Integration by Parts

1
7
Additional Topics in Integration

• Integration by Parts
• Integration Using Tables of Integrals
• Numerical Integration
• Improper Integrals
• Applications of Calculus to Probability

2
7.1

• Integration by Parts

3
The Method of Integration by Parts

• Integration by parts formula

4
Example

• Evaluate
• Solution
• Let u x and dv ex dx
• So that du dx and v ex
• Therefore,

Example 1, page 484
5
Guidelines for Integration by Parts

• Choose u an dv so that
• du is simpler than u.
• dv is easy to integrate.

6
Example

• Evaluate
• Solution
• Let u ln x and dv x dx
• So that and
• Therefore,

Example 2, page 485
7
Example

• Evaluate
• Solution
• Let u xex and
• So that and
• Therefore,

Example 3, page 486
8
Example

• Evaluate
• Solution
• Let u x2 and dv ex dx
• So that du 2xdx and v ex
• Therefore,

(From first example)
Example 4, page 486
9
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
• 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
10
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,
• So,

Applied Example 5, page 487
11
Applied Example Oil Production

• Solution
• Use integration by parts to evaluate the
  integral.
• Let and
• So that and
• Therefore,

Applied Example 5, page 487
12
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,
• Thus, the required production function is given by

Applied Example 5, page 487
13
7.2

• Integration Using Tables of Integrals

14
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.

15
Examples

• Use the table of integrals to find
• Solution
• We first rewrite
• Since is of the form , with a 3, b 1, and u
  x, we use Formula (5),
• obtaining

Example 1, page 493
16
Examples

• Use the table of integrals to find
• Solution
• We first rewrite 3 as , so that has the form
• with and u x.
• Using Formula (8),
• obtaining

Example 2, page 493
17
Examples

• Use the table of integrals to find
• Solution
• We can use Formula (24),
• Letting n 2, a ½, and u x, we have

Example 5, page 494
18
Examples

• Use the table of integrals to find
• Solution
• We have
• Using Formula (24) again, with n 1, a ½,
  and u x, we get

Example 5, page 494
19
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
• percent per year.
• If the prediction holds true, what will be the
  average mortgage rate over the 12 months?

Applied Example 6, page 495
20
Applied Example Mortgage Rates

• Solution
• The average mortgage rate over the next 12 months
  will be given by

Applied Example 6, page 495
21
Applied Example Mortgage Rates

• Solution
• We have
• Use Formula (1)
• to evaluate the first integral
• or approximately 6.99 per year.

Applied Example 6, page 495
22
7.3

• Numerical Integration

23
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
• 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.

24
The Trapezoidal Rule

• Consider the problem of finding the area under
  the curve of f(x) for the interval a, b

y
R
x
b
a
25
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
R1
R2
R3
R4
R5
R6
x
b
a
26
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
R1
R2
R3
R4
R5
R6
x
b
a
27
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
?x
f(x0)
R1
f(x1)
x
b
x0 a
x1
28
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
?x
f(x1)
f(x2)
R2
x
b
a
x1
x2
29
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
f(x2)
f(x3)
R3
x
b
a
x3
x2
30
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
f(x3)
f(x4)
R4
x
b
a
x3
x4
31
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
f(x4)
f(x5)
R5
x
b
a
x5
x4
32
The Trapezoidal Rule

• The area of each trapezoid is calculated by
  multiplying its base, ?x , by its average height

y
f(x5)
R6
f(x6)
x
b x6
a
x5
33
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
R1
R2
R3
R4
R5
R6
x
b
a
34
The Trapezoidal Rule

• Adding the areas R1 through Rn of the trapezoids
  yields the following rule

• Trapezoidal Rue

35
Example

• Approximate the value of using the trapezoidal
  rule with n 10.
• Compare this result with the exact value of the
  integral.
• Solution
• Here, a 1, b 2, an n 10, so
• and
• x0 1, x1 1.1, x2 1.2, x3 1.3, , x9
  1.9, x10 1.10.
• The trapezoidal rule yields

Example 1, page 500
36
Example

• Approximate the value of using the trapezoidal
  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
• Thus the trapezoidal rule with n 10 yields a
  result with an error of 0.000624 to six decimal
  places.

Example 1, page 500
37
Applied Example Consumers Surplus

• The demand function for a certain brand of
  perfume is given by
• 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
38
Applied Example Consumers Surplus

• Solution
• When p 60, we have
• 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
• 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
39
Applied Example Consumers Surplus

• Solution
• We can use the trapezoidal rule with a 0, b
  800, and n 10.
• and
• x0 0, x1 80, x2 160, x3 240, , x9
  720, x10 800.
• The trapezoidal rule yields

Applied Example 2, page 500
40
Applied Example Consumers Surplus

• Solution
• The trapezoidal rule yields
• Therefore, the consumers surplus is
  approximately 22,294.

Applied Example 2, page 500
41
Simpsons Rule

• Weve seen that the trapezoidal rule approximates
  the area under the curve by adding the areas of
  trapezoids under the curve

y
R1
R2
x
x0
x1
x2
42
Simpsons Rule

• The Simpsons 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
R
x
x0
x1
x2
43
Simpsons 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))
(x1, f(x1))
(x0, f(x0))
x
x0
x1
x2
44
Simpsons Rule

• Simpsons rule approximates the area under the
  curve of a function f(x) using a quadratic
  function
• Simpsons rule

45
Example

• Find an approximation of using Simpsons rule
  with n 10.
• Solution
• Here, a 1, b 2, an n 10, so
• Simpsons rule yields

Example 3, page 503
46
Example

• Find an approximation of using Simpsons rule
  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.
• Simpsons 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
47
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
48
Applied Example Cardiac Output

• The graph of c(t) shows the concentration of dye
  in a persons aorta, measured in 2-second
  intervals after 5 mg of dye have been injected

y
3.9
4.0
3.2
4 3 2 1
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1
0
x
2 4 6 8 10 12 14 16 18 20 22 24 26
Applied Example 4, page 504
49
Applied Example Cardiac Output

• The persons cardiac output, measured in liters
  per minute (L/min) is computed using the formula
• where D is the quantity of dye injected.

y
3.9
4.0
3.2
4 3 2 1
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1
0
x
2 4 6 8 10 12 14 16 18 20 22 24 26
Applied Example 4, page 504
50
Applied Example Cardiac Output

• Use Simpsons rule with n 14 to evaluate the
  integral and determine the persons cardiac
  output.

y
3.9
4.0
3.2
4 3 2 1
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1
0
x
2 4 6 8 10 12 14 16 18 20 22 24 26
Applied Example 4, page 504
51
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.
• Simpsons rule yields

Applied Example 4, page 504
52
Applied Example Cardiac Output

• Solution
• Therefore, the persons cardiac output is
• or approximately 6.0 L/min.

Applied Example 4, page 504
53
7.4

• Improper Integrals

54
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.

55
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
• if the limit exists.

56
Examples

• Evaluate if it converges.
• Solution
• Since ln b ? ?, as b ? ? the limit does not
  exist, and we conclude that the given improper
  integral is divergent.

Example 2, page 513
57
Examples

• Find the area of the region R under the curve y
  ex/2 for x ? 0.
• Solution
• The required area is shown in the diagram below

y
1
R
y ex/2
x
1 2 3
Example 3, page 514
58
Examples

• Find the area of the region R under the curve y
  ex/2 for x ? 0.
• Solution
• Taking b gt 0, we compute the area of the region
  under the curve y ex/2 from x 0 to x b,
• Then, the area of the region R is given by
• or 2 square units.

Example 3, page 514
59
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
• if the limit exists.

60
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 3 7
1 1
x
R
x 1
y e2x
Example 4, page 514
61
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 lt 1, compute
• Then, the area under the required region R is
  given by

Example 4, page 514
62
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
• are convergent.
• Then, the improper integral of f over ( ?, ?) is
  defined by

63
Examples

• Evaluate the improper integral
• and give a geometric interpretation of the
  result.
• Solution
• Take the number c to be zero and evaluate first
  for the interval ( ?, 0)

Example 5, page 515
64
Examples

• Evaluate the improper integral
• and give a geometric interpretation of the
  result.
• Solution
• Now evaluate for the interval (0, ?)

Example 5, page 515
65
Examples

• Evaluate the improper integral
• and give a geometric interpretation of the
  result.
• Solution
• Therefore,

Example 5, page 515
66
Examples

• Evaluate the improper integral
• and give a geometric interpretation of the
  result.
• Solution
• Below is the graph of y xex2, showing the
  regions of interest R1 and R2

y
1 1
R2
2 1
x
1 2
R1
Example 5, page 515
67
Examples

• Evaluate the improper integral
• 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 1
R2
2 1
x
1 2
R1
Example 5, page 515
68
Examples

• Evaluate the improper integral
• 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 1
R2
2 1
x
1 2
R1
Example 5, page 515
69
Examples

• Evaluate the improper integral
• and give a geometric interpretation of the
  result.
• Solution
• Thus, adding the areas of the two regions yields
  zero

y
1 1
R2
2 1
x
1 2
R1
Example 5, page 515
70
7.5

• Applications of Calculus to Probability

71
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.
• The total area R of the region under the graph of
  f is equal to 1

y
R
x
72
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.
• The probability that an observed value of the
  random variable x lies in the interval a, b is
  given by

y
R1
x
a
b
73
Examples

• Show that the function
• 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

Example 1, page 522
74
Examples

• Show that the function
• 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

Example 1, page 522
75
Examples

• Determine the value of the constant k so that the
  function
• is a probability density function on the
  interval 0, 5.
• Solution
• We compute
• Since this value must be equal to one, we find
  that

Example 2, page 522
76
Examples

• If x is a continuous random variable for the
  function
• compute the probability that x will assume a
  value between x 1 and x 2.
• Solution
• The required probability is given by

Example 2, page 522
77
Examples

• If x is a continuous random variable for the
  function
• 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
0.6 0.4 0.2
P(1 ? x ? 2) 7/125
x
1 2 3 4 5
Example 2, page 522
78
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
• where x denotes the life span of a light bulb.
• Determine the probability that a light bulb will
  have a life span of
• 500 hours or less.
• More than 500 hours.
• More than 1000 hours but less than 1500 hours.

Applied Example 3, page 523
79
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

Applied Example 3, page 523
80
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

Applied Example 3, page 523
81
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

Applied Example 3, page 523
82
Exponential Density Function

• The example we just saw involved a function of
  the form
• f(x) kekx
• where x ? 0 and k is a positive constant, with a
  graph
• 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 doctors
  office, etc.

y
k
f(x) kekx
x
83
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

84
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) kekx (0 ? x lt ?)
• 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
85
Applied Example Life Span of Light Bulbs

• Solution
• We compute
• Integrating by parts with u x and dv
  ekxdx
• so that
• We have

Applied Example 4, page 525
86
Applied Example Life Span of Light Bulbs

• Solution
• We have

Applied Example 4, page 525
87
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,
• Therefore,
• 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
88
Expected Value of an Exponential Density Function

• If a continuous random variable x is
  exponentially distributed with probability
  density function
• f(x) kekx (0 ? x lt ?)
• then the expected (average) value of x is given
  by

89
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.
• Find the probability density function associated
  with x.
• What is the probability that between 6 and 8
  minutes will elapse between successive arrivals
  of planes.
• What is the probability that the time between
  successive arrivals of planes will be more than
  15 minutes?

Applied Example 5, page 527
90
Applied Example Airport Traffic

• Solution
• Since x is exponentially distributed, the
  associated probability density function has the
  form
• f(x) kekx
• The expected value of x is 10, so
• Thus, the required probability density function
  is
• f(x) 0.1e0.1x

Applied Example 5, page 527
91
Applied Example Airport Traffic

• Solution
• The probability that between 6 and 8 minutes will
  elapse between successive arrivals is given by

Applied Example 5, page 527
92
Applied Example Airport Traffic

• Solution
• The probability that the time between successive
  arrivals will be more than 15 minutes is given

Applied Example 5, page 527
93
End of Chapter