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. Integration Using Tables of Integrals
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. Numerical Integration
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 a b

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 a b

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 a b

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
x0 = a x1 b

28. The Trapezoidal Rule
The area of each trapezoid is calculated by multiplying its base, x , by its average height: y x
f(x1) R2
f(x2) x a x1 x2 b

29. The Trapezoidal Rule
The area of each trapezoid is calculated by multiplying its base, x , by its average height: y
f(x2) R3
f(x3) x a x2 x3 b

30. The Trapezoidal Rule
The area of each trapezoid is calculated by multiplying its base, x , by its average height: y
f(x3) R4
f(x4) x a x3 x4 b

31. The Trapezoidal Rule
The area of each trapezoid is calculated by multiplying its base, x , by its average height: y
f(x4) R5
f(x5) x a x4 x5 b

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 a x5
b = x6

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 a b

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 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. 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 R1 R2 x x0 x1 x2

42. 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 R x x0 x1 x2

43. 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))
(x1, f(x1))
(x0, f(x0)) x x0 x1 x2

44. Simpson’s Rule
Simpson’s rule approximates the area under the curve of a function f(x) using a quadratic function:
Simpson’s rule

45. Example Find an approximation of using Simpson’s rule with n = 10.
Solution
Here, a = 1, b = 2, an n = 10, so
Simpson’s rule yields
Example 3, page 503

46. Example Find an approximation of using Simpson’s rule with n = 10.
Solution
Recall that the trapezoidal rule with n = 10 yielded an approximation of , with an error of from the value of ln 2 ≈ to six decimal places.
Simpson’s rule yields an approximation with an error of 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 person’s aorta, measured in 2-second intervals after 5 mg of dye have been injected: y
3.9
4.0 4 3 2 1
3.2
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1 x
Applied Example 4, page 504

49. Applied Example: Cardiac Output
The person’s 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 4 3 2 1
3.2
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1 x
Applied Example 4, page 504

50. Applied Example: Cardiac Output
Use Simpson’s rule with n = 14 to evaluate the integral and determine the person’s cardiac output. y
3.9
4.0 4 3 2 1
3.2
2.5
1.8
2.0
1.3
0.8
0.5
0.4
0.2
0.1 x
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.
Simpson’s rule yields
Applied Example 4, page 504

52. Applied Example: Cardiac Output
Solution
Therefore, the person’s 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 = e–x/ for x  0.
Solution
The required area is shown in the diagram below: y 1 R
y = e–x/2 x
1 2 3
Example 3, page 514

58. Examples
Find the area of the region R under the curve y = e–x/ 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,
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 – 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 < 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 = xe–x2, 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. Applications of Calculus to Probability
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
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) = ke–kx
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 doctor’s office, etc. y k
f(x) = ke–kx 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) = 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

85. Applied Example: Life Span of Light Bulbs
Solution
We compute
Integrating by parts with u = x and dv = e–kxdx
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) = ke–kx (0  x < )
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) = ke–kx
The expected value of x is 10, so
Thus, the required probability density function is
f(x) = 0.1e–0.1x
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