Recall the formula for local linearization
L( x)  y( x0 )  y' ( x0 )dx
y1  y0  f ( x0 , y0 )dx
where
dy
 f ( x, y )
dx
6.6 Euler’s Method
dy
 2x
dx
y0  1
dx  0.5
y1  y0  f ( x0 , y0 )dx
(0,1)
y1  1  (0)(.5)  1
(.5,1)
y2  y1  f ( x1, y1 )dx
y2  1  (1)(.5)  1.5
5
4
3
2
(1,1.5)
1
0
1
2
3
6.6 Euler’s Method
y3  y2  f ( x2 , y2 )dx
y3  1.5  (2)(.5)  2.5
5
(1.5,2.5)
3
y4  y3  f ( x3 , y3 )dx
y4  2.5  (3)(.5)  4
4
2
(2,4)
1
0
1
2
3
6.6 Euler’s Method
5
Exact Solution:
dy
 2x
dx
dy  2 x dx
 0,1 dx  0.5
4
3
y  x2  C
2
1
1 0C
y  x2  1
0
1
2
3
6.6 Euler’s Method
This is called Euler’s Method.
5
4
It is more accurate if a smaller
value is used for dx.
It gets less accurate as you
move away from the initial
value.
3
2
1
0
1
2
3
6.6 Euler’s Method
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.