CALCULUS II
Chapter 5
Definite Integral
b

a
 Area above  Area below 
  

f (t )dt 
 x  axis   x  axis 
Example
b

f (t )dt  5
4
3
a
2
http://www.youtube.com/watch?v=LkdodHMcBuc
Properties of the Definite Integral
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
Indefinite Integrals or Antiderivatives
You should distinguish carefully between definite and indefinite
integrals. A definite integral
is a number, whereas an
indefinite integral
is a function (or family of functions).
Antiderivative
An antiderivative of a function f is a function F such that
F  f
Ex.
An antiderivative of
is
F ( x)  3 x 2  2
since
F ( x)  f ( x).
f ( x)  6 x
Indefinite Integral
The expression:

f ( x)dx
read “the indefinite integral of f with respect to x,”
means to find the set of all antiderivatives of f.

Integral sign
f ( x)dx
x is called the variable
of integration
Integrand
Constant of Integration
Every antiderivative F of f must be of the form F(x) = G(x) + C,
where C is a constant.
Notice
 6 xdx  3x
2
C
Represents every possible antiderivative of 6x.
Power Rule for the Indefinite
Integral, Part I

Ex.

n 1
x
x dx 
 C if n  1
n 1
n
4
x
x dx 
C
4
3
Power Rule for the Indefinite
Integral, Part II

1
x dx  dx  ln x  C
x
1

Indefinite Integral of ex and bx
 e dx  e  C
b
 b dx  ln b  C
x
x
x
x
Sum and Difference Rules
  f  g  dx   fdx   gdx
x
x
  x  x  dx   x dx   xdx  3  2  C
3
2
Ex.
2
2
Constant Multiple Rule

Ex.


kf ( x)dx  k f ( x)dx (k constant)
4
4
x
x
2 x dx  2 x dx  2  C 
C
4
2
3

3
http://www.youtube.com/watch?v=Lb8QrUN6Nck
http://www.youtube.com/watch?v=aw_VM_ZDeIo
Table of Indefinite Integrals
http://video.google.com/videoplay?docid=-8662080079069101664#
http://www.youtube.com/watch?v=t3yq21QrGLY&feature=channel
Fundamental Theorem of Calculus
(part 1)
• If
b
F t 
is continuous for a  t  b , then







F
t
dt

F
b

F
a

a
Fundamental Theorem of Calculus
(part 2)
d x
f
t
dt

f
x





dx a
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
Visualization
Fundamental Theorem of Calculus
(part 2)
f ( x) continuous on [a, b]
x
A( x)   f (t ) dt
a
d
  A( x)   f ( x)
dx
The Fundamental
Theorem of Calculus
x
Ex.
If A( x)   3 t 4  5tdt , find A( x ).
a
A( x)  3 x 4  5 x
The Fundamental
Theorem of Calculus
x
Ex.
A( x)   3t dt
2
1
A(0), A(2), A(3) ?
d
 A(x) ?
dx
A(0), A(2), A(3) ?
First Fundamental Theorem:
d x

cos
x
cos
t
dt
dx 

d
x
sin t 
dx
d
dx

0
 sin x  sin    
d
sin x
dx
cos x
1. Derivative of an integral.
2. Derivative matches upper limit of
integration.
3. Lower limit of integration is a
constant.
1. Derivative of an integral.
d x 1
1
dt 
2
2

0
dx 1+t
1 x
2. Derivative matches upper limit of
integration.
3. Lower limit of integration is a
constant.
d x2
cos t dt

dx 0
 
d 2
cos x  x
dx
2
 
cos x 2  2 x
 
2 x cos x 2
The upper limit of integration does not match
the derivative, but we could use the chain
rule.
http://www.youtube.com/watch?v=PGmVvIglZx8
d 5
3t sin t dt

dx x
d x
  3t sin t dt
dx 5
3x sin x
The lower limit of integration is not a constant,
but the upper limit is.
We can change the sign of the integral and
reverse the limits.
Neither limit of integration is a constant.
d x2 1
dt
t

dx 2 x 2  e
We split the integral into two parts.
d 
1
1

dt  
dt 
 0
t
t
2x 2  e
dx  2  e

x2
0
2x
d  x2 1
1

dt  
dt 
 0
t
t
0 2e
dx  2  e

1
It does not matter what
constant we use!
(Limits are reversed.)
1
2x
2
(Chain rule2 is
used.)

2

2  2x 
2
x
2x
x
x
2e
2e
2e
2e
More Ex’s on the FTC
sin x
A( x) 
3
t
dt

2
5
x
A( x)   cos t dt
7
ln x
A( x) 

2
t  1 dt
3
Integration by Substitution
Method of integration related to chain rule differentiation. If u is a
function of x, then we can use the formula

 f 
fdx  
 du
 du / dx 

Integration by Substitution
Ex. Consider the integral:
 3x  x
2
3
 5 dx
pick u  x +5, then du  3x dx
du
 dx
2
3x
3

10
u du
u

C
10
Sub to get
Integrate
9
2
x


3
5
10


9
10
C
Back Substitute

Ex. Evaluate x 5x 2  7dx
du
Let u  5 x  7 then
 dx
10 x
1 1/ 2
2
x 5 x  7 dx 
u du
10
2


1 u
 
C
 10   3/ 2 
Pick u,
compute du
Sub in
3/ 2
5x


2
7
15

Integrate
3/ 2
C
Sub in
http://www.youtube.com/watch?v=LWp5pWj7s_8&feature=channel
dx
 x  ln x 
Ex. Evaluate
3
Let u  ln x then xdu  dx
dx
 x  ln x  
3
 u 3du
2
u

C
2
ln x 


2
2
C
http://www.youtube.com/watch?v=WDEZNX7arqY&feature=channel

Ex. Evaluate
e3t dt
3t
e 2
du
Let u  e +2 then 3t  dt
3e
e3t dt 1 1

du
3
t
e 2 3u
ln u

C
3
3t
ln e  2

C
3
3t


Examples
( x  5)
4
sin( 3 x)
1
x9
x6
x
2
x 8
x
34
x 8 e
4
xe
x2
9 x2
3
cos x
sin x
http://video.google.com/videoplay?docid=-6305369396691082890#
http://www.youtube.com/watch?v=mie4LtRH4q0&feature=channel
http://www.youtube.com/watch?v=xY9Sq0vQ9Bg&feature=channel
Shortcuts: Integrals of
Expressions Involving ax + b
Rule
  ax  b 
n
  ax  b 
dx 
1
 ax  b 
n 1
a(n  1)
C
1
dx  ln ax  b  C
a
e
ax  b
1 ax b
dx  e
C
a
c
ax  b
1 ax b
dx 
c
C
a ln c
 n  1
Evaluating the Definite Integral
Ex. Calculate

5
1

5
1
1 

 2 x   1 dx
x 

5
1 

2
 2 x   1 dx   x  ln x  x 1
x 

2
2
 5  ln 5  5  1  ln1  1

 
 28  ln5  26.39056

http://video.google.com/videoplay?docid=8180465335026223351#
Computing Area
Ex. Find the area enclosed by the x-axis, the vertical lines x = 0, x = 2 and
the graph of
y  2x
3
2
0 2x dx
2
0
3
2
Gives the area since 2x3
is nonnegative on [0, 2].
   
1 4
1 4 1 4
2 x dx  x
 2  0
2 0
2
2
3
Antiderivative
8
Fund. Thm. of Calculus
Examples
3
2
x
dx


1
cos
x
dx

0
2
2
1
dx
1 x
6
x
dx

2
0
3
e

1
 0.5 x
2
dx
xe
dx

x2
0
Substitution for Definite Integrals
Ex. Calculate 0 2 x  x  3 dx
1
2
1/ 2
let u  x  3 x
du
then
 dx
2x
2
 2x  x
1
0
2
 3x

1/ 2
Notice limits change
4
dx   u du
1/ 2
0
4
16
2 3/ 2

 u
3
3
0
http://video.google.com/videoplay?docid=-1473981413407960367#
The Definite Integral As a Total
If r(x) is the rate of change of a quantity
Q (in units of Q per unit of x), then the
total or accumulated change of the
quantity as x changes from a to b is
given by
b
a
Total change in quantity Q  r ( x)dx
The Definite Integral As a Total
Ex. If at time t minutes you are
traveling at a rate of v(t) feet per
minute, then the total distance traveled
in feet from minute 2 to minute 10 is
given by
10
2
Total change in distance  v(t )dt
Net or Total Change as the
Integral of a Rate
d
v(t )  s (t )   s(t )
dx
t2

s
(
t
)
dt

s
(
t
)

s
(
t
)
2
1

t1
Integral of a rate
of change
Total change over
[t1 , t2 ]
A honey bee makes several trips from the hive to a flower garden.
The velocity graph is shown below.
What is the total distance traveled by the bee?
200  200  200 100  700
700 feet
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
-100
200ft
100ft
What is the displacement of the bee?
200  200  200 100  100
100 feet towards the hive
100
ft
min 50
0
200ft
200ft
2
4
6
8
10
minutes
-50
-100
-200ft
-100ft
To find the displacement (position shift) from the velocity function, we just
integrate the function. The negative areas below the x-axis subtract from the
total displacement.
Displacement   V  t  dt
b
a
To find distance traveled we have to use absolute value.
Distance Traveled   V  t  dt
b
a
Find the roots of the velocity equation and integrate in pieces, just like when
we found the area between a curve and the x-axis. (Take the absolute value
of each integral.)
Or you can use your calculator to integrate the absolute value of the velocity
function.
2
Displacement:
1
1
2
1
0
1
2
13
2
4
5
1 1
1    2  1
2 2
2
-1
Distance Traveled:
velocity graph
-2
1 1
1   2  4
2 2
2
1
0
1
2
3
-1
-2
position graph
4
5
Examples
• A particle moves along a line so that its
velocity at time t is (in meters per second):
v(t )  t  t  6
2
• Find the displacement of the particle during
the period 1  t  4
• Find the total distance travelled during the
same period
Examples
• A factory produces bicycles at a rate of
95  t bicycles per week
(in t weeks)
• How many bicycles were produced from day 8
to 21?
Examples
• At 7 AM, water begins leaking from a tank at a
rate of
2  0.25t
gal / hour
(t is the number of hours after 7 AM)
• How much water is lost between 9 and 11
AM?
In the linear motion equation:
dS
 V t 
dt
V(t) is a function of time.
dS  V  t  dt
For a very small change in time, V(t) can be considered a
constant.
S  V  t  t
We add up all the small changes in S to get the total
distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3   t
S  V  t  t
We add up all the small changes in S to get the total
distance.
S  V1  t  V2  t  V3  t  
S  V1  V2  V3   t
k
S   Vn  t
n 1

S   Vn  t
n 1
S   V  t  dt
As the number of subintervals becomes infinitely large
(and the width becomes infinitely small), we have
integration.
This same technique is used in many different real-life problems.
Example 5:
National Potato Consumption
The rate of potato consumption for a
particular country was:
C  t   2.2  1.1t
where t is the number of years since 1970
and C is in millions of bushels per year.
For a small
t, the rate of consumption is constant.
The amount consumed during that short time is
C  t   t
Example 5:
National Potato Consumption
C  t   2.2  1.1t
The amount consumed during that short time is
C  t   t
We add up all these small amounts to get
the total consumption:
total consumption   C  t  dt
From the beginning of 1972 to the end of
1973:
1
t
2.2

1.1
dt
 2.2t 
1.1
2
ln1.1
4
4
 7.066
t
2
million
bushels
http://www.youtube.com/watch?v=N2sniUkxxek
http://www.youtube.com/watch?v=EY18ooh3GsA&feature=related
Review
• Recall derivatives of inverse trig functions
d
1 du
1
sin u 
, u 1
dx
1  u 2 dx
d
1 du
1
tan u 
dx
1  u 2 dx
d
1
du
1
sec u 
, u 1
2
dx
u u  1 dx
69
Integrals Using Same Relationships
du
u
 a 2  u 2  arcsin a  C
du
1
u
 a 2  u 2  a arctan a  C
du
1
u
 u u 2  a 2  a arcsec a  C
When given
integral problems,
look for these
patterns
70
Identifying Patterns
• For each of the integrals below, which inverse
trig function is involved?
4dx
 13  16 x 2

dx
9 x
2
x
dx
25 x 2  4
dx
 x 2  2 x  10
Hint: use
completing
the square
71
Warning
• Many integrals look like the inverse trig forms
• Which of the following are of the inverse trig
forms?

x dx
 1  x2
dx
 1  x2

x dx
1  x2
dx
1 x
2
If they are not,
how are they
integrated?
72