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7
Additional Topics in Integration
 Integration by Parts
 Integration Using Tables of Integrals
 Numerical Integration
 Improper Integrals
 Applications of Calculus to Probability
7.1
Integration by Parts
 x ln x dx   udv  uv   vdu
1 2
1 2 1
 x ln x   x    dx
2
2
x
1 2
1
 x ln x   xdx
2
2
The Method of Integration by Parts
 Integration by parts formula
 u dv  uv   v du
Example
 Evaluate
x
xe
 dx
Solution
 Let
u =x
 So that
du = dx
 Therefore,
and
and
dv = ex dx
v = ex
x
xe
 dx   udv
 uv   vdu
 xe x   e x dx
 xe x  e x  C
 ( x  1)e x  C
Example 1, page 484
Guidelines for Integration by Parts
 Choose u an dv so that
1. du is simpler than u.
2. dv is easy to integrate.
Example
 Evaluate
 x ln x dx
Solution
 Let
 So that
u = ln x
1
du  dx
x
and
and
dv = x dx
1
v  x2
2
 Therefore,
 x ln x dx   udv  uv   vdu

1 2
1
1
x ln x   x 2    dx
2
2
x
1 2
1
 x ln x   xdx
2
2
1 2
1 2
1 2
 x ln x  x  C  x (2 ln x  1)  C
2
4
4
Example 2, page 485
Example
xe x
 Evaluate 
dx
2
( x  1)
Solution
dv 
 Let
u = xex and
 So that
du  ( xe  e )dx
x
x
 e x ( x  1)dx
 Therefore,
1
dx
2
( x  1)
and
v
1
x 1
xe x
 ( x  1)2 dx   udv  uv   vdu
 1  x
x  1 
 xe 
 
 e ( x  1)dx
 x 1
 x 1
x
xe x
xe

  e x dx  
 ex  C
x 1
x 1
Example 3, page 486
Example
2 x
x
 e dx
 Evaluate
Solution
 Let
 So that
u = x2
du = 2xdx
and
and
dv = ex dx
v = ex
 Therefore,
2 x
x
 e dx   udv  uv   vdu
 x 2e x   e x (2 x )dx
 x 2e x  2  xe x dx
 xe x  2 ( x  1)e x   C
 e x ( x 2  2 x  2)  C
Example 4, page 486
(From first example)
Applied Example: Oil Production
 The estimated rate at which oil will be produced from an
oil well t years after production has begun is given by
R(t )  100te0.1t
thousand barrels per year.
 Find an expression that describes the total production of
oil at the end of year t.
Applied Example 5, page 487
Applied Example: Oil Production
Solution
 Let T(t) denote the total production of oil from the well at
the end of year t (t  0).
 Then, the rate of oil production will be given by T ′(t)
thousand barrels per year.
 Thus,
T (t )  R(t )  100te0.1t
 So,
T (t )   100te 0.1t dt
 100 te 0.1t dt
Applied Example 5, page 487
Applied Example: Oil Production
Solution
 Use integration by parts to evaluate the integral.
 Let
 So that
ut
and
du  dt
and
dv  e 0.1t dt
1 0.1t
v
e
0.1
 10e 0.1t
 Therefore,
T (t )  100 te0.1t dt  100  10te0.1t  10 e0.1t dt 


 100  10te 0.1t  100e 0.1t   C
 1000e 0.1t (t  10)  C
Applied Example 5, page 487
Applied Example: Oil Production
Solution
 To determine the value of C, note that the total quantity of
oil produced at the end of year 0 is nil, so T(0) = 0.
 This gives,
T (t )  1000e0.1t (t  10)  C
T (0)  1000e 0.1(0) (0  10)  C  0
 1000(10)  C  0
C  10,000
 Thus, the required production function is given by
T (t )  1000e0.1t (t  10)  10,000
Applied Example 5, page 487
7.2
Integration Using Tables of Integrals
udu
1

 a  bu b2 [a  bu  a ln a  bu ]  C

u
2
udu
2
 2 (bu  2a ) a  bu  C
a  bu 3b
u 2
a4
2
a  u du  (a  2u )  ln u  a 2  u 2  C
8
8
2
2
n au
 u e du 
1 n au n n 1 au
u e   u e du
a
a
A Table of Integrals
 We have covered several techniques for finding the
antiderivatives of functions.
 There are many more such techniques and extensive
integration formulas have been developed for them.
 You can find a table of integrals on pages 491 and 492 of
the text that include some such formulas for your benefit.
 We will now consider some examples that illustrate how
this table can be used to evaluate an integral.
Examples
 Use the table of integrals to find

Solution
 We first rewrite

2 x dx
3 x
2 x dx
x dx
 2
3 x
3 x
3  x is of the form a  bu , with a = 3, b = 1,
and u = x, we use Formula (5),
u du
2
 a  bu  3b2 (bu  2a) a  bu  C
obtaining
 2

x dx
2
 2
( x  2  3) 3  x   C
2
3 x
 3(1)

 Since

Example 1, page 493
4
( x  6) 3  x  C
3
Examples
 Use the table of integrals to find
2
2
x
3

x
dx

Solution
2
 We first rewrite 3 as ( 3) , so that
3  x 2 has the form
a 2  u2 with a  3 and u = x.
 Using Formula (8),
4
u
a
2
2
2
2
2
2
2
2
2
u
a

u
du

(
a

2
u
)
a

u

ln
u

a

u
C

8
8
obtaining
x
2
x
9
2
2
3  x dx  (3  2 x ) 3  x  ln x  3  x 2  C
8
8
Example 2, page 493
2
Examples
 Use the table of integrals to find
2 ( 1/2) x
x
 e dx
Solution
 We can use Formula (24),
n au
u
 e du 
1 n au n n 1 au
u e   u e du
a
a
 Letting n = 2, a = – ½, and u = x, we have
x e
2 ( 1/2) x
1 2 ( 1/2) x
2
dx  1 x e
 1  xe( 1/2) x dx
( 2 )
( 2 )
 2 x 2e( 1/2) x  4 xe( 1/2) x dx
Example 5, page 494
Examples
 Use the table of integrals to find
Solution
 We have
2 (1/2) x
x
 e dx
2 (1/2) x
2 ( 1/2) x
( 1/2) x
x
e
dx


2
x
e

4
xe
dx


 Using Formula (24) again, with n = 1, a = – ½, and u = x,
we get
x e
2 ( 1/2) x
2 ( 1/2) x
dx  2 x e
 1

1
( 1/2) x
( 1/2) x
 4  1 xe
 1 e
dx 
( 2 )
 ( 2 )

 2 x 2e( 1/2) x  8   xe( 1/2) x   e( 1/2) x dx 


2 ( 1/2) x
 2 x e

1 ( 1/2) x 
( 1/2) x
 8   xe
 1 e
C

( 2 )


 2e( 1/2) x ( x 2  4 x  8)  C
Example 5, page 494
Applied Example: Mortgage Rates
 A study prepared for the National Association of realtors
estimated that the mortgage rate over the next t months
will be
6t  75
r (t ) 
(0  t  24)
t  10
percent per year.
 If the prediction holds true, what will be the average
mortgage rate over the 12 months?
Applied Example 6, page 495
Applied Example: Mortgage Rates
Solution
 The average mortgage rate over the next 12 months will be
given by
12 6t  75
1
A
dt

0
12  0
t  10
12 75
1  12 6t

 
dt  
dt 
0 t  10
12  0 t  10

1 12 t
25 12 1
 
dt  
dt
2 0 t  10
4 0 t  10
Applied Example 6, page 495
Applied Example: Mortgage Rates
Solution
 We have
1 12 t
25 12 1
A 
dt  
dt
2 0 t  10
4 0 t  10
 Use Formula (1)
udu
1
 a  bu  b2 [a  bu  a ln a  bu ]  C
to evaluate the first integral
1
25
12
12
A  [10  t  10 ln(10  t )] 0  ln(10  t ) 0
2
4
1
25
 [(22  10 ln 22)  (10  10 ln10)]  [ln 22  ln10]
2
4
 6.99
or approximately 6.99% per year.
Applied Example 6, page 495
7.3
Numerical Integration
y
 f ( x0 )  f ( x1 ) 
R1  
 x
2

f ( x)
x
x 
R1
R2
R3
R4
R5
ba
n
R6
x
a
b
Approximating Definite Integrals
 Sometimes, it is necessary to evaluate definite integrals
based on empirical data where there is no algebraic rule
defining the integrand.
 Other situations also arise in which an integrable function
has an antiderivative that cannot be found in terms of
elementary functions. Examples of these are
1
f ( x)  e
h( x ) 
g ( x)  x e
ln x
 Riemann sums provide us with a good approximation of a
definite integral, but there are better techniques and
formulas, called quadrature formulas, that allow a more
efficient way of computing approximate values of definite
integrals.
x2
1/2 x
The Trapezoidal Rule
 Consider the problem of finding the area under the curve
of f(x) for the interval [a, b]:
y
f ( x)
R
x
a
b
The Trapezoidal Rule
 The trapezoidal rule is based on the notion of dividing the
area to be evaluated into trapezoids that approximate the
area under the curve:
y
f ( x)
R1
R2
R3
R4
R5
R6
x
a
b
The Trapezoidal Rule
 The increments x used for each trapezoid are obtained
by dividing the interval into n equal segments
(in our example n = 6):
y
x 
ba
n
f ( x)
x
R1
R2
R3
R4
R5
R6
x
a
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x0 )  f ( x1 ) 
R1  
x

2


y
f ( x)
x
f(x0)
R1
f(x1)
x
x0 = a
x1
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x1 )  f ( x2 ) 
R2  
x

2


y
f ( x)
x
f(x1)
R2
f(x2)
x
a
x1
x2
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x2 )  f ( x3 ) 
R3  
x

2


y
f ( x)
f(x2)
R3
f(x3)
x
a
x2
x3
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x3 )  f ( x4 ) 
R4  
x

2


y
f ( x)
R4
f(x3)
f(x4)
x
a
x3
x4
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x4 )  f ( x5 ) 
R5  
x

2


y
f ( x)
R5
f(x4)
f(x5)
x
a
x4
x5
b
The Trapezoidal Rule
 The area of each trapezoid is calculated by multiplying its
base, x , by its average height:
 f ( x5 )  f ( x6 ) 
R6  
x

2


y
f ( x)
R6
f(x5)
f(x6)
x
a
x5
b = x6
The Trapezoidal Rule
 Adding the areas R1 through Rn (n = 6 in this case) of the
trapezoids gives an approximation of the desired area of
the region R:
y
R  R1  R2  ...  Rn
f ( x)
R1
R2
R3
R4
R5
R6
x
a
b
The Trapezoidal Rule
 Adding the areas R1 through Rn of the trapezoids yields the
following rule:
✦ Trapezoidal Rue

b
a
f ( x )dx 
x
[ f ( x0 )  2 f ( x1 )  2 f ( x2 )  ...  2 f ( xn 1 )  f ( xn )]
2
ba
where x 
.
n
Example
 Approximate the value of

2
1
1
dx using the trapezoidal
x
rule with n = 10.
 Compare this result with the exact value of the integral.
Solution
 Here, a = 1, b = 2, an n = 10, so
b  a 2 1 1
x 


 0.1
n
10
10
and
x0 = 1, x1 = 1.1, x2 = 1.2, x3 = 1.3, … , x9 = 1.9, x10 = 1.10.
 The trapezoidal rule yields

2
1
1
0.1
 1   1   1 
 1  1
dx  [1  2    2    2   ...  2    ]  0.693771
x
2
 1.1   1.2   1.3 
 1.9  2
Example 1, page 500
Example
 Approximate the value of

2
1
1
dx using the trapezoidal
x
rule with n = 10.
 Compare this result with the exact value of the integral.
Solution
 By computing the actual value of the integral we get

2
1
1
2
dx  ln x 1  ln 2  ln1  ln 2  0.693147
x
 Thus the trapezoidal rule with n = 10 yields a result with
an error of 0.000624 to six decimal places.
Example 1, page 500
Applied Example: Consumers’ Surplus
 The demand function for a certain brand of perfume is
given by
p  D( x)  10,000  0.01x 2
where p is the unit price in dollars and x is the quantity
demanded each week, measured in ounces.
 Find the consumers’ surplus if the market price is set at
$60 per ounce.
Applied Example 2, page 500
Applied Example: Consumers’ Surplus
Solution
 When p = 60, we have
10,000  0.01x 2  60
10,000  0.01x 2  3,600
x 2  640,000
or x = 800 since x must be nonnegative.
 Next, using the consumers’ surplus formula with `p = 60
and `x = 800, we see that the consumers’ surplus is given by
CS  
800
0
10,000  0.01x 2 dx  (60)(800)
 It is not easy to evaluate this definite integral by finding an
antiderivative of the integrand.
 But we can, instead, use the trapezoidal rule.
Applied Example 2, page 500
Applied Example: Consumers’ Surplus
Solution
 We can use the trapezoidal rule with a = 0, b = 800,
and n = 10.
b  a 800  0 800
x 


 80
n
10
10
and
x0 = 0, x1 = 80, x2 = 160, x3 = 240, … , x9 = 720, x10 = 800.
 The trapezoidal rule yields
CS  
800
0
10,000  0.01x 2 dx  (60)(800)
80 

100  2 10,000  (0.01)(80)2  2 10,000  (0.01)(160)2  ...
2
...  2 10,000  (0.01)(720)2  10,000  (0.01)(800)2 

Applied Example 2, page 500
Applied Example: Consumers’ Surplus
Solution
 The trapezoidal rule yields
CS  
800
0
10,000  0.01x 2 dx  (60)(800)
 40(100  199.3590  197.4234  194.1546  189.4835
 183.3030  175.4537  165.6985
 153.6750  138.7948  60)  (60)(800)
 70,293.82  48,000
 $22,294.82
 Therefore, the consumers’ surplus is approximately
$22,294.
Applied Example 2, page 500
Simpson’s Rule
 We’ve seen that the trapezoidal rule approximates the
area under the curve by adding the areas of trapezoids
under the curve:
y
f ( x)
R1
x0
R2
x1
x2
x
Simpson’s Rule
 The Simpson’s rule improves upon the trapezoidal rule by
approximating the area under the curve by the area under
a parabola, rather than a straight line:
y
f ( x)
R
x0
x1
x2
x
Simpson’s Rule
 Given any three nonlinear points there is a unique parabola
that passes through the given points.
 We can approximate the function f(x) on [x0, x2] with a
quadratic function whose graph contain these three points:
y
(x2, f(x2))
f ( x)
(x1, f(x1))
(x0, f(x0))
x0
x1
x2
x
Simpson’s Rule
 Simpson’s rule approximates the area under the curve of
a function f(x) using a quadratic function:
✦ Simpson’s rule

b
a
x
f ( x )dx  [ f ( x0 )  4 f ( x1 )  2 f ( x2 )  4 f ( x3 )  2 f ( x4 )
3
   4 f ( xn1 )  f ( xn )]
ba
where x 
and n is even.
n
Example
 Find an approximation of

2
1
with n = 10.
Solution
 Here, a = 1, b = 2, an n = 10, so
1
dx using Simpson’s rule
x
b  a 2 1 1
x 


 0.1
n
10
10
 Simpson’s rule yields

2
1
1
0.1
dx  [ f (1)  4 f (1.1)  2 f (1.2)  4 f (1.3)  2 f (1.4)    4 f (1.9)  f (2)]
x
3
0.1 
 1   1   1   1 
 1  1

1  4    2    4    2      4    

3 
 1.1   1.2   1.3   1.4 
 1.9  2 
 0.693150
Example 3, page 503
Example
 Find an approximation of

2
1
1
dx using Simpson’s rule
x
with n = 10.
Solution
 Recall that the trapezoidal rule with n = 10 yielded an
approximation of 0.693771, with an error of 0.000624 from
the value of ln 2 ≈ 0.693147 to six decimal places.
 Simpson’s rule yields an approximation with an error of
0.000003 to six decimal places, a definite improvement
over the trapezoidal rule.
Example 3, page 503
Applied Example: Cardiac Output
 One method of measuring cardiac output is to inject 5 to
10 mg of a dye into a vein leading to the heart.
 After making its way through the lungs, the dye returns to
the heart and is pumped into the aorta, where its
concentration is measured at equal time intervals.
Applied Example 4, page 504
Applied Example: Cardiac Output
 The graph of c(t) shows the concentration of dye in a
person’s aorta, measured in 2-second intervals after 5 mg
of dye have been injected:
y
3.9
4.0
4
3.2
2.5
3
1.8
2.0
2
1.3
0.8
1
0.5
0.4
0
2
4
Applied Example 4, page 504
0.2
6
8
10
12
14
16
18
20 22
24
0.1
26
x
Applied Example: Cardiac Output
 The person’s cardiac output, measured in liters per minute
(L/min) is computed using the formula
60D
R  28
0 c(t )dt
where D is the quantity of dye injected.
y
3.9
4.0
4
3.2
2.5
3
1.8
2.0
2

28
0
1
1.3
c(t )dt
0.8
0.5
0.4
0
2
4
Applied Example 4, page 504
0.2
6
8
10
12
14
16
18
20 22
24
0.1
26
x
Applied Example: Cardiac Output
 Use Simpson’s rule with n = 14 to evaluate the integral and
determine the person’s cardiac output.
60D
R  28
 c(t )dt
0
y
3.9
4.0
4
3.2
2.5
3
1.8
2.0
2

28
0
1
1.3
c(t )dt
0.8
0.5
0.4
0
2
4
Applied Example 4, page 504
0.2
6
8
10
12
14
16
18
20 22
24
0.1
26
x
Applied Example: Cardiac Output
Solution
 We have a = 0, b = 28, an n = 14, and t = 2, so that
t0 = 0, t1 = 2, t2 = 4, t3 = 6, … , t14 = 28.
 Simpson’s rule yields

28
0
2
c(t )dt  [c(0)  4c(2)  2c(4)  4c(6)  ...  4c(26)  c(28)]
3
2
 [0  4(0)  2(0.4)  4(2.0)  2(4.0)
3
 4(4.4)  2(3.9)  4(3.2)  2(2.5)  4(1.8)
 2(1.3)  4(0.8)  2(0.5)  4(0.2)  0.1]
 49.9
Applied Example 4, page 504
Applied Example: Cardiac Output
Solution
 Therefore, the person’s cardiac output is
60D
60(5)
R  28

 6.0
49.9
c
(
t
)
dt

0
or approximately 6.0 L/min.
Applied Example 4, page 504
7.4
Improper Integrals
y
1
–2
R2
–1
1
R1
–1
2
x
Improper Integrals
 In many applications we are concerned with integrals that
have unbounded intervals of integration.
 These are called improper integrals.
 We will now discuss problems that involve improper
integrals.
Improper Integral of f over [a, )
✦ Let f be a continuous function on the unbounded
interval [a, ). Then the improper integral of f
over [a, ) is defined by


a
b
f ( x)dx  lim  f ( x)dx
if the limit exists.
b a
Examples
 Evaluate

Solution

2
1
dx if it converges.
x


2
b1
1
dx  lim  dx
b 2 x
x
 lim ln x
b
b
2
 lim(ln b  ln 2)
b
 Since ln b → , as b →  the limit does not exist, and we
conclude that the given improper integral is divergent.
Example 2, page 513
Examples
 Find the area of the region R under the curve y = e–x/2
for x  0.
Solution
 The required area is shown in the diagram below:
y
1
R
y = e–x/2
1
Example 3, page 514
2
3
x
Examples
 Find the area of the region R under the curve y = e–x/2
for x  0.
Solution
 Taking b > 0, we compute the area of the region under the
curve y = e–x/2 from x = 0 to x = b,
b
I (b)   e
0
 x /2
dx  2e
 x /2 b
0
 2eb/2  2
 Then, the area of the region R is given by
1
2
b
/2
b e
I (b)  lim(2  2e  b/2 )  2  2 lim
b
or 2 square units.
Example 3, page 514
Improper Integral of f over (– , b]
✦ Let f be a continuous function on the unbounded
interval (– , b]. Then the improper integral of f
over (– , b] is defined by

b

b
f ( x)dx  lim  f ( x)dx
if the limit exists.
a a
Example
 Find the area of the region R bounded above by the x-axis,
below by y = – e2x, and on the right, by the line x = 1.
Solution
 The graph of region R is:
y
1
–1
1
–1
x
R
x=1
–3
–7
Example 4, page 514
y = e2x
Example
 Find the area of the region R bounded above by the x-axis,
below by y = – e2x, and on the right, by the line x = 1.
Solution
 Taking a < 1, compute
1
1
1
1 2x
1 2 1 2a
2x
2x
 e  e
I (a )   [0  ( e )]dx   e dx  e
a
a
2
2
2
a
 Then, the area under the required region R is given by
 1 2 1 2a 
lim I (a )  lim  e  e 
a 
a  2
2


1 2
1 2a
 e  lim e
a  2
2
Example 4, page 514
1 2
 e
2
Improper Integral Unbounded on Both Sides
Improper Integral of f over (– , )
 Let f be a continuous function over the
unbounded interval (– , ).
 Let c be any real number and suppose both the
improper integrals

c

f ( x)dx
and


c
f ( x)dx
are convergent.
 Then, the improper integral of f over (– , ) is
defined by



c


c
f ( x)dx   f ( x)dx   f ( x)dx
Examples
 Evaluate the improper integral



 x2
xe dx
and give a geometric interpretation of the result.
Solution
 Take the number c to be zero and evaluate first for the
interval (– , 0):

0

xe
 x2
dx  lim

0
a  a
xe
 x2
1
 lim  e  x
a  2
dx
0
2
a
1
 1 1  a2 
 lim    e   
a 
2
 2 2

Example 5, page 515
Examples

 Evaluate the improper integral


 x2
xe dx
and give a geometric interpretation of the result.
Solution
 Now evaluate for the interval (0, ):


0
xe
 x2
b
dx  lim  xe
b 0
 x2
dx
b
 1  x2 
 lim   e 
b
 2
0
 1  b2 1  1
 lim   e   
b
2 2
 2
Example 5, page 515
Examples
 Evaluate the improper integral



 x2
xe dx
and give a geometric interpretation of the result.
Solution
 Therefore,



xe
 x2
Example 5, page 515
0
dx   xe

 x2

dx   xe
0
 x2
1 1
dx     0
2 2
Examples
 Evaluate the improper integral


 x2
xe dx

and give a geometric interpretation of the result.
Solution
–x2
 Below is the graph of y = xe , showing the regions of
interest R1 and R2:
y
1
–2
R2
–1
1
R1
–1
Example 5, page 515
2
x
Examples
 Evaluate the improper integral


 x2
xe dx

and give a geometric interpretation of the result.
Solution
 Region R1 lies below the x-axis, so its area is negative
(R1 = – ½):
y
1
–2
R2
–1
1
R1
–1
Example 5, page 515
2
x
Examples
 Evaluate the improper integral


 x2
xe dx

and give a geometric interpretation of the result.
Solution
 While the symmetrically identical region R2 lies above the
x-axis, so its area is positive (R2 = ½):
y
1
–2
R2
–1
1
R1
–1
Example 5, page 515
2
x
Examples
 Evaluate the improper integral


 x2
xe dx

and give a geometric interpretation of the result.
Solution
 Thus, adding the areas of the two regions yields zero:
1 1
R  R1  R2     0
2 2
y
1
–2
R2
–1
1
R1
–1
Example 5, page 515
2
x
7.5
Applications of Calculus to Probability
y
R1  P(a  x  b)
R1
x
a
b
Probability Density Functions
A probability density function of a random variable x in
an interval I, where I may be bounded or unbounded, is a
nonnegative function f having the following properties.
1. The total area R of the region under the graph of f is
equal to 1:
y
R  P(  x  )  1
R
x
Probability Density Functions
A probability density function of a random variable x in
an interval I, where I may be bounded or unbounded, is a
nonnegative function f having the following properties.
2. The probability that an observed value of the random
variable x lies in the interval [a, b] is given by
b
y
P(a  x  b)   f ( x)dx
a
R1  P(a  x  b)
R1
x
a
b
Examples
 Show that the function
2
f ( x) 
x ( x  1)
(1  x  4)
27
satisfies the nonnegativity condition of Property 1 of
probability density functions.
Solution
 Since the factors x and (x – 1) are both nonnegative, we see
that f(x)  0 on [1, 4].
 Next, we compute
4

4
1
2 4 2
2 1 3 1 2
2
( x  x )dx 
x ( x  1)dx 
x  x 


1
27
27
27  3
2 1
2  64
2  27 
  1 1 

  1
  8      

27  3
  3 2   27  2 
Example 1, page 522
Examples
 Show that the function
1 ( 1/3) x
f ( x)  e
(0  x  )
3
satisfies the nonnegativity condition of Property 1 of
probability density functions.
Solution
 First, f(x)  0 for all values of x in [0, ).
 Next, we compute
b1
1
( 1/3) x
( 1/3) x

lim
e
dx
e
dx
0 3
b 0 3
 lim  e
b
( 1/3) x

b
0
 lim  e( 1/3) b  1
b
1
Example 1, page 522
Examples
 Determine the value of the constant k so that the function
f ( x)  kx 2
is a probability density function on the interval [0, 5].
Solution
 We compute
5
5
5
125
k
2
2
3
0 kx dx  k 0 x dx  3 x 0  3 k
 Since this value must be equal to one, we find that
125
k 1
3
3
k
125
Example 2, page 522
Examples
 If x is a continuous random variable for the function
f ( x)  kx 2
compute the probability that x will assume a value
between x = 1 and x = 2.
Solution
 The required probability is given by
P(1  x  2)  
2
1
Example 2, page 522
2
1 3
3 2
1
7
x
x dx 

(8  1) 
125 1
125
125
125
Examples
 If x is a continuous random variable for the function
f ( x)  kx 2
compute the probability that x will assume a value
between x = 1 and x = 2.
Solution
 The graph of f showing P(1  x  2) is:
y
y
0.6
3 2
x
125
0.4
0.2
P(1  x  2) = 7/125
1
Example 2, page 522
2
3
4
5
x
Applied Example: Life Span of Light Bulbs
 TKK Inc. manufactures a 200-watt electric light bulb.
 Laboratory tests show that the life spans of these light
bulbs have a distribution described by the probability
density function
f ( x)  0.001e0.001x
(0  x  )
where x denotes the life span of a light bulb.
 Determine the probability that a light bulb will have a
life span of
a. 500 hours or less.
b. More than 500 hours.
c. More than 1000 hours but less than 1500 hours.
Applied Example 3, page 523
Applied Example: Life Span of Light Bulbs
Solution
a. The probability that a light bulb will have a life span of
500 hours or less is given by
P(0  x  500)  
500
0
 e
0.001e0.001x dx
0.001 x 500
0
 e 0.5  1
 0.3935
Applied Example 3, page 523
Applied Example: Life Span of Light Bulbs
Solution
b. The probability that a light bulb will have a life span of
more than 500 hours is given by

P( x  500)   0.001e0.001x dx
500
b
 lim  0.001e0.001x dx
b 500
 lim  e
b
0.001 x

b
500
 lim  e 0.001b  e 0.5 
b
 e 0.5
 0.6065
Applied Example 3, page 523
Applied Example: Life Span of Light Bulbs
Solution
c. The probability that a light bulb will have a life span of
more than 1000 hours but less than 1500 hours is given by
P(1000  x  1500)  
1500
1000
 e
0.001e0.001x dx
0.001 x 1500
1000
  e 1.5  e 1
 0.2231  0.3679
 0.1448
Applied Example 3, page 523
Exponential Density Function
 The example we just saw involved a function of the form
f(x) = ke–kx
where x  0 and k is a positive constant, with a graph:
y
k
f(x) = ke–kx
x
 This probability function is called an exponential density
function, and the random variable associated with it is
said to be exponentially distributed.
 Such variables are used to represent the life span of
electric components, the duration of telephone calls, the
waiting time in a doctor’s office, etc.
Expected Value
Expected Value of a Continuous Random Variable
 Suppose the function f defined on the interval [a, b]
is the probability density function associated with a
continuous random variable x.
 Then, the expected value of x is
b
E ( x)   xf ( x)dx
a
Applied Example: Life Span of Light Bulbs
 Show that if a continuous random variable x is
exponentially distributed with the probability density
function
f(x) = ke–kx
(0  x < )
then the expected value E(x) is equal to 1/k.
 Using this result and continuing with our last example,
determine the average life span of the 200-watt light bulb
manufactured by TKK Inc.
Applied Example 4, page 525
Applied Example: Life Span of Light Bulbs
Solution
 We compute


0
0
b
E ( x)   xf ( x)dx   kxe dx  k lim  xe kx dx
 Integrating by parts with u = x
 kx
b 0
and dv = e–kxdx
1
du  dx and v   e  kx
k
so that
 We have
 1  kx b 1 b  kx 
E ( x )  k lim   xe
  xe dx 
0
b
k 0
 k

Applied Example 4, page 525
Applied Example: Life Span of Light Bulbs
Solution
 We have
 1  kx b 1 b  kx 
E ( x )  k lim   xe
  xe dx 
0
b
k 0
 k

1  kx b 
 1  kb 1
 k (0)
 k lim   be  (0)e
 2e
0
b
k
k
 k

 1  kb 1  kb 1 
 k lim   be  2 e  2 
b
k
k 
 k
b 1
1 1
  lim kb  lim kb  lim1
b e
k b e
k b
Applied Example 4, page 525
Applied Example: Life Span of Light Bulbs
Solution
 Now, by taking a sequence of values of b that approaches
infinity (such as b = 10, 100, 1000, 10,000, … ) we see that,
for a fixed k,
b
lim kb  0
b e
Therefore,
b 1
1 1
E ( x )   lim kb  lim kb  lim1
b e
k b e
k b
1
1
   0    0   (1)
k
k
1

k
 Finally, since k = 0.001, we see that the average life span of
the TKK light bulbs is 1/(0.001) = 1000 hours.
Applied Example 4, page 525
Expected Value of an Exponential Density Function
 If a continuous random variable x is exponentially
distributed with probability density function
f(x) = ke–kx
(0  x < )
then the expected (average) value of x is given by
1
E ( x) 
k
Applied Example: Airport Traffic
 On a typical Monday morning, the time between
successive arrivals of planes at Jackson International
Airport is an exponentially distributed random variable x
with expected value of 10 minutes.
a. Find the probability density function associated with x.
b. What is the probability that between 6 and 8 minutes will
elapse between successive arrivals of planes.
c. What is the probability that the time between successive
arrivals of planes will be more than 15 minutes?
Applied Example 5, page 527
Applied Example: Airport Traffic
Solution
a. Since x is exponentially distributed, the associated
probability density function has the form
f(x) = ke–kx
✦ The expected value of x is 10, so
1
E ( x )   10
k
1
k
10
 0.1
✦ Thus, the required probability density function is
f(x) = 0.1e–0.1x
Applied Example 5, page 527
Applied Example: Airport Traffic
Solution
b. The probability that between 6 and 8 minutes will elapse
between successive arrivals is given by
8
P(6  x  8)   0.1e0.1x dx
6
 e
0.1 x 8
6
  e 0.8  e 0.6
 0.10
Applied Example 5, page 527
Applied Example: Airport Traffic
Solution
c. The probability that the time between successive arrivals
will be more than 15 minutes is given

P( x  15)   0.1e0.1x dx
15
b
 lim  0.1e0.1x dx
b 15
0.1 x b 

 lim  e
15 
b 

 lim(  e 0.1b  e 1.5 )
b
 e 1.5
 0.22
Applied Example 5, page 527
End of
Chapter