The Integral
chapter
6
• The Indefinite Integral
• Substitution
• The Definite Integral As a Sum
• The Definite Integral As Area
• The Definite Integral: The Fundamental
Theorem of Calculus
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Antiderivative
An antiderivative of a function f is a
function F such that
F  f
Ex. An antiderivative of f ( x )  6 x
is F ( x )  3 x 2  2
since F ( x )  f ( x ).
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
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
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
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  3 x
2
C
Represents every possible
antiderivative of 6x.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Power Rule for the Indefinite
Integral, Part I

Ex.

x dx 
n
x dx 
3
x
n 1
n 1
x
 C if n   1
4
C
4
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Power Rule for the Indefinite
Integral, Part II

1
x dx 
1
x
dx  ln x  C
Indefinite Integral of ex and bx


e dx  e  C
x
b dx 
x
x
b
x
C
ln b
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Sum and Difference Rules
f


Ex.

 g  dx 

x  x dx 
2


fdx 
x dx 
2

gdx
 xdx

x
3

3
x
2
C
2
Constant Multiple Rule

Ex.

kf ( x ) dx  k


f ( x ) dx
2 x dx  2 x dx  2
3
3
x
( k constant )
4
4
C 
x
4
C
2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Integral Example/Different Variable
Ex. Find the indefinite integral:

 u 7

2
 3 e   2 u  6  du
u



 3 e du  7
u
1
 u du  2  u
 3 e  7 ln u 
u
2
2
du 
 6 du
u  6u  C
3
3
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Position, Velocity, and Acceleration
Derivative Form
If s = s(t) is the position function of an
object at time t, then
Velocity = v =
ds
dt
Acceleration = a =
dv
dt
Integral Form
s (t ) 
 v ( t ) dt
v (t ) 
a
(
t
)
dt

Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Integration by Substitution
Method of integration related to chain
rule differentiation. If u is a function of
x, then we can use the formula

fdx 

f


 du / dx

 du

Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Integration by Substitution
Ex. Consider the integral:
 3x  x
2
3
5

9
dx
pick u  x +5, then du  3 x dx
3
2
du
3x

9
u du
Sub to get

u
10
C
10
Integrate

x
2
3
 dx
5

10
C
10
Back Substitute
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Evaluate

x 5 x  7 dx
2
Let u  5 x  7
2
du
then
 dx
10 x
x
5 x  7 dx 
2
1
 10 u
1/ 2
du
Pick u,
compute du
Sub in
3/2
 1  u

C

 10   3 / 2 

5x
2
7

Integrate
3/2
C
Sub in
15
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Evaluate
dx
 x  ln x  3
L et u  ln x
th en xd u  d x
dx
 x  ln x  3   u


u
3
du
2
2
C
 ln x 
2
2
C
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
3t
e dt
 e 3t  2
Ex. Evaluate
Let u  e + 2
3t
then
du
3e
3t
e dt
1
3t
 dt
1
 e 3 t  2  3  u du

ln u
C
3


ln e
3t
2

C
3
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Shortcuts: Integrals of
Expressions Involving ax + b
Rule
  ax  b 
n
  ax  b 
e
ax  b
c
ax  b
dx 
1
dx 
 ax  b 
n 1
a ( n  1)
dx 
1
C
n 
 1
ln ax  b  C
a
1
e
ax  b
C
a
dx 
1
c
ax  b
C
a ln c
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Riemann Sum
If f is a continuous function, then the left Riemann
sum with n equal subdivisions for f over the interval
[a, b] is defined to be
n 1
 f  x k  x
k 0
 f ( x 0 )  x  f ( x1 )  x  ...  f ( x n 1 )  x

 f ( x0 ) 
f ( x1 )  ...  f ( x n 1 )   x
where a  x 0  x1  ...  x n  b are the
subdivisions and  x  ( b  a ) / n.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Definite Integral
If f is a continuous function, the definite integral
of f from a to b is defined to be
b
n 1
f  x k  x

 f ( x ) dx  nlim

k 0
a
The function f is called the integrand, the numbers
a and b are called the limits of integration, and the
variable x is called the variable of integration.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Approximating the Definite
Integral
Ex. Calculate the Riemann sum for the
2
integral  x dx using n = 10.
2
0
n 1
9
 f  x k  x   x k
k 0
k 0
2
1
 
5
2
2
2
  (1 / 5)  (2 / 5)  ...  (9 / 5)  (1 / 5)


 2 .2 8
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Definite Integral
b

f ( x ) dx
a
is read “the integral, from a to b of f(x)dx.”
Also note that the variable x is a “dummy
variable.” b
b
 f ( x ) dx   f (t ) dt
a
a
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
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
T otal change in quantity Q 

r ( x ) dx
a
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
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
T otal change in distance 
 v ( t ) dt
2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Area Under a Graph
Width:  x 
(n rect.)
y  f ( x)
ba
n
a
b
Idea: To find the exact area under the graph
of a function.
Method: Use an infinite number of rectangles
of equal width and compute their area with a
limit.
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Approximating Area
Approximate the area under the graph of
f ( x)  2 x
2
on  0, 2 
using n = 4.
A   x  f ( x 0 )  f ( x1 )  f ( x 2 )  f ( x 3 ) 
1
1
 3 
A   f  0   f    f 1   f   
2
2
 2 
1
1
9 7
A  0   2   
2
2
2 2
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Area Under a Graph
y  f ( x)
a
b
f continuous, nonnegative on [a, b]. The area is
n 1
f  x k  x

n 
A rea  lim
k 0


b
f ( x ) dx
a
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Geometric Interpretation
(All Functions)
y  f ( x)
R1
a

b
R3
R2
b
f ( x ) dx  Area of R1 – Area of R2 + Area of R3
a
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Area Using Geometry
Ex. Use geometry to compute the integral
5
  x  1  dx
1
Area =4
Area = 2
5
  x  1  dx  4  2  2
1
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Fundamental Theorem of
Calculus
Let f be a continuous function on [a, b].
x
1. If A ( x ) 

f ( t ) dt , then A ( x )  f ( x ).
a
2. If F is any continuous antiderivative of
f and is defined on [a, b], then

b
f ( x ) dx  F ( b )  F ( a )
a
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
The Fundamental Theorem
of Calculus
x
Ex. If A ( x ) 

3
t  5 t dt , find A ( x ).
4
a
A ( x ) 
3
x  5x
4
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
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




 
 5  ln 5  5  1  ln 1  1
2
2

 2 8  ln 5  2 6 .3 9 0 5 6
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Substitution for Definite Integrals
Ex. Calculate 0 2 x  x
1
2
3

1/ 2
dx
let u  x  3 x
du
then
 dx
2x
2

1
0

2x x  3x
2

1/ 2
Notice limits change
dx 


4
u
1/ 2
du
0
2
3
4
u

3/2
0
16
3
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.
Computing Area
Ex. Find the area enclosed by the x-axis, the
vertical lines x = 0, x = 2 and the graph of
y  2x .
2

2
0
2
3
2 x dx
0
2 x dx 
3
Gives the area since 2x3 is
nonnegative on [0, 2].
1
2
2
x
4

0
Antiderivative
2   0 

2
2
1
4
1
4
8
Fund. Thm. of Calculus
Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.