Download Chapter 7: Principles of Integral Evaluation

140
Chapter 7: Principles of Integral Evaluation
Summary: The primary focus of this chapter is to introduce and explain more
advanced integration techniques. With all of the techniques that are introduced,
however, the focus is usually to try and find a way to use the basic integral rules (or
basic list of antiderivatives) that were introduced in Chapter 5. Many of the ideas
in this chapter are methods to use algebraic manipulation or special substitutions
so that integrals may be simplified into an integral resembling a basic integral rule.
IDEA: At this point the reader should review the integration techniques from
Chapter 5. Particular emphasis should be placed upon the basic antiderivatives (§5.2) and u-substitution (§5.3).
One very powerful method that is introduced in this chapter is the method of integration by parts. This method can simply be thought of as the product rule for
antiderivatives. Next various trigonometric integrals are considered and then some
special trigonometric substitutions that can be used to simplify a variety of integrals that involve square roots and differences and sums of squares. In the case of
rational integrands, the method of partial fraction decomposition may sometimes
be used to simplify the integrand. In cases where antiderivatives cannot be expressed using simple functions, the idea of numerical integration is considered for
definite integrals. Finally at the end of the chapter, integrals are considered that
may involve either infinite limits of integration or integrands that have an infinite
discontinuity (such as a vertical asymptote).
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. be familiar with the basic integration techniques (see §5.2 and §7.1)
2. be aware of the different resources for evaluating integrals (§7.1)
3. be able to use integration by parts and recognize when it will be useful
(§7.2)
4. manipulate integrands using trigonometric identities and formulas to obtain
more simple integrals (§7.3)
5. be able to use trigonometric substitution and recognize when it can be used
(§7.4)
141
basic antiderivatives
→ §5.2
142
6. use partial fraction expansion to evaluate integrals (§7.5)
7. use CAS and tables of integrals where appropriate (§7.6)
8. use numerical integration techniques where appropriate (§7.7 and also §5.4)
9. identify definite integrals as improper and determine whether they converge
(to a value) or diverge (§7.8)
7.1 An Overview of Integration Methods
PURPOSE: To review general integration techniques.
REVIEW:
1. basic antiderivatives (§5.2)
2. u-substitution (§5.2 and §5.9)
3. F.T.C (§5.6)
There are no new ideas in this section but this section does serve as a good checkpoint for the reader. There are several concepts and methods that the reader should
be comfortable with before attempting to learn the material in this chapter. First,
this section should serve as a reminder to the reader of all of the integration rules
that have been introduced in the text so far. It is also a good idea at this point to review the integration techniques that have been introduced so far. This essentially
means to review the basic integral formulas that are related to polynomials, logarithms, exponential functions, trigonometric functions, hyperbolic functions and
inverse trigonometric or hyperbolic functions. Also, the reader should be comfortable with the idea of u-substitution both in indefinite integrals and with definite
integrals. At this point, evaluating definite integrals using the Fundamental Theorem of Calculus should seem more natural.
IDEA: The emphasis throughout this chapter will be to use tables, transformations and other techniques to evaluate integrals that do not qualify as any of
the “basic” integrals seen in Chapter 5.
A free CAS available online is
Maxima:
→ maxima.sourceforge.net
The emphasis in this chapter is going to be on using tables and transformation
methods. For example, appropriate transformation methods can simplify an integral so that the reader only has to use the basic integral rules that were introduced
in Chapter 5. Take some time now to locate the integral tables at the beginning and
end of the book. Before starting the chapter the reader may also want to identify a
CAS that they can use (for example, Maple, Mathematica and Derive are all good
choices). This is important for two reasons. First, using a CAS can be a valuable
tool for checking your answers as you progress through the material. On the other
hand, there are some problems that are intended to be done using a CAS.
Checklist of Key Ideas:
review basic integration techniques
technology; CAS
tables of integrals
transformation methods
143
7.2 Integration by Parts
PURPOSE: To develop a versatile integration technique that allows many integrals to be rewritten involving a simpler integrand.
Integration by parts can be thought of as the product rule for antiderivatives. In
fact, the product rule for derivatives gives a good way to remember this formula.
Here is the differential version of the product rule.
d(uv) = u dv + v du
Then the following is the integral version.
Z
(u dv + v du) =
Solve for
Z
Z
Z
u dv +
Z
v du = uv
u dv to find the integration by parts formula.
u dv = uv −
Z
v du
IDEA: Integration by parts is really just an antiderivative version of the product
rule.
This simple expression of the integration by parts formula can be deceptive, however. The variables u and v are actually supposed to represent functions of x. So
the differential form of the product rule could really be written as follows.
d u(x)v(x) = u(x) v′ (x)dx + v(x) u′ (x)dx
Then the integral version would be as shown below.
Z
u(x)v′ (x) dx +
Z
v(x)u′ (x) dx = u(x)v(x)
So the integration by parts formula is as follows.
Z
u(x)v′ (x) dx = u(x)v(x) −
Z
v(x)u′ (x) dx
CAUTION: Do not forget the constant of integration with indefinite integrals.
Another important thing is to not forget the constant of integration. As long as
there is still an indefinite integral involved, there should be a C as a part of the antiderivative. On the other hand, it is important to remember with definite integrals
that even the uv term needs to be evaluated at the limits of integration.
Zb
a
x=b Zb
u(x)v (x) dx = u(x)v(x) − v(x)u′ (x) dx
′
x=a
a
Note:
u = u(x)
dv = v′ (x) dx
144
IDEA: Integration by parts may allow a complicated integrand to be rewritten
as a more basic integrand.
R
The basic idea behind integration by parts is that the integral u dv may not be a
function that can be integrated by our basic techniques.
But through an appropriate
R
choice of u and dv, the resulting integral of v du may be easier to handle. The
toughest part about this method is picking u and dv. Finding du and v can be
accomplished by taking the derivative of u and integrating dv respectively.
IDEA: du is found by taking the derivative of u. v is found by integrating dv.
LIATE Method:
pick u towards the top
⇓
L - logarithmic
I - inverse trigonometric
A - algebraic, polynomial
T - trigonometric
E - exponential
⇓
pick dv towards the bottom
The LIATE method tries to make picking u and dv easier by following a particular
pattern. For example, the types of functions represented by LIATE (Logarithmic
– Inverse Trigonometric – Algebraic – Trigonometric – Exponential) can be used
as a guide for picking u and dv (see margin to the left). Algebraic functions really
stand for polynomials or rational functions.
tabular integration by parts
Tabular integration by parts can simplify the process of using integration by
parts. After u = u(x) and dv = v′ (x)dx are initially picked, tabular integration
by parts creates two columns. In the “u” column, derivatives of u are calculated.
In the “dv” column, dv terms are integrated over and over. This process should
simply be continued until the derivative of u either is zero or it repeats itself as the
original u. Then the appropriate terms should be matched from each column with
the appropriate signs and a constant
of integration added to the end. If the original
R
u and dv are repeated, then u dv will need to be solved for algebraically.
IDEA: Integration by parts can be tried on any integral such as
letting u = f (x) and dv = dx.
R
f (x) dx by
Another difficult thing to decide is whether or not to even use integration by parts.
It is a very versatile method and very often just trying it can lead to an integral
with a basic
formula. One choice for u and dv that can always be tried with the
R
integral f (x) dx is u = f (x) and dv = dx. On the other hand, there are times
when integration by parts must be used. Some obvious examples are when there
is a product of functions of different types that does not give rise to a basic integral
form.
Integration by parts also gives rise to the following useful trigonometric reduction
formulas.
Z
1
n−1
sinn x dx = − sinn−1 x cos x +
n
n
Z
sinn−2 x dx for n ≥ 2
n−1
1
sin x cosn−1 x +
cosn−2 x dx for n ≥ 2
n
n
These formulas are used in the next section on trigonometric integrals (see §7.3).
Z
cosn x dx =
Z
145
Checklist of Key Ideas:
reverse of product rule
guidelines for integration by parts; choosing u and dv
LIATE
tabular integration by parts
integration by parts and definite integrals
reduction formulas
7.3 Integrating Trigonometric Functions
PURPOSE: To develop integration techniques for integrands that
are comprised of powers or products of trigonometric functions.
Some integrals are comprised solely of trigonometric functions. These are integrals where powers of trigonometric functions occur either by themselves or as
products. Most often the techniques that are used here are reduction techniques.
The goal is to transform the integrand into a simpler expression for which the
basic methods from Chapter 5 can be applied. Sometimes this takes the form of
using reduction formulas or other times, an appropriate trigonometric identity can
be used with u-substitution to simplify the integrand.
reduction formulas
see §5.2
IDEA: The goal is to reduce or transform an integrand to the point where basic
antiderivative techniques may be used.
In many cases, trigonometric identities and definitions can be used to simplify a
given integral. For example, the integral
Z
sin2 x cos2 x dx
does not have a simple antiderivative and u-substitution does not work directly.
However, the Pythagorean identity and the cosine reduction formula can be used
on this integral. Using sin2 x = 1 − cos2 x leads to the following.
Z
(cos2 x − cos4 x) dx
Then the trigonometric reduction formulas (see §7.2) can be used to evaluate these
two integrals.
Z
cos2 x dx =
1
1
sin x cos x +
2
2
Z
1
dx = (sin x cos x + x) +C
2
Pythagorean Identity
sin2 u + cos2 u = 1
146
R
cos4 x dx=
=
3
1
sin x cos3 x +
4
4
Z
cos2 x dx
3
1
sin x cos3 x + (sin x cos x + x) +C
4
8
These are then combined to give the antiderivative below.
1
1
x
sin2 x cos2 x dx = − sin x cos3 x + sin x cos x + +C
4
8
8
Z
IDEA: Integrals involving trigonometric functions may often be evaluated in
more than one way.
There is often more than one way
Z to integrate trigonometric integrals. For example, consider again the integral
sin2 x cos2 x dx. By using a double-angle identity
for sine we have
sin2 x cos2 x = (sin x cos x)2 =
1 2
1
sin (2x) = (1 + cos (4x))
4
8
and then it is a simple matter to integrate as follows.
1
8
Note:
Many of these techniques make
use of basic trig definitions,
double angle-identities, and the
Pythagorean identity.
Z
(1 + cos (4x)) dx =
x sin (4x)
+
+C
8
32
General techniques are given for the following types of integrals in this section:
Z
sinm x cosn x dx
Z
tann x dx
Z
sin (mx) cos (nx) dx
Z
secn x dx
Z
tanm x secn x dx
where m and n are positive integers.
Checklist of Key Ideas:
reduction formulas
simplifying using trigonometric identities
products of sines and cosines to integer powers
integrating powers of tangent and secant
trigonometric identities and u-substitution
147
7.4 Trigonometric Substitutions
PURPOSE: To develop integration techniques to handle integrands that involve square roots.
Some integrals involve square root terms where u-substitution is ineffective. In
some of these cases, a special type of substitution called a trigonometric substitution may be used to simplify the square root so that the integral has a more
simple form. The idea with each trigonometric substitution is to make use of the
Pythagorean identity or one of its forms in order to simplify the radical.
Three Forms of
the Pythagorean identity:
sin2 u + cos2 u = 1
tan2 u + 1 = sec2 u
1 + cot2 u = csc2 u
IDEA: Trigonometric substitution involves u-substitution combined with the
Pythagorean identity.
√
For integrals that have a2 − x2 , a substitution of x = a sin u will often work. Notice that dx = a cos u du. This substitution can be remembered since this square
root looks similar to the one which is involved in the integral for sin−1 x. (Incidentally, a substitution of x = a cos u should also work.) The substitution will get
rid of the square root in the following way.
p
√
a2 − x2 = pa2 − a2 sin2 u
= a√ 1 − sin2 u
= a cos2 u
= ±a cos u
√
For integrals that have a2 + x2 or a2 + x2 , a substitution of x = a tan u will often work. Simplification will come from an alternative form of the Pythagorean
identity:
√
a2 − x2 → try x = a sin u
a2 + x2 → try x = a tan u
1 + tan2 x = sec2 x
This substitution x = a tan u can be remembered since the form a2 + x2 looks similar to the integral that is related to tan−1 x.
Completing the square on some quadratic terms can lead to terms that look like
a2 + (x − h)2 . The appropriate substitution should be x − h = a tan u and then
dx = a sec2 u du.
√
For integrals that have x2 − a2 , a substitution of x = a sec u will often work.
Simplification will come by rearranging the version of the Pythagorean identity
used above.
1 + tan2 x = sec2 x −→ sec2 x − 1 = tan2 x
This substitution x = a sec u can be remembered since the form
similar to the integral that relates to sec−1 x.
√
x2 − a2 looks
√
x2 − a2 → try x = a sec u
148
Checklist of Key Ideas:
method of trigonometric substitution
√
integrals involving a2 − x2
√
integrals involving a2 + x2
√
integrals involving x2 − a2
integrals involving irreducible quadratic denominators, a(x − h)2 + c2
7.5 Integrating Rational Functions by Partial Fractions
PURPOSE: To develop a technique to assist in integrating rational functions.
When a proper rational function is in an integral, it may sometimes be rewritten
using partial fraction decomposition. This amounts to taking a fraction and
writing it as a sum of fractions that do not have a common denominator. Often the
resulting pieces or partial fractions are simpler and easier to handle in an integral.
Partial Fraction Decomposition:
1. write as a proper rational function
2. factor the denominator
3. solve for constants in numerator
To be able to use partial fraction decomposition requires three things. First, the
rational function should be written as a proper rational function (degree of the
polynomial on the bottom is greater than the top). Second, the denominator of the
rational expressions should be factored. Last, after the partial fraction decomposition is performed, the values of the resulting constants in the numerators need to
be determined.
The quadratic equation is useful for
factoring.
If ax2 + bx + c = 0 then
√
−b ± b2 − 4ac
r1,2 =
2a
⇓
ax2 + bx + c = (x − r1 )(x − r2 )
There are several cases considered here. Each case depends upon the types of
factors in the denominator.
1. distinct linear factors
A
B
ax + b
=
+
(x − r1 )(x − r2 ) x − r1 x − r2
2. repeated linear factors
ax + b
A
B
=
+
(x − r)2
x − r (x − r)2
ax2 + bx + c
A
B
C
=
+
+
(x − r)3
x − r (x − r)2 (x − r)3
149
3. distinct quadratic factors
b3 x3 + b2 x2 + b1 x + b0
A1 x + A0
B1 x + B0
=
+
(a2 x2 + a1 x + a0 )(c2 x2 + c1 x + c0 ) a2 x2 + a1 x + a0 (c2 x2 + c1 x + c0 )
4. repeated quadratic factors
A1 x + A0
B1 x + B0
b3 x3 + b2 x2 + b1 x + b0
=
+
(a2 x2 + a1 x + a0 )2
a2 x2 + a1 x + a0 (a2 x2 + a1 x + a0 )2
There are two ways to evaluate the constants in the numerators after partial
fractions have been determined based upon the factors of the denominator. A new
numerator involving the new, unknown coefficients can be found by recombining
the partial fraction terms into one rational term. Then several equations for the
coefficients can be obtained by equating the numerators of the terms with the
full denominators. Then the coefficients can either be determined by trying to
match the coefficients of the terms with the corresponding powers of x or by
substituting different values for x into the equations.
Checklist of Key Ideas:
partial fraction decomposition
proper rational function
determining coefficients by matching terms
determining coefficients by substitution
linear factors
quadratic factors
quadratic formula
using long division (or synthetic division) when possible
7.6 Using Computer Algebra Systems and Tables of
Integrals
PURPOSE: To introduce alternative methods for evaluating integrals.
There are an infinite variety of integrals that can arise. It will not always be a
simple process to find a substitution or transformation that will allow a simple
antiderivative to be found. Using a table of integrals can shorten the effort required to evaluate a given integral. Likewise, a Computer Algebra System (CAS)
can greatly shorten the time required to evaluate an integral. It is still useful to
evaluating constants:
1. substitution
2. matching terms
150
understand how integration works, however, as sometimes the results that are returned by a table of integrals or CAS may seem more mysterious then the original
integral.
CAUTION: Sometimes no perfect match or no possible match for an integrand
can be found → u-substitution may help.
When using a table of integrals, sometimes a perfect match cannot be found for
the integral that is being considered. In some cases, an appropriate u-substitution
can be used to find a form in the tables that does match. One example of this is
integrals that involve fractional powers of x. In these cases, a good substitution to
try is u = x1/r where r is the LCD of all of the fractional powers of x. This can
have the effect of removing all of the fractional powers of x so that other methods
may then be employed.
Checklist of Key Ideas:
be familiar with what is in the table of integrals
perfect matches and imperfect matches
using substitution to create a match
integrals involving fractional powers of x
rational trigonometric functions
understanding and using integration with CAS
7.7 Numerical Integration; Simpson’s Rule
PURPOSE: To develop techniques for approximating the value of
a definite integral.
see also §5.4
midpoint method
Definite integrals will often arise that cannot be evaluated using the Fundamental
Theorem of Calculus since a simple antiderivative is not known for the integrand.
In these cases, numerical integration may be the only way to find a value for the integral. Some forms of numerical integration have already been encountered: left
endpoint approximations, right endpoint approximations and midpoint approximations. More sophisticated methods are introduced here.
The midpoint method has already been introduced (i.e., the rectangle method
with heights of rectangles determined at the midpoints of the intervals). This is
sometimes called the tangent line approximation because it is equivalent to drawing a tangent line to a curve at the midpoint of each subinterval. The trapezoidal
area underneath this tangent line is the same as the area of the rectangle determined by the midpoint method. This is due to the symmetry of the tangent line at
the midpoint of the subinterval.
151
The trapezoidal approximation is the average of the left and right endpoint approximations. In essence, each subinterval is approximated by the area of a trapezoid. The trapezoid is formed by connecting the function values at the left and
right endpoints of the subinterval with a straight line. Then the area of the resulting trapezoid is used to approximate the area under the curve on the subinterval.
trapezoidal method
Simpson’s rule is the result of drawing a quadratic function over the interval and
then finding the area under the quadratic function on the interval. The quadratic
function that is used is required to go through the function values at the left endpoint, right endpoint and the midpoint of the subinterval. As it turns out, the
following is true.
Simpson’s rule
1
1
Sn = (Ln/2 + 4Mn/2 + Rn/2 ) = (Tn/2 + 2Mn/2 )
6
3
Usually Simpson’s Rule is applied on an even number of intervals so that the
midpoints occur at the nodes of the partition that is being used and extra function
values do not need to be calculated. Thus, in the formula given above, n and n/2
should be integers.
Checklist of Key Ideas:
trapezoidal approximations
midpoint approximations; tangent line approximation
Simpson’s Rule
absolute errors; |EM |, |ET | and |ES |
overestimates and underestimates
error bounds
7.8 Improper Integrals
PURPOSE: To evaluate integrals with discontinuous integrands
or infinite limits of integration.
When an infinite discontinuity (such as a vertical asymptote) or an infinite limit of
integration appears within a definite integral then we have an improper integral.
In some of these cases, a value for the integral can be found. Then the definite
integral is said to converge. In cases where an improper integral does not evaluate
to some finite number, then the integral is said to diverge.
To evaluate improper integrals, the definite integral is rewritten using a dummy
variable such as b. There are two situations that may arise: if the limits of integration are infinite or if the integrand is discontinuous in some way (particularly
at the endpoints of the interval). In the case of infinite limits of integration, the
converge vs. diverge
Types of Improper Integrals:
1. infinite limit of integration
2. discontinuous integrand
152
limits are rewritten using a dummy variable (such as b). The integral is evaluated
with the dummy variable and then the limit is taken as the dummy variable (i.e.,
b) approaches the appropriate limiting value. For example, the following integral
has an infinite upper limit of integration.
Z b
0
e−x dx = lim
Z b
b→∞ 0
e−x dx
IDEA: First, evaluate the integral with a limit of integration of b. Then take the
limit involving b.
The definite integral is first evaluated using a limit of integration of b and then the
limit involving b is evaluated. If the limit exists, then the improper integral is said
to converge. Otherwise, the integral is said to diverge.
In the case where the integrand is discontinuous, the point of discontinuity may
occur either at an endpoint or in the middle of the interval. If the discontinuity occurs at an endpoint then a dummy variable is again used in place of the appropriate
limit of integration. The following integral has an integrand that is discontinuous
as it approaches the lower limit of integration.
x=1
Z 1
Z 1
√ 1/2 −1/2
−1/2
= lim 2 − 2 b = 2
x
dx = lim
x
dx = lim 2x b→0+ b
0
b→0+
x=b
b→0+
Again, the integral is evaluated first with the dummy variable, then the limit is
evaluated. If the limit exists and is finite then the improper integral converges,
otherwise it diverges.
IDEA: More than one dummy variable and more than one integral may be
required if
1. there are two infinite limits of integration.
2. there is a discontinuity in the middle of the interval.
discontinuity in middle of interval
If a discontinuity occurs in the middle of the interval then the integral needs to
be written as two integrals and two different limits taken. If either of the resulting
integrals diverges then the whole improper integral will diverge. For example,
Z 1
1
−1
x
dx = lim
Z a
1
a→0− −1
x
dx + lim
Z 1
1
b→0+ b
x
dx
which diverges because both integrals diverge. Notice that in writing two different
integrals, two different dummy variables are used.
A similar situation can also arise with infinite limits of integration. For example,
Z ∞
−∞
2
e−x dx = lim
Z 0
a→−∞ a
2
e−x dx + lim
Z b
b→∞ 0
2
e−x dx
Again, notice that two dummy variables are used, one for each infinite limit of
integration. In this example, the fact that x = 0 was chosen as the “middle” limit
of integration was arbitrary, although the same number had to be used in both
integrals.
153
Checklist of Key Ideas:
improper integrals; infinite discontinuities; infinite limits of integration
rewriting as a proper integral and taking a limit
convergence and divergence
applications to arc length and surface area
154
Chapter 7 Sample Tests
1
cos9 x +C
9
1
(b) − cos9 x +C
9
Section 7.1
1. Evaluate
Z
(a)
(8 − 2x)5 dx.
(8 − 2x)6
+C
6
−(8 − 2x)6
(b)
+C
6
−(8 − 2x)6
(c)
+C
12
−(8 − 2x)6
+C
(d)
3
Z √
2. Evaluate
4x + 9 dx.
(c) 9 cos9 x +C
(a)
1
√
+C
8 4x + 9
2
+C
(b) √
4x + 9
1
(c) (4x + 9)3/2 +C
6
1
(d) (4x + 9)3/2 +C
2
(d) −9 cos9 x +C
7.
Z
Z
(c)
(d) e
8.
Z
ex dx
5 + ex
x
+C
+C
=
5ex
+C
5 + ex
(b) 5 ln (5 + ex ) +C
(a)
9.
Z
(c) 2 ln |4 − x2 | +C
1 x2
e +C
2
cos x
sin x e
dx =
(a) cos x ecos x +C
(b) −ecos x +C
(c) − sin x
ecos x +C
(d) −xecos x +C
8
cos x sin x dx =
4x dx
=
4 − x2
(b) −2 ln |4 − x2 | +C
2
2
ex
+C
5 + ex
(a) 4 ln |4 − x4 | +C
x ex dx =
(c) ex +C
2
1
(d) x2 ex +C
2
6.
4
√
(d)
2
Z
e
(a) 3 cos (x2 ) +C
(b) 2ex +C
5.
3e x
+C
4x3/2
(c) ln (5 + ex ) +C
(a)
Z
(b)
3x sin (x2 ) dx =
(c) −6 cos (x2 ) +C
3
(d) − cos (x2 ) +C
2
4.
e x
+C
2x3/2
√
x
(b) 6 cos (x2 ) +C
Z
√
(a)
√
(a)
3.
√
e x
√ dx =
2 x
(d) 4 ln |4 − x2 | +C
10.
Z
4 sinh2 x cosh x dx =
(a)
4
sinh3 x +C
3
(b) 12 sinh3 x +C
(c) 4 sinh2 x +C
(d) 2 sinh2 x +C
11. Answer true or false. In evaluating
for u is u = x2 .
Z
12. Answer true or false. In evaluating
choice for u would be sin x.
2
x 7x dx a good choice
Z
cos6 x sin x dx a good
155
13. Answer true or false. In evaluating
choice for u would be ex + 8.
14. Answer true or false. In evaluating
for u would be sin x.
Z
15. Answer true or false. In evaluating
choice for u would be x4 .
Z
1
1
cos (2x) − x2 cos (2x) +C
2
4
1
1
(b) x sin (2x) + cos (2x) − x2 cos (2x) +C
4
2
1
(c) x sin (2x) + cos (2x) − x2 cos (2x) +C
2
1
(d) 2x sin (2x) + cos (2x) − x2 cos (2x) +C
2
ex (ex + 8) dx a good
(a) x sin (2x) +
4 sin x
dx a good choice
cos x
Z
x3 cos (x4 ) dx a good
6.
Z
Section 7.2
1.
Z
(a)
2.
Z
7.
(d)
8.
(d) 3x3 e3x +C
x cos (9x) dx =
x2 sin (9x)
+C
18
sin (9x)
(b)
+C
9
cos (9x) x sin (9x)
+
+C
(c)
81
9
cos (9x) x cos (9x)
(d)
+
+C
81
9
(a)
4.
Z
2x sin x dx =
9.
Z
5e4x sin (3x) dx =
(a) 5e4x 3 sin (3x) + cos (3x) +C
(b) 5e4x 3 sin (3x) − cos (3x) +C
5
(c) e4x 3 sin (3x) − 3 cos (3x) +C
7
1
(d) e4x 4 sin (3x) − 3 cos (3x) +C
5
Z 1
4xe6x dx =
(a) 223.9
(b) 224.2
(c) 224.7
(d) 225.1
10.
Z 3
x2 ln x dx =
1
(a) 7.00
(b) 2 sin x − 2 cos x +C
(b) 7.03
(d) 2 cos x + 2x cos x +C
(d) 6.92
(c) 2 cos x − 2x cos x +C
5.
2x2 sin (2x) dx =
√
cos−1 (6x)
+C
6
0
(a) 2 sin x − 2x cos x +C
Z
sin−1 (6x) dx =
1 − 36x2
+C
6
√
1 − 36x2
+C
(b) x cos−1 (6x) +
6
(c) 6 cos−1 (6x) +C
(c) 27e3x +C
3.
(d) x2 ln (2x) − x2 +C
(a) x sin−1 (6x) +
e3x
(9x2 − 6x + 2) +C
3
(b) xe3x +C
Z
Z
9x2 e3x dx =
(a)
1 2
x2
x ln (2x) − +C
4
4
5 2
5x2
(b) x ln (2x) −
+C
2
4
(c) 5x2 ln (2x) − 5x2 +C
(a)
x e6x dx =
e6x
(6x − 1) +C
6
e6x
(6x − 1) +C
(b)
36
1
(c) x2 e6x +C
2
e6x
(d)
+C
6
5x ln (2x) dx =
(c) 6.96
11.
Z 3
1
cos (ln x) dx =
156
1
3x
+ sin (2x) + sin 4x +C
2
8
3x 3
1
− sin (2x) + cos3 x sin x +C
(b)
2
4
8
3
3
(c) sin (2x) + sin x cos x +C
4
3x 3
+ cos (2x) + sin3 x cos x +C
(d)
2
4
(a) 1.57
(a)
(b) 1.52
(c) 1.48
(d) 1.42
12. Answer true or false.
Z π /4
x sin (2x) dx = 1/4.
0
13. Answer true or false.
Z 3π /4
14. Answer true or false.
Z 1
π /4
x tan x dx = 1.
5.
Z
15. Answer true or false.
0
4xe−3x dx = 0.
Section 7.3
1.
Z
13
cos x sin x dx =
cos14 x
+C
14
cos14 x
+C
(b) −
14
cos14 x
(c)
+C
13
cos14 x
+C
(d) −
13
(a)
2.
Z
2 sin2 (5x) dx =
sin (10x)
(a) x −
+C
10
(b) 2x − 2 sin (5x) +C
6.
Z
ln |cos (9x)|
+C
9
ln |cos (9x)|
(b) −
+C
9
tan2 (9x)
+C
(c)
18
tan2 (9x)
+C
(d) −
18
7. Z
Answer true or false.
sin (6x) cos (4x) dx = −
8.
Z
− sin3 (2x) dx =
(a)
(b)
(c)
(d)
4.
Z
1
1
cos (2x) + sin3 (2x) +C
2
6
1
1
cos (2x) − cos3 (2x) +C
2
6
1
1
− cos (2x) + cos3 (2x) +C
2
6
1
1
− cos (2x) − sin3 (2x) +C
2
6
4 cos4 x dx =
sec2 (5x + 8) dx =
tan (5x + 8)
+C
5
tan (5x + 8)
+C
(b)
5x + 8
tan (5x + 8)
(c) −
+C
5
tan (5x + 8)
(d) −
+C
5x + 8
(a)
sin (5x)
+C
5
sin (5x)
(d) x −
+C
10
Z
tan (9x) dx =
(a)
(c) x −
3.
x sin (16x)
−
+C
8
128
x cos (16x)
(b) −
+C
8
128
x sin (8x)
(c) −
+C
8
128
x cos (8x)
(d) −
+C
8
128
(a)
ln (x2 + 20) dx = 1.
0
Z 1
sin2 (4x) cos2 (4x) dx =
9.
Z
csc (6x) dx =
ln |tan (3x)|
+C
6
ln |tan (3x)|
(b) −
+C
6
ln |tan (6x)|
+C
(c)
6
ln |tan (6x)|
(d) −
+C
12
(a)
cos (2x) cos (10x)
−
+C
4
20
157
10. Answer true or false.
11.
Z
Z
tan11 x sec2 x dx =
tan12 x
+C
12
(b) 1.388
tan x sec5 x dx =
(c) 3.271
(a) sec5 x +C
(d) 1.398
sec6 x
(b)
+C
6
sec5 x
+C
(c)
5
(d) sec6 x +C
12. Answer
−
(a) 1.381
true
4.
Z 3
2
(−x)2
dx
p
=
(−x)2 − 1
(a) 14.941
(b) 0.077
(c) 0.093
or
Z
false.
cot6 (4x)
+C
24
13. Answer true or false.
(d) 17.01
cot5 (4x) csc2 (4x) dx =
5.
Z 2
4dx
√
=
4
x x2 + 5
(a) 2.003
1
Z π /4
tan2 (6x) dx = 1.00.
(b) 1.684
0
14. Answer true or false.
Z π /4
(c) 1.692
sec2 (x) dx = 0.
0
15. Answer true or false.
Z π /3
−π /3
(d) 1.698
tan (2x) dx = 0.
6.
Z 1
0
x3 dx
=
(9 + x2 )5/2
(a) 0.00086
Section 7.4
1.
Z p
(c) 0.00041
9 − x2 dx =
√
x
x 9 − x2
+ sin−1
+C
6
9
√
x 9 − x2 9 −1 x (b)
+ sin
+C
2
2
3
√
x
x 9 − x2
+ 2 sin−1
+C
(c)
2
3
√
x
x 9 − x2
(d)
+ 3 sin−1
+C
2
3
Z
dx
√
=
2.
5 − x2
√ sin−1 5 x
(a)
+C
5
−1
sin (5x)
(b)
+C
5
√ !
5x
+C
(c) sin−1
5
√
√ !
5 −1
5x
(d)
sin
+C
5
5
(a)
3.
(b) 0.00031
Z 0
−1
3ex
p
4 − 2e2x dx =
(d) 0.00082
7.
Z 4√ 2
3x − 2 dx
x
(a) 4.72
=
1
(b) 4.79
(c) 3.12
(d) 3.17
8.
Z 2π
π
cos θ d θ
p
=
4 − sin2 θ
(a) 1
(b) 0
(c) −1
(d) 4
9.
Z
dx
=
x2 + 3x + 9
√ !
√
√
2 3
−1 2 3 x + 3 3
tan
+C
(a)
9
9
2x + 3
2
+C
(b) tan−1
3
3
2
(c) tan−1 (2x + 3) +C
3
158
(d)
2
tan−1
11
2x + 3
11
10. Z
Answer true or false.
Z
dx
dx
√
= p
2
x − 4x + 2
(x − 2)2 − 2
11. Z
Answer true or false.
Z
Z
Z
dx
dx
dx
dx
=
−7
+
2
2
x
3
x − 7x + 3
x
Z 1
dx
√
=
12.
0 2 5x − x2
(a) 0.46
(b) 0.47
(c) 1
(d) 0
13.
Z 2
1
3
2
+
x−1 x
3. Which of the following is the partial fraction decomposition
3x2 − 2x + 2
of 2
?
(x + 2)(x − 1)
(d)
+C
dx
=
3x2 + 9x + 4
1
2
+
x2 + 2 x − 1
2
1
+
(b) 2
x +2 x−1
2x
1
(c) 2
+
x +2 x−1
x
2
(d) 2
+
x +2 x−1
Z
4x + 10
dx =
4.
x2 + 5x − 6
(a)
(a) 2 ln |x2 + 5x − 5| +C
(b) 2 ln |x2 + 5x − 6| +C
(a) 0.021
(c) 2 ln |x + 6| + ln |x − 1| +C
(b) 0.034
(d) 2 ln |x + 2| + ln |x + 3| +C
(c) 0.043
(d) 0.053
14. Answer true or false.
5.
Z 2
0
15. Answer true or false.
Z π
0
√
4x 16x − 1 dx = 90.9.
cos x sin x
1. Which of the following is the the partial fraction decomposi5x + 10
tion of
?
(x − 2)(x + 3)
(b) 2 ln |x2 + 1| + ln |x + 2| +C
(c) 2 tan−1 x + ln |x + 2| +C
(d) tan−1 x + ln |x + 2| +C
6. Answer true or false.
Z
ln |x + 2| + ln |x + 4| +C
7. Answer true or false.
ln |x − 2| +C
(a)
2. Which of the following is the partial fraction decomposition
5x − 2
?
of 2
x −x
1
2
(a)
+
x−1 x
1
2
+
(b)
x−1 x
3
2
(c)
+
x−1 x
x2 + 2x + 5
dx =
x3 + 2x2 + x + 2
(a) ln |4x2 + 1| + ln |2x + 4| +C
p
1 − sin2 x dx = 0.
Section 7.5
4
1
+
x+3 x−2
2
3
(b)
+
x+3 x−2
4
1
(c)
+
x+3 x−2
3
2
(d)
+
x+3 x−2
Z
8. Answer true or false.
ln |x2 + 2| +C
9.
Z
Z
x2 + 2x + 1
x3
dx =
+ x2 + x +
(x + 2)(x + 4)
3
Z
dx
= ln |x + 3| +
(x + 3)(x − 2)
2x3 + 4x2 + 2x + 2
dx = tan−1 x +
(x2 + 4)(x2 + 2)
x2 + 4
dx =
(x − 1)3
(a) ln |x − 1| −
5
2
+C
−
x − 1 2(x − 1)2
(b) ln |x − 1| +C
x3
+ 2x + ln3 |x − 1| +C
3
x3
1
(d)
+C
+ 2x −
3
2(x − 1)2
(c)
10. Z
Answer true or false.
x3 + x + 3
dx = ln |x| + ln |x + 5| +C
x(x + 5)
159
11. Answer true or false.
Z
12. Z
Answer true or false.
2x + 1
dx = ln |x2 + 2| + ln |x + 2| +C
(x2 + 2)(x + 2)
13. Z
Answer true or false.
x2 − 3x − 17
3
dx = ln |x + 7| − tan−1 (x/2) +C
2
(x + 7)(x2 + 4)
14. Z
Answer true or false.
dx
1
= tan−1 x + tan−1 (x/2) +C
2
(x2 + 1)(x2 + 4)
15. Answer true or false.
Z
x6 1
+ +C
6 x
1
ln (3x)
+C
−
(b) x6
6
36
dx
= ln3 |x − 6| +C
(x − 6)3
(a)
x6 ln (3x)
1
−
+C
6
36
x6 1
(d)
− +C
6
x
Z
√
5.
4 x ln x dx =
(c)
16
8x3/2
ln x − x3/2 +C
3
9
8 3/2
3/2
(b) 4x ln x − x +C
3
16
8x3/2
ln x −
+C
(c)
3
9
8x3/2
(d)
ln x +C
3
x dx
= ln |x + 3| +C
(x + 3)2
(a)
Section 7.6
1.
Z
6x
dx =
4x + 3
(a)
(b)
(c)
(d)
2.
Z
3 3
+ ln |4x + 3| +C
2 8
3 3
− ln |4x + 3| +C
2 8
x2
6 ln |4x + 3| + +C
2
3x 9
− ln |4x + 3| +C
2
8
(c)
(d)
Z
1
4
−
ln |4 − 5x| +C
25(4 − 5x) 25
2
ln |4 − 5x| +C
25
sin (3x) sin (7x) dx =
1
1
sin (4x) −
sin (10x) +C
8
20
1
1
(b) sin (7x) −
sin (3x) +C
8
20
1
1
(c) cos (7x) − sin (3x) +C
7
3
1
1
(d) cos (7x) + sin (3x) +C
7
3
(a)
4.
Z
Z
7.
Z
16x
dx =
(4 − 5x)2
2
(a) ln |4 − 5x| +C
5
64 + 16(4 − 5x) ln |4 − 5x|
+C
(b)
25(4 − 5x)
3.
6.
x5 ln (3x) dx =
e4x cos (2x) dx =
sin (2x)
4x cos (2x)
+C
+
(a) e
3
2
e4x
(b)
cos (2x) − sin (2x) +C
20
e4x
2 cos (2x) + sin (2x) +C
(c)
10
e4x
3 cos (2x) + 2 sin (2x) +C
(d)
6
e−3x sin (4x) dx =
e−3x
− 3 sin (4x) − 4 sin (4x) +C
5
e−3x
− 3 cos (4x) + 4 sin (4x) +C
(b)
5
e−3x
3 cos (4x) + 4 sin (4x) +C
(c)
25
−e−3x
3 sin (4x) + 4 cos (4x) +C
(d)
25
Z
dx
√
8.
=
2
x 2x2 + 6
−6x
(a) √
+C
2x2 + 6
6x
(b) √
+C
2x2 + 6
√
2x2 + 6
+C
(c)
6x
√
− 2x2 + 6
+C
(d)
6x
(a)
160
9.
Z
3. Use n = 4 subintervals with the midpoint rule to approximate
ln (4x + 2) dx =
the integral
2
(a) 2 ln (4x + 2) +C
(a) 2.5257
(4x + 2) ln (4x + 2)
− x +C
4
(c) 4x ln (4x + 2) − 4x +C
(b) 2.5430
(c) 2.5621
ln2 (4x + 2)
(d)
+C
2
(d) 2.5745
10. Answer true or false. For
11. Answer true or false.
12. Answer true or false.
13.
√
Z
Z
Z
Z
x ln (6 − 2x2 ) dx a good choice
√ √ cos x dx = 2 sin x +C.
√
e
x
√
dx = e
x
√
( x + 1) +C
Z 1
0
|x − 1/3| dx.
(a) 0.2733
(b) 0.2708
(d) 0.2917
5. Use 2n = 6 subintervals with Simpson’s rule to approximate
the given integral:
Z 5
1
dx.
1 x
(a) 1.6131
2
(3x − 14)(x + 7)3/2 +C
15
(b) (x − 7)3/2 +C
2
(c) (x − 7)3/2 + x +C
3
5
2
(d) (x + 7)5/2 − (x − 7)3/2 +C
3
2
14. Answer true or false. The area enclosed by y =
π
y = 0, x = 0 and x = 4 is 128
3 .
15. Answer true or false.
4. Use n = 5 subintervals to approximate the following integral
using the trapezoidal rule:
(c) 0.2867
x x + 7 dx =
(a)
ln (x + 1) dx.
0
(b)
for u is 6 − 2x2 .
Z 3
Z x
3
(b) 1.6094
(c) 1.7351
√
16 − x2 ,
1
dt
√
= + x.
3
t 3t − 5
(d) 1.6436
6. Use Simpson’s rule with 2n = 8 subintervals to approximate
the given integral:
Z 4
1
|3x − 4| dx.
(a) 10.875
(b) 10.833
Section 7.7
(c) 10.828
(d) 10.750
1. Use n = 10 to approximate the integral by the midpoint rule:
Z 1
2x3/2 dx
0
(a) 0.804
(b) 0.799
(c) 4.98
(d) 8.04
2. Use n = 10 to approximate the integral by the midpoint rule:
Z 1
(x3 + 1) dx
0
(a) 3.49
7. Use the midpoint rule with n = 5 subintervals to approximate
the given integral:
Z 1
(x3 + 4) dx.
0
(a) 4.24500
(b) 4.24875
(c) 4.24219
(d) 4.26000
8. Use n = 6 subintervals to approximate the integral by the
Z 3
dx
trapezoidal rule:
3
1 x +1
(a) 0.31546
(b) 1.249
(b) 0.32546
(c) 1.257
(c) 0.31398
(d) 3.51
(d) 0.31888
161
9. Use the midpoint rule with n = 5 subintervals to approximate
Z 1
dx
the integral:
.
−1 1 + x2
(a) 1.57746
(b) 1.55747
(c) 1.58118
Section 7.8
1. Answer true or false.
Z 8
dx
is an improper integral.
x−4
Z 4
dx
is an improper integral.
2. Answer true or false.
0 x−1
0
3. Answer true or false.
(d) 1.57542
4.
Z ∞
dx
x4
1
10. Use Simpson’s rule with 2n = 6 subintervals to approximate
Z 2p
x3 + 1 dx.
the integral:
Z 2
−∞
e5x dx is an improper integral.
=
(a) 1/3
0
(b) 1/6
(a) 3.23200
(c) 1/2
(b) 3.24594
(d) Diverges.
(c) 3.25988
5.
Z ∞
dx
√ =
x
6
(d) 3.24109
(a) 1/2
11. Use the trapezoidal rule with n = 6 subintervals to approximate the integral:
Z 5
(b) 1/6
x sin (1/x) dx.
(c) 2
2
(d) Diverges.
(a) 2.95032
(b) 2.95071
6.
(c) 2.94954
−∞
e6x dx =
(a) −1/6
(d) 2.94730
(b) 1/6
12. Use Simpson’s Rule with 2n = 6 to approximate the integral:
Z 4
Z 0
(c) 6
(d) Diverges.
x sin (1/x) dx.
1
7.
(a) 2.877695
Z 0
1
√
dx
1 − x2
(b) 2.877027
(a) 3/2
(c) 2.871765
(b) −3/2
(c) 0
(d) 2.880575
(d) −π /2
13. Answer true or false. If the trapezoidal rule is used with
n = 38 subintervals to approximate the integral
then |ET
Z 2
0
| ≤ 1 × 10−3 .
14. Answer true or false.
=
2
e−x dx,
8.
Z π /2
cot x dx =
0
(a) 0
If the midpoint rule is used with
n = 19 subintervals to approximate the integral
Z 2
0
then |EM | ≤ 1 × 10−5 .
2
e−x dx,
(b) 1
(c) −30.08
(d) Diverges.
15. Answer true or false. If Simpson’s rule is used with 2n = 8
subintervals to approximate the integral
|ES | ≤ 1 × 10−3 .
Z 2
0
2
e−x dx, then
9. Answer true or false.
Z 3
0
10. Answer true or false.
Z ∞
1
x−2 dx diverges.
x−2 dx diverges.
162
11. Answer true or false.
Z 1
ln (2x) dx diverges.
(d)
0
12. Answer true or false.
Z ∞
cos x dx diverges.
6. Answer true or false.
0
13. Answer true or false.
Z ∞
0
e−6x dx diverges.
7.
14. Answer true or false.
Z 0
e−5x dx diverges.
15. Answer true or false.
Z 4
x−3 dx = 1 − lim
−∞
−∞
Z
b→−∞
−2
= 1.
b2
sinh12 x cosh x dx =
(a)
Z
1
sinh13 x +C
13
choice for u is 4ex + 5.
4.
Z
x
1 p
x 9 − x2 + 9 sin−1
+C
2
9
x
9
1 p
+C
(b) x 9 − x2 + sin−1
2
2
3
p
1
x
+C
(c) x 9 − x2 − 9 sin−1
2
9
p
9
1
x
(d) x 9 − x2 − sin−1
+C
2
2
3
Z 1
2 dx
√
=
10.
0
6x − x2
(a) 1.68
(b) 1.66
(c) 1.72
(d) 1.84
Z
3x cos x dx =
x
x
e (4e + 5)dx, a good
2
1
+
is the partial fraction dex+4 x−8
3x − 12
.
composition of
(x + 4)(x − 8)
11. Answer true or false.
12.
Z
(c) 9 tan−1 x + 2 ln |x + 2| +C
(d) 3 cos x + 3x cos x +C
(a) e3x 3 sin (6x) + 6 cos (6x) +C
(b) e3x 3 sin (6x) − 6 cos (6x) +C
(c) −e3x sin (6x) − 2 cos (6x) +C
dx =
(b) 9 ln |x2 + 1| + 2 ln |x + 2| +C
(c) 3 cos x − 3x sin x +C
e3x sin (6x) dx =
2x2 + 9x + 20
x3 + 2x2 + x + 2
(a) ln |9x2 + 1| + ln |2x + 4| +C
(b) 3 sin x + 3x sin x +C
5.
tan (6x) dx =
1
ln |cos (6x)| +C
6
1
(b) − ln |cos (6x)| +C
6
1
(c)
tan2 (6x) +C
12
1
(d) − tan2 (6x) +C
12
8. Z
Answer true or false.
(a) 3 cos x + 3x sin x +C
Z
x cot x dx = 0.
(a)
2x dx
√
=
9 − x4
x
+C
(a) sin−1
3
2
x
(b) sin−1
+C
3
x
+C
(c) cos−1
3
2
x
(d) cos−1
+C
3
3. Answer true or false. In evaluating
5π /4
1
cos (14x)
sin (8x) cos (6x) dx = − cos (2x) −
+C
4
28
Z p
9 − x2 dx =
9.
(b) 13 sinh13 x +C
1
sinh11 x +C
(c)
11
(d) 11 sinh11 x +C
2.
Z
Z 7π /4
(a)
Chapter 7 Test
1.
e3x
sin (6x) − 2 cos (6x) +C
15
13.
Z
(d) 3 tan−1 x + 2 ln |x + 2| +C
x8 ln2 (x) dx =
x9
81 ln2 (x) − 18 ln (x) + 2 +C
729
1
x9 ln2 (x)
−
+C
(b)
9
81
(a)
163
(c)
x9 ln2 (x)
+C
9
17. Use 2n = 10 subdivisions to approximate the integral by
Simpson’s Rule:
x9 ln2 (x) 1
(d)
− +C
9
9
midpoint rule:
(cos x + 1) dx
(a) 4.1667
(b) 4.1892
(c) 4.1995
(d) 4.2001
18.
Z ∞
0
0
e−5x dx =
(a) 1.8424
(a) 1/5
(b) 1.8422
(b) −1/5
(c) 5
(c) 1.8420
(d) Diverges.
(d) 1.8418
16. Use n = 10 subdivisions to approximate the integral by the
trapezoid rule:
Z 2
1
7
(x − 1) dx
19.
Z 3
0
√
dx
9 − x2
(b) −π /2
(b) 31.32
(c) 0
(d) 31.20
=
(a) π /2
(a) 31.24
(c) 31.29
(x5 + 4) dx.
0
14. Z
Answer true or false.
x2 x sin (4x) cos (4x)
x sin (4x) dx =
−
−
.
4
4
16
15. Use n = 10 subdivisions to approximate the integral by the
Z 1
Z 1
(d) Diverges.
20. Answer true or false.
Z ∞
4
e2x dx diverges.
164
Chapter 7: Answers to Sample Tests
Section 7.1
1. c
9. b
2. c
10. a
3. d
11. true
4. a
12. false
5. b
13. true
6. b
14. false
7. d
15. true
8. c
2. a
10. a
3. c
11. b
4. a
12. true
5. c
13. false
6. b
14. false
7. a
15. false
8. d
2. a
10. true
3. b
11. c
4. a
12. true
5. a
13. false
6. b
14. false
7. true
15. true
8. a
2. c
10. true
3. c
11. false
4. b
12. a
5. d
13. c
6. a
14. true
7. a
15. true
8. b
2. c
10. false
3. c
11. false
4. b
12. false
5. c
13. true
6. false
14. false
7. false
15. false
8. false
2. b
10. true
3. a
11. false
4. b
12. false
5. a
13. a
6. c
14. false
7. d
15. false
8. d
2. b
10. d
3. c
11. c
4. c
12. b
5. a
13. true
6. d
14. false
7. a
15. true
8. b
2. true
10. false
3. true
11. false
4. a
12. true
5. d
13. false
6. b
14. true
7. d
15. false
8. d
2. b
10. a
18. a
3. true
11. true
19. a
4. a
12. c
20. true
5. d
13. a
6. false
14. false
7. b
15. d
8. true
16. c
Section 7.2
1. b
9. b
Section 7.3
1. b
9. a
Section 7.4
1. b
9. a
Section 7.5
1. a
9. a
Section 7.6
1. d
9. b
Section 7.7
1. b
9. a
Section 7.8
1. true
9. true
Chapter 7 Test
1. a
9. b
17. a