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Reform Calculus: Part II
Marcel B. FinanArkansas Tech University
c©All Rights Reserved
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PREFACE
This supplement consists of my lectures of a sophomore-level
mathematicsclass offered at Arkansas Tech University. The lectures
are designed to ac-company the textbook ”Calculus: Single and
Multivariable” written by theHarvard Consortium.This book has been
written in a way that can be read by students. That is,the text
represents a serious effort to produce exposition that is
accessible toa student at freshmen or sophomore levels.This
supplement is a continuation of the previous calculus book. The
lec-tures cover Chapters 7, 8, 9, 10, and 11. These chapters are
well suited for a4-hour one semester course of a second course
Calculus.
Marcel B. FinanMay 2003
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Contents
40 Integration by Substitution 5
41 The Method of Integration by Parts 10
42 Tables of Integrals 16
43 Integration by Partial Fractions. Trigonometric
Substitu-tions 20
44 Numerical Approximation of Definite Integrals 29
45 Simpson’s Rule and Error Estimates 38
46 Improper Integrals 45
47 Comparison Tests for Improper Integrals 54
48 Areas and Volumes 60
49 Solids of Revolution- Arc Length 69
50 Density and Center of Mass 81
51 Applications to Physics 92
52 Applications to Economics 99
53 Continuous Random Variables: Distribution Function andDensity
Function 108
54 The Median and Mean 117
55 Geometric Series 124
56 Convergence of Sequences and Series 130
57 Tests for Convergence 140
58 Power Series 150
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59 Approximations by Taylor’s Polynomials 156
60 The Error in Taylor Polynomial Approximations:
Taylor’sTheorem 162
61 Taylor Series 166
62 Constructing New Taylor Series from Known Ones 176
63 Introduction to Fourier Series 182
64 The Definition of a Differential Equation 189
65 Slope Fields 193
66 Euler’s Method 200
67 Seperation of Variables 204
68 Differential Equations: Growth and Decay Models 208
69 First Order Differential Equations Models 215
70 The Logistic Model 222
71 Second-Order Differential Equations Models 227
72 Second-Order Homogeneous Linear Differential Equations
233
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40 Integration by Substitution
The purpose of this section is to evaluate the integral∫f
′(g(x))g′(x)dx. (1)
This is done by letting u = g(x). Then du = g′(x)dx. (See
Section 23). Hence,we have the following∫
f ′(g(x))g′(x)dx =∫
f ′(u)du = f(u) + C = f(g(x)) + C (2)
The above procedure is referred to as the method of integration
by sub-stitution. Thus, when the integrand looks like a compound
function timesthe derivative of the inside then try using
substitution to integrate. Notealso that this method of
antidifferentiation reverses the chain rule of differ-entiation.The
following examples illustrate the use of this method.
Example 40.1Find
∫3x2 cos x3dx.
Solution.Let u(x) = x3. Then du = 3x2dx and therefore∫
3x2 cos x3dx =∫
cos udu = sin u + C = sin x3 + C.
The method of substitution works even when the derivative of the
inside ismissing a constant factor as shown in the next couple of
examples.
Example 40.2Find
∫xex
2+1dx.
Solution.Letting u(x) = x2 + 1 then du = 2xdx. Thus,∫
xex2+1dx = 1
2
∫2xex
2+1dx= 1
2
∫eudu = e
u
2+ C
= 12ex
2+1 + C.
You may wonder why 12
∫eudu = 1
2eu + C and not 1
2
∫eudu = 1
2(eu + C) =
eu
2+ C
2. The convention is always to add C to whatever antiderivative
we
have calculated.
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Example 40.3Find
∫x3√
x4 + 5dx.
Solution.Let u = x4 + 5. Then du = 4x3dx. Thus,∫
x3√
x4 + 5dx = 14
∫4x3
√x4 + 5dx
= 14
∫ √udu = 1
4· u
3232
+ C
= 16(x4 + 5)
32 + C
In some situations, the integrand consists of a fraction having
a function inthe denominator and its derivative in the numerator.
This leads to a naturallogarithm as seen in the next two
examples.
Example 40.4Find
∫ exex+1
dx.
Solution.Let u = ex + 1. Then du = exdx. Thus,∫ ex
ex+1dx =
∫ duu
= ln |u|+ C= ln |ex + 1|+ C.
Example 40.5Find
∫tan xdx.
Solution.Since tan x = sin x
cos xthen letting u = cos x we find that du = − sin xdx and
therefore ∫tan xdx = −
∫ duu
= − ln |u|+ C = − ln | cos x|+ C.
Next, we discuss the evaluation of a definite integral using the
techniqueof substitution. From ( 2) we have that f(g(x)) is an
antiderivative of thefunction f ′(g(x))g′(x). Applying the
Fundamental Theorem of Calculus wecan write ∫ b
af ′(g(x))g′(x)dx = f(g(x))|ba = f(g(b))− f(g(a)).
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If we let u = g(x) then the previous formula reduces to
∫ ba
f ′(g(x))g′(x)dx = f(g(b))− f(g(a)) =∫ g(b)
g(a)f ′(u)du.
Warning: When evaluating definite integrals, there is no
constant of inte-gration in the final answer.
Example 40.6Compute
∫ 20 xe
x2dx.
Solution.Let u(x) = x2. Then du = 2xdx, u(0) = 0, and u(2) = 4.
Thus,
∫ 20
xex2
dx =1
2
∫ 40
eudu =eu
2
∣∣∣∣40
=1
2(e4 − 1).
Example 40.7
Compute∫ π
40
tan3 xcos2 x
dx.
Solution.Let u = tan x. Then du = dx
cos2 x, u(0) = 0, and u(π
4) = 1. Thus,
∫ π4
0
tan3 x
cos2 xdx =
∫ 10
u3du =u4
4
∣∣∣∣∣1
0
=1
4.
Example 40.8
Find∫ √
1 +√
xdx.
Solution.Let u = 1 +
√x. Then du = dx
2√
x= dx
2(u−1) or dx = 2(u− 1)du. Thus,
∫ √1 +
√xdx =
∫2√
u(u− 1)du =∫(2u
√u− 2
√u)du
=∫(2u
32 − 2u 12 )du
= 2u5252
− 2u3232
+ C
= 45(1 +
√x)
52 − 4
3(1 +
√x)
32 + C.
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Practice Problems
Exercise 40.1Find
∫x2(1 + 2x3)2dx.
Exercise 40.2Find
∫x(x2 − 4) 72 dx.
Exercise 40.3Find
∫ 1√4−xdx.
Exercise 40.4Find
∫sin x(cos x + 5)7dx.
Exercise 40.5Find
∫x2ex
3+1dx.
Exercise 40.6Find
∫ (ln x)2x
dx.
Exercise 40.7Find
∫ ex+1ex+x
dx.
Exercise 40.8Find
∫ 1+ex√x+ex
dx.
Exercise 40.9Find
∫ x+1x2+2x+19
dx.
Exercise 40.10Find
∫ x cos (x2)√sin (x2)
dx.
Exercise 40.11Compute
∫ π2
0 e− cos xdx.
Exercise 40.12Compute
∫ 81
e3√x
3√x2
dx.
Exercise 40.13Compute
∫ 41
cos√
x√x
dx.
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Exercise 40.14Compute
∫ 1−1
11+x2
dx.
Exercise 40.15Compute
∫ 31
dx(x+7)2
.
Exercise 40.16Compute
∫ 21
sin xx
dx.
Exercise 40.17Find the exact area under the graph of f(x) =
xex
2between x = 0 and x = 2.
Exercise 40.18Find the exact area enclosed by the graph of y =
4
x, the x-axis and the lines
x = 2 and x = 4.
Exercise 40.19Suppose
∫ 10 f(x)dx = 3. Calculate the following:
(a)∫ 0.50 f(2x)dx (b)
∫ 10 f(x− 1)dx (c)
∫ 1.51 f(3− 2x)dx.
Exercise 40.20If t is in years since 1990, the population P, of
the world in billions can bemodeled by P (t) = 5.3e0.014t.(a) What
does this model give the world population in 1990? in 2000?(b) Use
the Fundamental Theorem of Calculus to find the average
populationof the world during the 1990s.
Exercise 40.21Decide whether the following statements are true
or false. Give an explana-tion for your answer.(a)
∫f ′(x) cos (f(x))dx = sin(f(x)) + C.
(b)∫ 1
f(x)dx = ln |f(x)|+ C.
(c)∫
x sin (5− x2)dx can be evaluated using substitution.(d)
∫sin7 θ cos6 θdθ can be written as a polynomial with cos θ as
the variable.
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41 The Method of Integration by Parts
The integration by parts formula is an antidifferentiation
method which re-verses the product rule of differentiation. To see
this, let u and v be twodifferentiable functions. Then the product
rule asserts that the product func-tion uv is also differentiable
and its derivative is given by
d
dx(uv) = u
dv
dx+
du
dxv.
This says that an antiderivative of u dvdx
+ v dudx
is the function uv. In terms ofindefinite integrals we have∫
(udv
dx+
du
dxv)dx = uv + C. (3)
But the differential of u is du = dudx
dx and that of v is dv = dvdx
dx. Hence, (3)becomes ∫
udv +∫
vdu = uv
or ∫udv = uv −
∫vdu (4)
Formula (4) is known as the integration by parts formula.
Example 41.1Find
∫xexdx.
Solution.Let u = x and dv = exdx. Then du = dx and v =
∫exdx = ex. Note that in
finding v we did not include the constant of integration. We
will write theconstant C in the answer of
∫udv. Now, substituting in formula (4) to obtain∫
xexdx = xex −∫
exdx= xex − ex + C.
Remark 41.1If we chose u = ex and dv = xdx then we would have du
= exdx and v = x
2
2.
In this case, formula (4) yields∫xexdx =
x2
2ex −
∫ x22
exdx
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and the second integral is definitely worse than the one to the
left of theformula we had in the previous example, i.e.
∫xexdx. It follows that for the
method to be useful it is important to choose u and dv in such a
way to makethe integral on the right easier to find then the
integral on the left.
Example 41.2Find
∫x sin xdx.
Solution.Let u = x and dv = sin x. Then du = dx and v = − cos x.
Substituting informula (4) to obtain∫
x sin xdx = −x cos x−∫
(− cos x)dx= −x cos x +
∫cos xdx
= −x cos x + sin x + C.
There are examples which don’t look like good candidates for
integrationby parts because they don’t appear to involve products,
but for which themethod works well. Such examples often involves ln
x or the inverse trigono-metric functions.
Example 41.3Calculate
∫ 51 ln xdx.
Solution.Let u = ln x and dv = dx. Then du = dx
xand v = x. Substituting in formula
(4) we obtain ∫ 51 ln xdx = x ln x|51 −
∫ 51 x
1xdx
= x ln x|51 −∫ 51 dx
= x ln x|51 − x|51= 5 ln 5− ln 1− (5− 1)= 5 ln 5− 4.
Example 41.4Find
∫x3 ln xdx.
Solution.Let u = ln x and dv = x3dx. Then du = dx
xand v = x
4
4. Substituting in
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formula (4) to obtain∫x3 ln xdx = x
4
4ln x−
∫ x44
1xdx
= x4
4ln x− 1
4
∫x3dx
= x4
4ln x− 1
4x4
4+ C
= x4
4ln x− 1
16x4 + C.
Sometimes evaluating an integral might require integration by
parts morethan once as the following example shows.
Example 41.5Find
∫x2 cos xdx.
Solution.Let u = x2 and dv = cos xdx. Then du = 2xdx and v = sin
x. Substitutingin formula (4) to obtain∫
x2 cos xdx = x2 sin x−∫
2x sin xdx= x2 sin x− 2
∫x sin xdx
= x2 sin x− 2(−x cos x + sin x) + C= x2 sin x + 2x cos x− 2 sin
x + C
where we have used Example 41.2 to evaluate∫
x sin xdx.
The following example illustrates a very useful technique: Use
integration byparts to transform the integral into an expression
containing another copyof the same integral, possibly multiplied by
a coefficient, then solve for theoriginal integral.
Example 41.6Find
∫sin2 xdx.
Solution.Method I:Using a half-angle formula to write sin2 x =
1−cos 2x
2. In this case the problem
reduces to integrating 1−cos 2x2
. That is,∫sin2 xdx = 1
2
∫(1− cos 2x)dx
= 12
∫dx− 1
2
∫cos 2xdx
= 12x− 1
4
∫cos udu
= 12x− 1
4sin u + C
= 12x− 1
4sin 2x + C
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where the substitution u = 2x is used to evaluate the
integral∫
cos 2xdx.
Method II:We now use integration by parts formula to evaluate
the given integral. Letu = sin x and dv = sin xdx. Then du = cos x
and v = − cos x. Substitutingin formula (4) to obtain∫
sin2 xdx = − sin x cos x−∫− cos2 xdx
= − sin x cos x +∫
cos2 xdx.
Using the trigonometric identities sin 2x = 2 sin x cos x and
cos2 x+sin2 x = 1we can rewrite the right-hand side of the above
integral as∫
sin2 xdx = −12sin 2x +
∫(1− sin2 x)dx
= −12sin 2x +
∫dx−
∫sin2 xdx
= −12sin 2x + x−
∫sin2 xdx
Moving the right integral to the left side to obtain
2∫
sin2 xdx = −12
sin 2x + x
and finally dividing both sides by 2 to obtain∫sin2 xdx = −1
4sin 2x +
x
2+ C.
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Practice Problems
Exercise 41.1Find
∫ 10 arctan xdx.
Exercise 41.2Find
∫x2e5xdx.
Exercise 41.3Find
∫x3 ln xdx.
Exercise 41.4Find
∫x2 sin xdx.
Exercise 41.5Find
∫cos2 xdx.
Exercise 41.6Find
∫ ln xx2
dx.
Exercise 41.7Find
∫x5 ln 5xdx.
Exercise 41.8Find
∫ x√5−xdx.
Exercise 41.9Find
∫(ln x)2dx.
Exercise 41.10Find
∫x arctan x2dx.
Exercise 41.11Find
∫x5 cos x3dx.
Exercise 41.12Evaluate
∫ 100 xe
−xdx.
Exercise 41.13Evaluate
∫ 50 ln (1 + x)dx.
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Exercise 41.14Evaluate
∫ 10 arcsin xdx.
Exercise 41.15Find
∫ex sin xdx.
Exercise 41.16Compute
∫cos2 θdθ in two different ways.
Exercise 41.17Find the exact value of the area under the first
arch of f(x) = x sin x.
Exercise 41.18Derive the following formula∫
cosn xdx =1
ncosn−1 x sin x +
n− 1n
∫cosn−2 xdx.
Exercise 41.19Let f be twice differentiable function such that
f(0) = 6, f(1) = 5, andf ′(1) = 2. Evaluate the integral
∫ 10 xf
′′(x)dx.
Exercise 41.20Suppose that F (a) represents the area under the
graph of y = x2e−x betweenx = 0 and x = a > 0.(a) Find a formula
for F (a).(b) Is F increasing or decreasing function?(c) Is F
concave up or down for 0 < a < 2?
Exercise 41.21In describing the behavior of an electron, we use
wave functions Ψ1, Ψ2, Ψ3, · · ·of the form
Ψn(x) = Cn sin (nπx), n = 1, 2, 3, · · ·where x is the distance
from a fixed point and Cn is a positive constant.(a) Find C1 so
that Ψ1 satisfies∫ 1
0(Ψ1(x))
2dx = 1.
(b) For any integer n, find Cn so that∫ 10
(Ψn(x))2dx = 1.
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42 Tables of Integrals
If the standard integration techniques presented previously fail
to yield anantiderivative, the last measure of despair are integral
tables. These tablesbasically consist of collections of functions
together with their antiderivatives.In order to use them, you may
have to re-write the integrand function firstin a standard form
listed in the table. A short table of common formulas isgiven in
the back cover of your textbook. The first example shows how
toapply an integral with no manipulations of any kind.
Example 42.1Find
∫cos (2x) cos (7x)dx.
Solution.Using formula II-11 with a = 2 and b = 7 we have∫
cos (2x) cos (7x)dx =1
45(7 cos (2x) sin (7x)− 2 sin (2x) cos (7x)) + C.
The following two examples use reduction formulas.
Example 42.2rm Find
∫(x2 − 3x + 2)e−4xdx.
Solution.Using formula III-14 with p(x) = x2 − 3x + 2 and a = −4
to obtain∫
(x2−3x+2)e−4xdx = −14(x2−3x+2)e−4x− 1
16(2x−3)e−4x− 1
32e−4x +C.
Example 42.3Find
∫sin4 xdx.
Solution.Using the reduction formula IV-17 we find∫
sin4 xdx = −14sin3 x cos x + 3
4
∫sin2 xdx
= −14sin3 x cos x− 3
8sin x cos x + 3
8x + C.
Remember that some integrals can be evaluated without the use of
an integraltable as shown in the next example.
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Example 42.4Find
∫cos3 x sin4 xdx.
Solution.This integral can be evaluated without the use of a
table of integral. Indeed,let u = sin x then du = cos xdx and∫
cos3 x sin4 xdx =∫(1− sin2 x) cos x sin4 xdx
=∫(1− u2)u4du =
∫(u4 − u6)du
= u5
5− u7
7+ C
= sin5 x5− sin7 x
7+ C.
Example 42.5Find
∫sin2 x cos2 xdx.
Solution.Using the trigonometric identity cos2 x + sin2 x = 1
and Formula IV-17 wehave∫
sin2 x cos2 xdx =∫
sin2 x(1− sin2 x)dx=
∫sin2 xdx−
∫sin4 xdx
= (−12sin x cos x + 1
2x)− (−1
4sin3 x cos x + 3
4
∫sin2 xdx)
= −12sin x cos x + 1
2x + 1
4sin3 x cos x− 3
4(−1
2sin x cos x + 1
2x) + C
= −12sin x cos x + 1
2x + 1
4sin3 x cos x + 3
8sin x cos x− 3
8x + C
= −18sin x cos x + 1
4sin3 x cos x + 1
8x + C.
The integrands in the following require some manipulations to
fit the entriesin the table.
Example 42.6 (Factoring)Find
∫ 1x2+4x+3
dx.
Solution.Factoring the denominator and then using Formula V-26
we obtain∫ 1
x2 + 4x + 3dx =
∫ 1(x + 1)(x + 3)
dx =1
2(ln |x + 1| − ln |x + 3|) + C.
Example 42.7 (Long division)Find
∫ x2+1x2−1dx.
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Solution.Using the long division of polynomials we find
x2 + 1
x2 − 1= 1 +
2
x2 − 1= 1 +
2
(x− 1)(x + 1).
This with Formula V-26 yield
∫ x2 + 1x2 − 1
dx = x + 2[1
2(ln |x− 1| − ln |x + 1|) + C = x + ln
∣∣∣∣x− 1x + 1∣∣∣∣+ C.
Example 42.8 (Completing the square)Find
∫ 1x2+4x+5
dx.
Solution.Completing the square we find x2 + 4x + 5 = (x + 2)2 +
1. Now, lettingu = x + 2 we can write∫ 1
x2 + 4x + 5dx =
∫ 1u2 + 1
du.
Finally, using Formula V-24 we find∫ 1x2 + 4x + 5
dx =∫ 1
u2 + 1du = arctan u + C = arctan (x + 2) + C.
Example 42.9 (Using substitution)Find
∫xe2x
2cos (2x2)dx.
Solution.Let u = 2x2. Then du = 4xdx so that∫
xe2x2
cos (2x2)dx =1
4
∫eu cos udu.
Now, applying Formula II-9 to the last integral we find∫xe2x
2
cos (2x2)dx =1
8e2x
2
(cos (2x2) + sin (2x2)) + C.
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Practice Problems.
Exercise 42.1Find
∫x5 ln xdx.
Exercise 42.2Find
∫sin x cos4 xdx.
Exercise 42.3Find
∫sin (3x) sin (5x)dx.
Exercise 42.4Find
∫sin3 (3x) cos3 (3x)dx.
Exercise 42.5Find
∫x4e3xdx.
Exercise 42.6Find
∫ x2x2+4
dx.
Exercise 42.7Find
∫sin3 xdx.
Exercise 42.8Find
∫ dx4−x2 .
Exercise 42.9Find
∫ 1x2+4x+4
dx.
Exercise 42.10Suppose n is a positive integer and Ψn = Cn sin
(nπx) is the wave functionused in describing the behavior of an
electron. If n and m are differentintegers, find ∫ 1
0Ψn(x) ·Ψm(x)dx.
Exercise 42.11Decide whether the following statements are true
or false. Give an explana-tion for your answer.(a)
∫ 1x2+4x+5
dx involves a natural logarithm.
(b)∫ 1
x2+4x−5dx involves an arctangent.
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43 Integration by Partial Fractions. Trigono-
metric Substitutions
Method of Integration by Partial FractionsThe method of
integration by partial fractions is a technique for
integratingrational functions, i.e. functions of the form
R(x) =P (x)
Q(x)
where P (x) and Q(x) are polynomials.The idea consists of
writing the rational function as a sum of simpler frac-tions called
partial fractions. This can be done in the following way:
Step 1. Use long division to find two polynomials r(x) and q(x)
such that
P (x)
Q(x)= q(x) +
r(x)
Q(x).
Note that if the degree of P (x) is smaller than that of Q(x)
then q(x) = 0and r(x) = P (x).
Step 2. Write Q(x) as a product of factors of the form (ax + b)n
or (ax2 +bx+c)n where ax2+bx+c is irreducible, i.e. ax2+bx+c = 0
has no real zeros.
Step 3. Decompose r(x)Q(x)
into a sum of partial fractions in the followingway:(1) For each
factor of the form (x− α)k write
A1x− α
+A2
(x− α)2+ · · ·+ Ak
(x− α)k,
where the numbers A1, A2, · · · , Ak are to be determined.(2)
For each factor of the form (ax2 + bx + c)k write
B1x + C1ax2 + bx + c
+B2x + C2
(ax2 + bx + c)2+ · · ·+ Bkx + Ck
(ax2 + bx + c)k,
where the numbers B1, B2, · · · , Bk and C1, C2, · · · , Ck are
to be determined.
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Step 4. Multiply both sides by Q(x) and simplify. This leads to
an ex-pression of the form
r(x) = a polynomial whose coefficients are combinations of Ai,
Bi, and Ci.
Finally, we find the constants, Ai, Bi, and Ci by equating the
coefficients oflike powers of x on both sides of the last
equation.
Example 43.1Decompose into partial fractions R(x) = x
3+x2+2x2−1 .
Solution.Step 1. x
3+x2+2x2−1 = x + 1 +
x+3x2−1 .
Step 2. x2 − 1 = (x− 1)(x + 1).Step 3. x+3
(x+1)(x−1) =A
x+1+ B
x−1 .
Step 4. Multiply both sides of the last equation by (x− 1)(x +
1) to obtain
x + 3 = A(x− 1) + B(x + 1).
Expand the right hand side, collect terms with the same power of
x, andindentify coefficients of the polynomials obtained on both
sides:
x + 3 = (A + B)x + (B − A).
Hence, A + B = 1 and B −A = 3. Adding these two equations gives
B = 2.Thus, A = −1 and so
x3 + x2 + 2
x2 − 1= x + 1− 1
x + 1+
2
x− 1.
Now, after decomposing the rational function into a sum of
partial fractionsall we need to do is to integrate expressions of
the form A
(x−α)n orBx+C
(ax2+bx+c)n.
Example 43.2Find
∫ 1x(x−3)dx.
Solution.We write
1
x(x− 3)=
A
x+
B
x− 3.
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Multiply both sides by x(x− 3) and simplify to obtain
1 = A(x− 3) + Bx
or1 = (A + B)x− 3A.
Now equating the coefficients of like powers of x to obtain −3A
= 1 andA + B = 0. Solving for A and B we find A = −1
3and B = 1
3. Thus,∫ 1
x(x−3)dx = −13
∫ dxx
+ 13
∫ dxx−3
= −13ln |x|+ 1
3ln |x− 3|+ C
= 13ln∣∣∣x−3
x
∣∣∣+ C.Example 43.3Find
∫ 3x+6x2+3x
dx.
Solution.We factor the denominator and split the integrand into
partial fractions:
3x + 6
x(x + 3)=
A
x+
B
x + 3.
Multiplying both sides by x(x + 3) to obtain
3x + 6 = A(x + 3) + Bx= (A + B)x + 3A
Equating the coefficients of like powers of x to obtain 3A = 6
and A+B = 3.Thus, A = 2 and B = 1. Finally,∫ 3x + 6
x2 + 3xdx = 2
∫ dxx
+∫ dx
x + 3= 2 ln |x|+ ln |x + 3|+ C.
Example 43.4Find
∫ x2+1x(x+1)2
dx.
Solution.We factor the denominator and split the integrand into
partial fractions:
x2 + 1
x(x + 1)2=
A
x+
B
x + 1+
C
(x + 1)2.
22
-
Multiplying both sides by x(x + 1)2 and simplifying to
obtain
x2 + 1 = A(x + 1)2 + Bx(x + 1) + Cx= (A + B)x2 + (2A + B + C)x +
A.
Equating coefficients of like powers of x we find A = 1, 2A + B
+ C = 0 andA + B = 1. Thus, B = 0 and C = −2. Now integrating to
obtain
∫ x2 + 1x(x + 1)2
dx =∫ dx
x− 2
∫ dx(x + 1)2
= ln |x|+ 2x + 1
+ C.
Example 43.5Find
∫ 2x2−x−1(x2+1)(x−2)dx.
Solution.We first write
2x2 − x− 1(x2 + 1)(x− 2)
=Ax + B
x2 + 1+
C
x− 2.
Multiply both sides by (x2 + 1)(x− 2) and simplify
2x2 − x− 1 = (Ax + B)(x− 2) + C(x2 + 1)= (A + C)x2 + (−2A + B)x−
2B + C.
Equating coefficients of like powers of x
A + C = 2, − 2A + B = −1, − 2B + C = −1.
Solving for A, B, and C we find A = B = C = 1. Thus∫
2x2−x−1(x2+1)(x−2)dx =
∫ x+1x2+1
dx +∫ dx
x−2=
∫ xx2+1
dx +∫ dx
x2+1+∫ dx
x−2= 1
2ln (x2 + 1) + arctan x + ln |x− 2|+ C.
Trigonometric Substitutions
This section deals with integrands involving terms like√
x2 − a2,√
x2 + a2,and
√a2 − x2.
• Integrands involving√
a2 − x2,−a ≤ x ≤ a.
23
-
For each x in the interval [−a, a] there is a θ in the interval
[−π2, π
2] such that
x = a sin θ. Thus, using the substitution x = a sin θ,−π2≤ θ ≤
π
2to obtain
√a2 − x2 =
√a2(1− sin2 θ)
=√
a2 cos2 θ = a cos θ
where we have used the Pythagorean identity cos2 θ + sin2 θ = 1.
Note thatcos θ ≥ 0 since −π
2≤ θ ≤ π
2. It is important to point out here that by con-
structing a right triangle with one of the angle being θ then
the hypotenuseof the triangle has length a, the opposite side has
length x and the adjacent
side has length√
a2 − x2. It follows that cos θ =√
a2−x2a
. See Figure 106.
Figure 106
Example 43.6Find
∫ 1√4−x2 dx.
Solution.Let x = 2 sin θ,−π
2≤ θ < π
2. Then
√4− x2 =
√4− 4 sin2 θ =
√4 cos2 θ = 2 cos θ.
Moreover, dx = 2 cos θdθ. It follows that∫ 1√4− x2
dx =∫ 2 cos θ
2 cos θdθ = θ + C = arcsin
(x
2
)+ C.
• Integrands involving√
a2 + x2
In this case, we let x = a tan θ with −π2
< θ < π2. Such a substitution leads
to √a2 + x2 =
√a2 + a2 tan2 θ =
√a2(1 + tan2 θ) = a sec θ
since 1 + tan2 θ = sec2 θ and sec θ > 0 for −π2
< θ < π2.
24
-
Remark 43.1Letting θ be the angle of a right triangle with
opposite side x, adjacent side
a, and hypotenuse√
a2 + x2 we find sec θ =√
a2+x2
a. See Figure 107.
Figure 107
Example 43.7Find
∫ 1√x2+9
dx.
Solution.Let x = 3 tan θ with −π
2θ < π
2. Then
√x2 + 9 =
√9 + 9 tan2 θ =
√9(1 + tan2 θ) = 3 sec θ.
Moreover, dx = 3 sec2 θdθ. Thus,
∫ 1√x2 + 9
dx =∫ 3 sec2 θ
3 sec θdθ =
∫sec θdθ = ln | sec θ + tan θ|+ C.
Now, considering a triangle with acute angle θ, opposite side x,
and adjacent
side 3 we see that sec θ =√
9+x2
3and tan θ = x
3. Thus,
∫ 1√x2 + 9
dx = ln
∣∣∣∣∣√
9 + x2
3+
x
3
∣∣∣∣∣+ C.• Integrands Involving
√x2 − a2, x ≥ a or x ≤ −a.
Here, we let x = a sec θ with 0 ≤ θ < π, θ 6= π2
so that
√x2 − a2 =
√a2(sec2 θ − 1) =
√a2 tan2 θ = a |tan θ| .
25
-
Remark 43.2Letting θ be the angle of a right triangle with
opposite side
√x2 − a2, adja-
cent side a, and hypotenuse x we find tan θ =√
x2−a2a
. See Figure 108.
Figure 108
Example 43.8Find
∫ 1x√
x2−1dx.
Solution.Let x = sec θ, 0 ≤ θ ≤ π, θ 6= π
2. Then dx = sec θ tan θdθ and
x2 − 1 = sec2 θ − 1 = tan2 θ.
Thus,∫ 1x√
x2 − 1dx =
∫ sec θ tan θsec θ| tan θ|
dθ = ±∫
dθ = ±θ + C =∣∣∣sec−1x∣∣∣+ C.
Completing the Square to Use Trigonometric Substitution
Example 43.9Find
∫ 1√x2+6x+25
dx.
Solution.Completing the square we find x2 + 6x + 25 = (x + 3)2 +
16. So let x + 3 =4 tan θ,−π
2θ < π
2. Then dx = 4 sec2 θdθ and
√x2 + 6x + 25 =
√16 sec2 θ =
4 sec θ. Thus, ∫ 1√x2+6x+25
dx =∫ 4 sec2 θ
4 sec θdθ =
∫sec θdθ
= ln | sec θ + tan θ|+ C= ln |
√x2+6x+25
4+ x+3
4|+ C.
26
-
We can summarize the above substitutions in the following
table
expression substitution identity√a2 − u2 u = a sin θ 1− sin2 θ =
cos2 θ√a2 + u2 u = a tan θ 1 + tan2 θ = sec2 θ√u2 − a2 u = a sec θ
sec2 θ − 1 = tan2 θ
27
-
Practice Problems
Exercise 43.1Find
∫ 1x2+4x+3
dx.
Exercise 43.2Find
∫ 14−x2 dx.
Exercise 43.3Find
∫ 11+(x+2)2
dx.
Exercise 43.4Find
∫ 1x2+4x+5
dx.
Exercise 43.5Find
∫ 1x2+4x+4
dx.
Exercise 43.6Find
∫ 1x2−1dx.
Exercise 43.7Find
∫ 1x2−xdx.
Exercise 43.8Find
∫ 3x+1x2−3x+2dx.
Exercise 43.9Evaluate
∫ 13x−x2 dx.
Exercise 43.10Evaluate
∫ x2+1x2−3x+2dx.
Exercise 43.11Evaluate
∫ x3x2+3x+2
dx.
Exercise 43.12Find
∫ 1x2+5x+4
dx.
Exercise 43.13Find numbers A, B and C such that
1
(x2 + 6x + 14)(x− 1)=
Ax + B
x2 + 6x + 14+
C
x− 1.
Exercise 43.14Write the fraction 1
e2x−4ex+3 as the sum of two fractions.
28
-
44 Numerical Approximation of Definite In-
tegrals
Sometimes the integral of a function cannot be expressed with
elementaryfunctions, i.e. polynomial, trigonometric, exponential,
logarithmic, or asuitable combination of these. For example, the
functions sin (x2) and sin x
x
don’t have simple antiderivatives, i.e. it is not possible to
write down a sim-ple analytic formula in terms of elementary
fucntions. In cases like these westill can find an approximate
value for the integral of such functions on aninterval.
Left- and Right-hand Riemann SumsWe already know that we can
approximate a definite integral
∫ ba f(x)dx by a
left- or right-hand Riemann sum for some finite n:
LEFT (f, n) =n−1∑i=0
f(xi)∆x
and
RIGHT (f, n) =n∑
i=1
f(xi)∆x
where x0, x1, · · · , xn are n+1 equally spaced points in [a, b]
with x0 = a, xn =b, and ∆x = b−a
n.
Example 44.1Approximate
∫ 51
1xdx using a left- and right-hand Riemann sum with 4 inter-
vals, i.e. n = 4.
Solution.We first construct the following chart.
xi 1 2 3 4 5f(xi) 1
12
13
14
15
Thus,LEFT ( 1
x, 4) = (f(1) + f(2) + f(3) + f(4))∆x
= (1 + 12
+ 13
+ 14) · 1
= 2512≈ 2.0833
29
-
andRIGHT ( 1
x, 4) = (f(2) + f(3) + f(4) + f(5))∆x
= (12
+ 13
+ 14
+ 15) · 1
= 7760≈ 1.2833.
For comparison, we know that the exact value of the integral we
are seekingto approximate is ∫ 5
1
1
xdx = ln 5 ≈ 1.6094.
Thus, for n = 4 the left- and right-hand Riemann sums give poor
approx-imations. Figures 109(a) and 109(b) illustrate why the left
and right rulesare so inaccurate.
Figure 109
By drawing pictures of the geometric regions involved one can
see easily that
LEFT (f, n) ≤∫ b
af(x)dx ≤ RIGHT (f, n), when f(x) is increasing
and
RIGHT (f, n) ≤∫ b
af(x)dx ≤ LEFT (f, n), when f(x) is decreasing.
So if f(x) is increasing then LEFT (f, n) is an underestimate
while RIGHT (f, n)is an overestimate. Similarly, when f is
decreasing RIGHT (f, n) is an un-derestimate and LEFT (f, n) is an
overestimate.
30
-
Midpoint ApproximationInstead of using the left- or
right-endpoints we use the midpoint. Recall thatthe midpoint of an
interval [x, y] is given by the midpoint formula x+y
2. The
midpoint approximation is given by
MID(f, n) =n∑
i=1
f(
xi−1 + xi2
)∆x.
Example 44.2Approximate
∫ 51
1xdx with the midpoint rule using 4 intervals.
Solution.Let mi denote the midpoint of the interval [xi−1, xi].
Then
mi 1.5 2.5 3.5 4.5f(mi)
23
25
27
29
Thus,
MID( 1x, 4) = (f(1.5) + f(2.5) + f(3.5) + f(4.5))∆x
= 23
+ 25
+ 27
+ 29
≈ 1.5746.
This answer is fairly close to the exact answer. The midpoint
rule approxi-mates with rectangles on each subdivision that are
partly above and partlybelow the graph, so the errors tend to
balance out. See Figure 110.
31
-
Figure 110
Note that if the graph of f(x) is concave down on [a, b] then
MID(f, n) isan overestimate of
∫ ba f(x)dx since the midpoint rectangle and the trapezoid
constracted by drawing the tangent line to the graph at the
midpoint havethe same area since the triangles ABC and CEF are
equal. (See Figure 111)
Figure 111
On the other hand, if f(x) is concave up on [a, b] then MID(f,
n) is anunderestimate of
∫ ba f(x)dx. See Figure 112.
Figure 112
32
-
Trapezoid RuleThere is no reason why we should necessarily use
rectangles to approximate∫ ba f(x)dx. We could use the function
values of both endpoints of the interval
and approximate the interval by trapezoids instead. Recall that
the area ofa trapezoid with base [xi−1, xi] and sides f(xi−1) and
f(xi) is given by
f(xi−1) + f(xi)
2·∆x.
Thus, the trapezoid approximation is given by
TRAP (f, n) =n∑
i=1
f(xi−1) + f(xi)
2·∆x = LEFT (f, n) + RIGHT (f, n)
2.
Example 44.3Approximate
∫ 51
1xdx with the trapezoid rule using 4 intervals.
Solution.Using the following chart
xi 1 2 3 4 5f(xi) 1
12
13
14
15
we can write
TRAP ( 1x, 4) = (f(1)+f(2)
2+ f(2)+f(3)
2+ f(3)+f(4)
2+ f(4)+f(5)
2)∆x
= 34
+ 512
+ 724
+ 940
≈ 1.6833
This answer is fairly close to the exact answer.
Note that if the graph of f(x) is concave up then the area of
each trapezoid islarger than the area under the graph so that TRAP
(f, n) is an overestimateof∫ ba f(x)dx. See Figure 113(a). On the
other hand, if the graph of f(x) is
concave down then the area fo each trapezoid is smaller than the
area underthe graph so that TRAP (f, n) is an underestimate of the
definite integral.
33
-
See Figure 113(b).
Figure 113
It follows from the above discussions that:
(i) If f(x) is concave up and increasing on [a,b] then
LEFT (f, n) ≤ MID(f, n) ≤∫ b
af(x)dx ≤ TRAP (f, n) ≤ RIGHT (f, n).
(ii) If f(x) is concave up and decreasing on [a,b] then
RIGHT (f, n) ≤ MID(f, n) ≤∫ b
af(x)dx ≤ TRAP (f, n) ≤ LEFT (f, n).
(iii) If f(x) is concave down and increasing on [a,b] then
LEFT (f, n) ≤ TRAP (f, n) ≤∫ b
af(x)dx ≤ MID(f, n) ≤ RIGHT (f, n).
(iv) If f(x) is concave down and decreasing on [a,b] then
RIGHT (f, n) ≤ TRAP (f, n) ≤∫ b
af(x)dx ≤ MID(f, n) ≤ LEFT (f, n).
34
-
Practice Problems
In Exercises 1 - 4 sketch the area given by the following
approximationsto∫ ba f(x)dx. Identify each approximation as an
overestimate or an underes-
timate.(a) LEFT (f, 2) (b) RIGHT (f, 2) (c) TRAP (f, 2) (d)
MID(f, 2)
Exercise 44.1
Exercise 44.2
Exercise 44.3
35
-
Exercise 44.4
Exercise 44.5(a) Estimate
∫ 10
11+x2
dx by subdividing the interval [0, 1] into eight parts using:(i)
The left Riemann sum.(ii) The right Riemann sum.(iii) the trapeziod
rule.(b) Since the exact value of the integral is π
4, you can estimate the value of π
using part (a). Explain why your first estimate is too large and
your secondestimate is too small.
Exercise 44.6(a) Find LEFT (x2 + 1, 2) and RIGHT (x2 + 1, 2)
for
∫ 40 (x
2 + 1)dx.(b) Illustrate your answers to part (a) graphically. Is
each approximation anunderestimate or an overestimate?
Exercise 44.7(a) Find MID(x2 + 1, 2) and TRAP (x2 + 1, 2)
for
∫ 40 (x
2 + 1)dx.(b) Illustrate your answers to part (a) graphically. Is
each approximation anunderestimate or an overestimate?
Exercise 44.8Using the table, estimate the total distance
traveled from time t = 0 to timet = 6 using LEFT, RIGHT, and
TRAP.
t 0 1 2 3 4 5 6v 3 4 5 4 7 8 11
Exercise 44.9Using the figure below, order the following
approximations to the integral
36
-
∫ 30 f(x)dx and its exact value from smallest to largest: LEFT
(f, n), RIGHT (f, n), MID(f, n)
and TRAP (f, n).
Exercise 44.10(a) Values for f(x) are in the table. Which of the
four approximation methodsin this section is most likely to give
the best estimate of
∫ 120 f(x)dx? Estimate
the integral using this method.
x 0 3 6 9 12f(x) 100 97 90 78 55
(b) Assume f(x) si continuous with no critical points or points
of inflectionon the interval 0 ≤ x ≤ 12. Is the estimate found in
part (a) an overestimateor an underestimate?Explain.
Exercise 44.11Decide whether the following statements are true
or false. Give an explana-tion for your answer. f is assumed to be
continuous on [2, 6].(a) If n = 10, then the subdivision size is ∆x
= 1
10.
(b) If we double the value of n, we make ∆x half as large.(c)
LEFT(f, 10) ≤ RIGHT(f, 10).(d) As n approaches infinity, LEFT(f,n)
approaches zero.
(e) LEFT(f, n)− RIGHT(f, n) = f(6)−f(2)n
.(f) Doubling n decreases the difference LEFT (f, n) − RIGHT (f,
n) by ex-actly the factor 1
2.
(g) If LEFT (f, n) = RIGHT (f, n) for all n then f is a constant
function.
37
-
45 Simpson’s Rule and Error Estimates
The trapezoid rule discussed in the previous section uses line
segments toapproximate the graph of the integrand, i.e. we
approximate the area underthe graph by using trapezoids. In this
section, we will introduce an approx-imation technique, known as
simpson’s rule, that approximates the graphof the integrand by
using parabolas instead (i.e. functions with equationsy = Ax2 + Bx
+ C.).The process starts by dividing the interval [a, b] into n
equal subintervalseach of length ∆x = b−a
nusing the partition points a = x0 < x1 < x2 <
· · · < xn = b. On the interval [xi−1, xi] we want to
approximate f(x) by aquadratic function,i.e.
f(x) ≈ Ax2 + Bx + Csuch that
f(xi−1) = Ax2i−1 + Bxi−1 + C
f(xi) = Ax2i + Bxi + C
f(mi) = Am2i + Bmi + C
where mi is the midpoint of [xi−1, xi]. Thus,∫ xixi−1
f(x)dx ≈∫ xixi−1
(Ax2 + Bx + C)dx
=[
A3x3 + B
2x2 + Cx
]xixi−1
= A3(x3i − x3i−1) + B2 (x
2i − x2i−1) + C(xi − xi−1)
= A3(xi − xi−1)(x2i + xixi−1 + x2i−1) + B2 (xi − xi−1)(xi +
xi−1) + C(xi − xi−1)
= ∆x3
[A(x2i + xixi−1 + x
2i−1) +
32B(xi + xi−1) + 3C
].
But
f(xi−1) + 4f(mi) + f(xi) = Ax2i−1 + Bxi−1 + C + 4A
(xi−1+xi
2
)2+ 4B
(xi−1+xi
2
)+ 4C + Ax2i + Bxi + C
= 2[A(x2i + xixi−1 + x
2i−1) +
32B(xi−1 + xi) + 3C
].
It follows that∫ xixi−1
f(x)dx ≈ ∆x3
(f(xi−1)
2+ 2f(mi) +
f(xi)
2
).
Hence, ∫ ba f(x)dx ≈ 13
∑ni=1
(f(xi−1)+f(xi)
2∆x + 2f(mi)∆x
)= 2·MID(f,n)+TRAP (f,n)
3.
38
-
We denote the expression on the right-hand side by SIMP (f,
n).
Example 45.1Use Simpson’s rule to approximate the value of
π.
Solution.First recall that
∫ 10
11+x2
dx = arctan x|10 = π4 . Let n = 2 in Simpson’s rule sothat ∆x =
1
2, x0 = 0, x1 =
12, x2 = 1, m1 =
14, and m2 =
34. Thus,∫ 1
01
1+x2dx ≈ SIMP (f, 2)
= ∆x3
(f(x0)2
+ 2f(m1) +f(x1)
2+ f(x1)
2+ 2f(m2) +
f(x2)2
)= 1
6(1
2+ 2 · 16
17+ 2
5+ 2
5+ 2 · 16
25+ 1
4)
≈ 0.7854.
This produces the approximation π = 4∫ 10
11+x2
dx ≈ 3.141568.
Error EstimatesWe next discuss error estimates of the five
numerical methods discussed sofar. We define
Error = Exact V alue− Approximate V alue.
If the error is negative then the method produces an
overestimate while ifthe error is positive the error produces an
underestimate.In general, one does not know the exact error, for
otherwise one can find theexact value. Often, we work on finding an
upper bound on the error and getan idea of how much work is
involved in making the error smaller and thusobtaining better
estimation.
• Error in Left and Right RulesWith a bit of work, it can be
shown that∣∣∣∣∣
∫ ba
f(x)dx− LEFT (f, n)∣∣∣∣∣ ≤ M(b− a)22n
and ∣∣∣∣∣∫ b
af(x)dx−RIGHT (f, n)
∣∣∣∣∣ ≤ M(b− a)22nwhere M is the largest value of |f ′(x)| on
the interval [a, b]. In words, theabsolute value of the error in
either the left-hand rule or the right-hand rule
39
-
is bounded by a constant multiplied by 1n. Thus, doubling the
number of in-
tervals will decrease the error by a factor of 12. Also, since M
= max{|f ′(x)| :
a ≤ x ≤ b} then the error depends on how steeply the graph of f
rises orfalls. See Figure 114.
Figure 114
Example 45.2Let A =
∫ 101
1xdx ≈ 2.302585. Then by the definition of LEFT(f,n)and
RIGHT(f,n) we have
LEFT (f, n) =∑n−1
i=0 f(xi)∆x=
∑n−1i=0
1xi
9n
=∑n−1
i=01
1+ 9n·i ·
9n
=∑n−1
i=09
n+9i.
Using a TI-83, one can find LEFT(f,n) for any n. For example,
LEFT (f, 160)can be found by typing
sum(seq(9/(160 + 9N), N, 0, 159, 1))
and then hit enter.Similarly,
RIGHT (f, n) =n∑
i=1
9
n + 9i.
40
-
Thus, we have the following table
n RIGHT ( 1x, n) LEFT ( 1
x, n) A−RIGHT ( 1
x, n) A− LEFT ( 1
x, n)
10 1.960214 2.770214 0.342371 −0.46762920 2.116477 2.521477
0.186108 −0.21889240 2.205491 2.407991 0.097094 −0.10540680
2.253003 2.354253 0.049582 −0.051668160 2.277534 2.328159 0.025052
−0.025574320 2.289994 2.315307 0.012591 −0.012722
In agreement with the result above, the errors in the right-hand
rule or left-hand rule approximations decrease by a factor of,
roughly, 1
2when we double
the number of intervals. Also, note that the left errors are all
negative sincethe function 1
xis decreasing so that LEFT is an overestimate. Similar
remark
for the right errors.
• Error in the Trapezoid RuleIt can be shown that the absolute
value of the error in the trapezoid rule isbounded by ∣∣∣∣∣
∫ ba
f(x)dx− TRAP (f, n)∣∣∣∣∣ ≤ K(b− a)3n2
where K is the largest value of |f ′′(x)| in the interval [a,
b]. Thus, if wedouble the number of intervals then we should expect
the error to decreaseby a factor of 1
4.
Example 45.3Let A =
∫ 101
1xdx. Then by the definition of TRAP(f,n) we have
TRAP (f, n) =∑n
i=1f(xi−1)+f(xi)
2·∆x
=∑n
i=112
(1
xi−1+ 1
xi
)· 9
n
=∑n
i=112
(n
n+9(i−1) +n
n+9i
)· 9
n.
Using a TI-83 we have the table
n TRAP ( 1x, n) A− TRAP ( 1
x, n)
10 2.365214 −0.06262920 2.318977 −0.01639240 2.306741
−0.00415680 2.303628 −0.001043160 2.302846 −0.000265320 2.302650
−0.000065
41
-
We see that the errors in the trapezoid rule approximations are
significallysmaller than the corresponding errors for the left-hand
and right-hand ruleapproximations. Moreover, in agreement with the
result above, the errors inthe trapezoid rule approximations
decrease by a factor of, roughly, 1
4when
we double the number of intervals. Note that the errors are all
negativedue to the fact that the function 1
xis concave up so the trapezoid rule is an
overestimate. Note also, that the trapezoid rule converges to
the value ofthe definite integral at a significantly faster rate
than do the left-hand ruleor the right-hand rule.
• Error in Midpoint RuleAn anlaysis of error similar to ones
discussed before shows that∣∣∣∣∣
∫ ba
f(x)dx−MID(f, n)∣∣∣∣∣ ≤ C · 1n2 ,
where C depends on |f ′′(x)|. Hence, doubling the number of
intervals willdecrease the error by, roughly, a factor of 1
4. Moreover, a more careful exam-
ination of the error would show that there is a sense in which
it is typicallyon the order of 1
2the size of the error of the trapezoid rule.
Example 45.4Let A =
∫ 101
1xdx. Then by the definition of MID(f,n) we have
MID(f, n) =∑n
i=1 f(mi)∆x=
∑ni=1
1mi· 9
n
=∑n
i=12
xi−1+xi· 9
n
=∑n
i=118
2n+18i−9 .
Using a TI-83 we have the following table
n MID( 1x, n) A−MID( 1
x, n)
10 2.272740 0.02984520 2.294504 0.00808140 2.300515 0.00207080
2.302064 0.000521160 2.302455 0.000130320 2.302552 0.000033
Notice that errors in the midpoint rule approximations decrease
by a factorof, roughly, 1
4when we double the number of intervals. Note that the
errors
42
-
are all positive since the function 1x
is concave up so that the midpoint ruleis an underestimate. Note
also, that the error in each approximation isapproximately 1
2of the corresponding error for the trapezoid rule.
Remark 45.1The errors in either the midpoint rule of the
trapezoid rule not only dependon n but also on the size of f ′′ and
hence on the concavity of f. See Figure 115.
Figure 115
• Error of Simpson’s RuleIt may be shown that the absolute value
of the error using Simpson’s rule isbounded by a constant multiple
of 1
n4, resulting in a dramatic improvement
over both the trapezoid and midpoint rules. For Simpson’s rule,
doublingthe number of intervals decreases the error by, roughly
1
16, a general fact that
we can see some evidence in the following example. The constant
dependson the size of the fourth derivative of f.
Example 45.5Let A =
∫ 101
1xdx. Then by the definition of the Simpson’s rule we have
SIMP (f, n) =2MID(f, n) + TRAP (f, n)
3.
Using Examples 45.3 and 45.4 we have the following table
n SIMP ( 1x, n) A− SIMP ( 1
x, n)
10 2.303565 −0.00098020 2.302662 −0.00007740 2.302590
−0.00000580 2.302585 0.000000160 2.302455 0.000130320 2.302552
0.000033
43
-
Practice Problems
Exercise 45.1(a) Compute SIMP(2) for
∫ 40 (x
2 + 1)dx.(b) Use the Fundamental Theorem of Calculus to find
∫ 40 (x
2 + 1)dx exactly.(c) What is the error in SIMP (x2 + 1, 2) for
this integral?
Exercise 45.2In this problem you will investigate the behavior
of the errors in the approx-imation of the integral ∫ 2
1
1
xdx ≈ 0.6931471806 · · ·
(a) For n = 2, 4, 8, 16, 32, 64, 128 intervals, find the left
and right approxima-tions and the absolute values of the errors in
each. How do the errors changeif n is doubled?(b) For the values of
n in part (a), compute the midpoint and trapezoid ap-proximations
and the absolute value of the errors in each. How do the
errorschange if n is doubles?(c) For n = 2, 4, 8, 16, 32 intervals,
compute Simpson’s error approximationsand the absolute values of
the errors in each. How do the errors change if nis doubled?
Exercise 45.3(a) What is the exact value of
∫ 20 (x
3 + 3x2)dx?(b) Find SIMP (x3 + 3x2, n) for n = 2, 4, 100. What
do you notice?
Exercise 45.4(a) What is the exact value of
∫ 40 e
xdx?(b) Find LEFT (ex, 2), RIGHT (ex, 2), TRAP (ex, 2), MID(ex,
2), and SIMP (ex, 2).Compute the error for each.(c) Repeat part (b)
with n = 4.(d) For each rule in part (b), as n goes from n = 2 to n
= 4, does the errorgo down approximately as you would
expect?Explain.
44
-
46 Improper Integrals
A very common mistake among students is when evaluating the
integral∫ 1−1
1xdx. A non careful student will just argue as follows∫ 1
−1
1
xdx = [ln |x|]1−1 = 0.
Unfortunately, that’s not the right answer as we will see below.
The impor-tant fact ignored here is that the integrand is not
continuous at x = 0.Recall that the definite integral
∫ ba f(x)dx was defined as the limit of a left-
or right Riemann sum. We noted that the definite integral is
always well-defined if:(a) f(x) is continuous on [a, b],(b) and if
the domain of integration [a, b] is finite.Improper integrals are
integrals in which one or both of these conditionsare not met,
i.e.,
(1) The interval of integration is infinite:
[a,∞), (−∞, b], (−∞,∞),
e.g.: ∫ ∞1
1
xdx.
(2) The integrand has an infinite discontinuity at some point c
in the interval[a, b], i.e. the integrand is unbounded near c :
limx→c
f(x) = ±∞.
e.g.: ∫ 10
1
xdx.
An improper integral may not be well defined or may have
infinite value. Inthis case we say that the integral is divergent.
In case an improper integralhas a finite value then we say that it
is convergent.We will consider only improper integrals with
positive integrands since theyare the most common.
• Unbounded Intervals of Integration
45
-
The first type of improper integrals arises when the domain of
integration isinfinite. In case one of the limits of integration is
infinite, we define∫ ∞
af(x)dx = lim
b→∞
∫ ba
f(x)dx
or ∫ b−∞
f(x)dx = lima→−∞
∫ ba
f(x)dx.
If both limits are infinite, then we choose any number c in the
domain of fand define ∫ ∞
−∞f(x)dx =
∫ c−∞
f(x)dx +∫ ∞
cf(x)dx.
In this case, the integral is convergent if and only if both
integrals on theright converge.
Example 46.1Does the integral
∫∞1
1x2
dx converge or diverge?
Solution.We have∫ ∞
1
1
x2dx = lim
b→∞
∫ b1
1
x2dx = lim
b→∞[−1
x]b1 = lim
b→∞(−1
b+ 1) = 1.
In terms of area, the given integral represents the area under
the graph off(x) = 1
x2from x = 1 and extending infinitely to the right. The above
im-
proper integral says the following. Let b > 1 and obtain the
area shown inFigure 116.
Figure 116
46
-
Then∫ b1
1x2
dx is the area under the graph of f(x) from x = 1 to x = b. As
bgets larger and larger this area is close to 1.
Example 46.2Does the improper integral
∫∞1
1√xdx converge or diverge?
Solution.We have∫ ∞
1
1√xdx = lim
b→∞
∫ b1
1√xdx = lim
b→∞[2√
x]b1 = limb→∞
(2√
b− 2) = ∞.
So the improper integral is divergent.
Remark 46.1In general, some unbounded regions have finite areas
while others have infi-nite areas. This is true whether a region
extends to infinity along the x-axisor along the y-axis or both,
and whether it extends to infinity on one or bothsides of an axis.
For example the area under the graph of 1
x2is finite whereas
that under the graph of 1√x
is infinite. This has to do with how fast each
function approaches 0 as x →∞. The function 1x2
approaches 0 more rapidlythan that of 1√
x.
The following example generalizes the results of the previous
two examples.
Example 46.3Determine for which values of p the improper
integral
∫∞1
1xp
dx diverges.
Solution.Suppose first that p = 1. Then∫∞
11xdx = limb→∞
∫ b1
1xdx
= limb→∞[ln |x|]b1 = limb→∞ ln b = ∞
so the improper integral is divergent.Now, suppose that p 6= 1.
Then∫∞
11xp
dx = limb→∞∫ b1 x
−pdx
= limb→∞[x−p+1
−p+1 ]b1
= limb→∞(b−p+1
−p+1 −1
−p+1).
47
-
If p < 1 then −p + 1 > 0 so that limb→∞ b−p+1 = ∞ and
therefore the im-proper integral is divergent. If p > 1 then−p+1
< 0 so that limb→∞ b−p+1 = 0and hence the improper integral
converges:∫ ∞
1
1
xpdx =
−1−p + 1
.
Example 46.4For what values of c is the improper integral
∫∞0 e
cxdx convergent?
Solution.We have ∫∞
0 ecxdx = limb→∞
∫ b0 e
cxdx = limb→∞1cecx|b0
= limb→∞1c(ecb − 1) = −1
provided that c < 0. Otherwise, i.e. if c ≥ 0, then the
improper integral isdivergent.
Remark 46.2The previous two results are very useful and you may
want to memorizethem.
Example 46.5Show that the improper integral
∫∞−∞
11+x2
dx converges.
Solution.Splitting the integral into two as follows:∫ ∞
−∞
1
1 + x2dx =
∫ 0−∞
1
1 + x2dx +
∫ ∞0
1
1 + x2dx.
Now, ∫ 0−∞
11+x2
dx = lima→−∞∫ 0a
11+x2
dx = lima→−∞ arctan x|0a= lima→−∞(arctan 0− arctan a) = −(−π2 )
=
π2.
Similarly, we find that∫∞0
11+x2
dx = π2
so that∫∞−∞
11+x2
dx = π2
+ π2
= π.
• Unbounded IntegrandsSuppose f(x) is unbounded at x = a, that
is limx→a+ f(x)dx = ∞. Then wedefine ∫ b
af(x)dx = lim
t→a+
∫ bt
f(x)dx.
48
-
Similarly, if f(x) is unbounded at x = b, that is limx→b− f(x)dx
= ∞. Thenwe define ∫ b
af(x)dx = lim
t→b−
∫ ta
f(x)dx.
Now, if f(x) is unbounded at an interior point a < c < b
then we define∫ ba
f(x)dx = limt→c−
∫ ta
f(x)dx + limt→c+
∫ bt
f(x)dx.
If both limits exist then the integral on the right-hand side
converges. Ifone of the limits does not exist then we say that the
improper integral isdivergent.
Example 46.6Show that the improper integral
∫ 10
1√xdx converges.
Solution.Indeed, ∫ 1
01√xdx = limt→0+
∫ 1t
1√xdx = limt→0+ 2
√x|1t
= limt→0+(2− 2√
t) = 2.
In terms of area, we pick an a > 0 as shown in Figure
117:
Figure 117
Then the shaded area is∫ 1a
1√xdx. As a approaches 0 from the right, the area
approaches the value 2.
Example 46.7Investigate the convergence of
∫ 20
1(x−2)2 dx.
49
-
Solution.We deal with this improper integral as follows∫ 2
01
(x−2)2 dx = limt→2−∫ t0
1(x−2)2 dx = limt→2− −
1(x−2) |
t0
= limt→2−(− 1t−2 −12) = ∞.
So that the given improper integral is divergent.
Example 46.8Investigate the improper integral
∫ 1−1
1xdx.
Solution.We first write ∫ 1
−1
1
xdx =
∫ 0−1
1
xdx +
∫ 10
1
xdx.
On one hand we have,∫ 0−1
1xdx = limt→0−
∫ t−1
1xdx = limt→0− ln |x||t−1
= limt→0− ln |t| = ∞.
This shows that the improper integral∫ 0−1
1xdx is divergent and therefore the
improper integral∫ 1−1
1xdx is divergent.
• Improper Integrals of Mixed TypesNow, if the interval of
integration is unbounded and the integrand is un-bounded at one or
more points inside the interval of integration we can eval-uate the
improper integral by decomposing it into a sum of improper
integralwith finite interval but where the integrand is unbounded
and an improperintegral with an infinite interval. If each
component integrals converges, thenwe say that the original
integral converges to the sum of the values of thecomponent
integrals. If one of the component integrals diverges, we say
thatthe entire integral diverges.
Example 46.9Investigate the convergence of
∫∞0
1x2
dx.
Solution.Note that the interval of integration is infinite and
the function is undefinedat x = 0. So we write∫ ∞
0
1
x2dx =
∫ 10
1
x2dx +
∫ ∞1
1
x2dx.
50
-
But ∫ 10
1
x2dx = lim
t→0−
∫ 1t
1
x2dx = lim
t→0−−1
x|1t = lim
t→0−(1
t− 1) = ∞.
Thus,∫ 10
1x2
dx diverges and consequently the improper integral∫∞0
1x2
dx diverges.
51
-
Practice Problems
Exercise 46.1Investigate the convergence of
∫∞0
xex
dx.
Exercise 46.2Investigate the convergence of
∫ 0−∞
ex
ex+1dx.
Exercise 46.3Investigate the convergence of
∫∞−∞
1x2+25
dx.
Exercise 46.4Investigate the convergence of
∫ 40
1√16−x2 dx.
Exercise 46.5Investigate the convergence of
∫ 10
x4+1x
dx.
Exercise 46.6Investigate the convergence of
∫ 63
dx(4−x)2 .
Exercise 46.7Find the area under the curve y = xe−x for x ≥
0.
Exercise 46.8Given that
∫∞−∞ e
−x2dx =√
π, calculate the exact value of∫ ∞−∞
e−(x−a)2
b dx.
Exercise 46.9Consider the improper integral ∫ ∞
exp ln xdx.
For what values of p does the integral converge or diverge? What
is the valueof the integral when it converges?
Exercise 46.10Consider the improper integral ∫ e
0xp ln xdx.
For what values of p does the integral converge or diverge? What
is the valueof the integral when it converges?
52
-
Exercise 46.11The gamma function is defined for all x > 0 by
the rule
Γ(x) =∫ ∞0
tx−1e−tdt.
(a) Find Γ(1) and Γ(2).(b) Integrate by parts with respect to t
to show that, for positive n,
Γ(n + 1) = nΓ(n).
(c) Find a simple expression for Γ(n) for positive integers.
53
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47 Comparison Tests for Improper Integrals
Sometimes it is difficult to find the exact value of an improper
integral byantidifferentiation, for instance the integral
∫∞0 e
−x2dx. However, it is stillpossible to determine whether an
improper integral converges or diverges.The idea is to compare the
integral to one whose behavior we already know,such us
• the p-integral∫∞1
1xp
dx which converges for p > 1 and diverges otherwise;• the
integral
∫∞0 e
cxdx which converges for c < 0;• the integral
∫ 10
1xp
dx which converges for p < 1 and diverges otherwise.
The comparison method consists of the following:
Theorem 47.1Suppose that f and g are continuous and 0 ≤ g(x) ≤
f(x) for all x ≥ a. Then
(a) if∫∞a f(x)dx is convergent, so is
∫∞a g(x)dx
(b) if∫∞a g(x)dx is divergent, so is
∫∞a f(x)dx.
This is only common sense: if the curve y = g(x) lies below the
curve y =f(x), and the area of the region under the graph of f(x)
is finite, then ofcourse so is the area of the region under the
graph of g(x). For a proof ofthis theorem see Exercise
47.11.Similar results hold for the other types of improper
integrals.
Example 47.1Determine whether
∫∞1
1√x3+5
dx converges.
Solution.For x ≥ 1 we have that x3 + 5 ≥ x3 so that
√x3 + 5 ≥
√x3. Thus,
1√x3+5
≤ 1√x3
. Letting f(x) = 1√x3
and g(x) = 1√x3+5
then we have that
0 ≤ g(x) ≤ f(x). From the previous section we know that∫∞1
1
x32dx is con-
vergent, a p-integral with p = 32
> 1. By the comparison test,∫∞1
1√x3+5
dx isconvergent.
The next question is to estimate such a convergent improper
integral.
54
-
Example 47.2Estimate the value of
∫∞1
1√x3+5
dx with an error of less than 0.01.
Solution.We want to find b such that∣∣∣∣∣
∫ ∞1
1√x3 + 5
dx−∫ b1
1√x3 + 5
dx
∣∣∣∣∣ < 0.01.But ∫ ∞
1
1√x3 + 5
dx =∫ b1
1√x3 + 5
dx +∫ ∞
b
1√x3 + 5
dx.
Thus, the problem is to find b such that∣∣∣∣∣∫ ∞
b
1√x3 + 5
dx
∣∣∣∣∣ < 0.01.From the example above, we have∫ ∞
b
1√x3 + 5
dx <∫ ∞
b
1√x3
dx =2√b.
So it suffices to choose b such that 2√b
< 0.01 or b > 40, 000, say for exampleb = 45000. In this
case,∫ ∞
1
1√x3 + 5
dx ≈∫ 45,0001
1√x3 + 5
dx = 1.69824.
Example 47.3Investigate the convergence of
∫∞4
dxln x−1 .
Solution.For x ≥ 4 we know that ln x− 1 < ln x < x. Thus,
1
ln x−1 >1x. Let g(x) = 1
x
and f(x) = 1ln x−1 . Thus, 0 ≤ g(x) ≤ f(x). Since
∫∞4
1xdx =
∫∞1
1xdx−
∫ 41
1xdx
which is divergent since∫∞1
1xdx is divergent being a p-integral with p = 1.
By the comparison test∫∞4
dxln x−1 is divergent.
Example 47.4Investigate the convergence of the improper
integral
∫∞1
sin x+3√x
dx.
55
-
Solution.We know that −1 ≤ sin x ≤ 1. Thus 2 ≤ sin x + 3 ≤ 4.
Since x ≥ 1 then2√x≤ sin x+3√
x≤ 4√
x. Note that the two integrals
∫∞1
2√xdx and
∫∞1
4√xdx are
both divergent being a multiple of a p-integral with p = 12
< 1. If we letg(x) = sin x+3√
xand f(x) = 4√
xthen we have no conclusion since
∫∞1 g(x)dx may
or may not converge and still∫∞1 g(x)dx ≤
∫∞1 f(x)dx. Now if we let g(x) =
2√x
and f(x) = sin x+3√x
then by the comparison test∫∞1
sin x+3√x
is divergent
since∫∞1 f(x)dx ≥
∫∞1 g(x)dx and
∫∞1 g(x)dx is divergent.
Example 47.5Investigate the convergence of
∫∞1 e
− 12x2dx.
Solution.If x ≥ 2 then x
2≥ 1. Multiply both sides of this inequality by x ≥ 2 to
obtain
12x2 ≥ x. Now, multiply both sides of this last inequality by −1
to obtain−1
2x2 ≤ −x and therefore e− 12x2 ≤ e−x since the function ex is an
increasing
function. Thus, ∫ ∞1
e−12x2dx =
∫ 21
e−12x2dx +
∫ ∞2
e−12x2dx.
But ∫ 21
e−12x2dx ≈ 0.34
and ∫ ∞2
e−12x2dx ≤
∫ ∞2
e−xdx ≤∫ ∞0
e−xdx
so since∫∞0 e
−xdx is convergent then∫∞2 e
− 12x2dx is convergent. In conclu-
sion,∫∞1 e
− 12x2dx is convergent.
Sometimes it is laborious to find a convenient function f(x)
with g(x) ≤ f(x),but we may know that g(x) is no larger than a
constant multiple of f(x) forlarge enough x, and this is good
enough. The most powerful test of this formin the course is this
version of the limit comparison test:
Theorem 47.2Let f(x) and g(x) be two positive and continuous
functions on [a,∞).
56
-
(a)limx→∞f(x)g(x)
= 0, or
(b)limx→∞f(x)g(x)
= L, where L is a finite positive number, or
(c)limx→∞f(x)g(x)
= ∞, then
(a) If∫∞a g(x)dx converges, then so does
∫∞a f(x)dx.
(b)∫∞a g(x)dx converges if and only if
∫∞a f(x)dx does.
(c) If∫∞a g(x)dx diverges, then so does
∫∞a f(x)dx.
Proof.(a) Suppose that limx→∞
f(x)g(x)
= 0. Let � > 0 be given. Then there is a b > a
such that f(x)g(x)
< � for all x ≥ b. Thus, f(x) < �g(x) for all x ≥ b. By
thecomparison test, if
∫∞a g(x)dx is convergent so does
∫∞a f(x)dx.
(b) Now, suppose that limx→∞f(x)g(x)
= L, where L is a finite positive constant.Let � < L. Then
there is a constant b > a such that for all x ≥ b we
have∣∣∣∣∣f(x)g(x) − L
∣∣∣∣∣ < �.That is,
L− � < f(x)g(x)
< L + �.
Thus, for x ≥ b we have (L − �)g(x) < f(x) < (L + �)g(x).
Now the resultfollows from the comparison test.(c) Finally, suppose
that limx→∞
f(x)g(x)
= ∞. Then given there is b > a suchthat f(x)
g(x)≥ 1 for all x ≥ b. That is, g(x) ≤ f(x) for all x ≥ b.
Therefore, if∫∞
a f(x)dx converges, then∫∞a g(x)dx converges.
Remark 47.1The Comparison Test and Limit Comparison Test also
apply, modified asappropriate, to other types of improper
integrals.
Example 47.6Show that the improper integral
∫∞1
11+x2
dx is convergent.
Solution.Since the integral
∫∞1
dxx2
is convergent (p-integral with p > 1) and since
limx→∞1
1+x2
1x2
= limx→∞x2
x2+1= 1 then by the limit comparison test we have
that∫∞1
dxx2+1
is also convergent.
57
-
Practice Problems.
Exercise 47.1Investigate the convergence of
∫∞1
x2
x4+1dx.
Exercise 47.2Investigate the convergence of
∫∞1
x2+1x3+3x+2
dx.
Exercise 47.3Investigate the convergence of
∫∞1
1e5x+2
dx.
Exercise 47.4Investigate the convergence of
∫∞1
dxx2+x
.
Exercise 47.5Investigate the convergence of
∫ 10
dx√x3+x
.
Exercise 47.6Investigate the convergence of
∫∞1
2+cos xx2
dx.
Exercise 47.7For what values of p does the integral
∫∞2
dxx(ln x)p
converge or diverge?
Exercise 47.8For what values of p does the integral
∫ 21
dxx(ln x)p
converge or diverge?
Exercise 47.9Find the value of a (to three decimal places) that
makes∫ ∞
−∞ae−
x2
2 dx = 1.
Exercise 47.10In Planck’s Radiation Law, we encounter the
integral∫ ∞
1
dx
x5(e1x − 1)
.
(a) Graph the functions y = x + 1 and y = ex. Conclude from the
graph that1 + x ≤ ex for all x.(b) Replacing x by 1
xin (a), show that for all x
e1x − 1 > 1
x.
(c) Use the comparison test to show that the original integral
converges.
58
-
Exercise 47.11Let f(x) ≥ 0 for all x ≥ a.(a) Show that the
sequence an =
∫ na f(x)dx is increasing.
(b) Suppose that∫ ba f(x)dx ≤ M for all b ≥ a, where M > 0.
Show that∫∞
a f(x)dx is convergent. Hint: You need to use the following
result fromSection 56: If an is an increasing sequence of positive
numbers such thatan ≤ M for all n ≥ 1 then liman exists.(c) Proof
the comparison test of improper integrals.
59
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48 Areas and Volumes
In this section we illustrate how the definite integral can be
used to computethe the area of a region and the volume of certain
solids. For example, theapproach to finding the area of a region
will be to think of the region as ap-proximated by small elements,
each of which is so geometrically simple thatits area can be
calculated directly, for example the slices might be
rectangles,circles or triangles. Next, the areas of each of these
elements are added toobtain a Riemann sum. The limit of such
Riemann sum gives the desiredarea. The same idea applies in finding
the volume of a solid. We refer to thisprocess as the method of
slicing.
Finding Areas by The Method of SlicingWhen calculating the area
of a region using Riemann sums, slice the regioninto thin pieces in
which the geometry is so simple that the area can beestimated.
Example 48.1Use horizontal slices to find the area of an
isosceless triangle with vertices at(0, 0), (5, 5) and (10, 0).
Solution.We slice the triangle into n horizontal slices, each
slice being approximatelya rectangle of width ∆y = yi − yi−1 and
length wi. See Figure 118.
Figure 118
To find wi, note that the triangles ABC and AB’C’ are similar
triangles sothat
wi10
=5− yi
5.
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-
Solving for wi, we find wi = 10− 2yi. Thus, the area of each
piece is approx-imately (10− 2yi)∆y so that
Total Area ≈n∑
i=1
(10− 2yi)∆y.
Letting n →∞ we get
Total Area =∫ 50
(10− 2y)dy = 10y − y2∣∣∣50
= 25 square units.
Example 48.2Use horizontal slices to set up a definite integral
representing the area of asemicircle of radius 7cm.
Solution.For simplicity, we assume that the circle is centered
at (0, 0). We slice the
semicircle into n thin slices of width ∆y = yi − yi−1 and length
2√
49− y2i .See Figure 119.
Figure 119
Thus, the area of a slice is approximately 2√
49− y2i ∆y so that
Total Area ≈n∑
i=1
2√
49− y2i ∆y.
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Taking n →∞ we obtain
Total Area =∫ 70 2√
49− y2dy= 2 · 1
2
[y√
49− y2 + 40 arcsin (y7)]70
= 492π cm2
• Finding Volumes by SlicingWhen calculating the volume of a
solid using Riemann sums, slice the solidinto thin pieces in which
the geometry is so simple that the volume can beestimated.
Example 48.3Use vertical slicing to find the volume of a cone
with height 5 cm and radiusof base 5 cm.
Solution.We consider a cone centered at the origin and with axis
the x-axis.We divide the cone into n thin disks each of thickness
∆x = xi − xi−1 andradius yi = xi. See Figure 120.
Figure 120
Thus, the volume of a slice is approximately πx2i ∆x so that
Total V olume ≈n∑
i=1
πx2i ∆x.
Letting n →∞ to obtain
Total V olume =∫ 50
πx2dx = πx3
3
∣∣∣∣∣5
0
=125
3π cm3.
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Example 48.4The Great Pyramid of Egypt has a square base with
side 755 feet long andheight 410 feet. Find the volume of the Great
Pyramid in cubic feet.
Solution.The pyramid may be thought of as being made up of
layers parallel to thebase. Each layer is a thin rectangular box
with square base and with thick-ness ∆z. Figure 121 illustrates a
slice of the pyramid.
Figure 121
The following figure illustrates a triangular cross-section of a
typical layerwith a plane perpendicular at the center of the
layer.
Figure 122
Let si denote the length of the base of a typical layer, then
the similartriangles shown above imply that
si755
=410− zi
410,
where zi denotes the height above the horizontal that the center
of the layerlies. Solving for si one sees that the length of the
rectangular box is given
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-
by the formula
si = 755−755
410zi.
The total volume is approximated by adding the volumes of the n
layers
V ≈n∑
i=1
[755− 755410
zi]2∆z.
Letting n →∞ we obtain
V =∫ 4100 [755− 755410z]
2dz = (755410
)2∫ 4100 (410− z)2dz
= (755410
)2[− (410−z)3
3]4100
= 4103
(755410
)2 ≈ 78 million ft3
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Practice problems
In Exercises 1 - 5, write a Riemann sum and then use a definite
integralrepresenting the area of the region, using the strip shown.
Evaluate theintegral exactly.
Exercise 48.1
Exercise 48.2
Exercise 48.3
Exercise 48.4
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Exercise 48.5
In Exercises 6 - 9, write a Riemann sum and then a definite
integral repre-senting the volume of the region, using the slice
shown. Evaluate the integralexactly.
Exercise 48.6
Exercise 48.7
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Exercise 48.8
Exercise 48.9
Exercise 48.10Find the volume of a sphere of radius r by
slicing.
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Exercise 48.11A rectangular lake is 150 km long and 3 km wide.
The vertical cross-sectionthrough the lake in Figure 123 shows that
the lake is 0.2 km deep at the cen-ter. Set up and evaluate a
definite integral giving the total volume of the lake.
Figure 123
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49 Solids of Revolution- Arc Length
In this section we will discuss the use of definite integrals in
solving problemsin geometry such as finding the
• volume of a solid of known cross-section,• volume of a solid
of revolution,• arc length of a curve in the plane.
Recall the process of slicing which consists of the following
steps:
◦ Divide the solid, region, or curve into small pieces whose
volume, area,or length can be easily approximated.
◦ Add the volumes, areas, or lengths of all the pieces. (Thus
obtaininga Riemann sum) that approximates the total volume, total
area, or totallength.
◦ Take the limit of the Riemann sum from the previous step as n
→ ∞.This gives us a definite integral that gives the total volume,
total area, ortotal length.
• Volume of a Solid with Known Cross SectionWhen we take a plane
perpendicualr to a given solid then the common regionbetween the
plane and the solid is knwon as a cross section. By knowncross
sections we mean cross sections such as circles, squares,
rectangles, orringsLet R be a solid lying alongside some interval
[a, b] of the x-axis. See Figure124.
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-
Figure 124
Divide the interval into n equal subintervals with mesh points
x0 = a < x1 <x2 < · · · < xn−1 < xn = b. The planes
that are perpendicular to the x-axisat the points x0, x1, x2 · · ·
, xn divide the solid into n slices. Since the crosssection of R
changes little along a subinterval [xi−1, xi], the slab
positionedalongside that subinterval can be considered a cylinder
of height ∆x andwhose base has area A(xi). So the volume of the
slice is
∆Vi ≈ A(xi)∆x.
The approximate total volume of the solid is
V =n∑
i=1
∆Vi ≈n∑
i=1
A(xi)∆x.
Once again we recognize a Riemann sum at the right. Letting n →
∞ weobtain the so-called Cavalieri’s principle:
V =∫ b
aA(x)dx.
Of course, the formula can be applied to any axis. For instance
if a solid liesalongside some interval [a, b] on the y-axis, the
formula becomes
V =∫ b
aA(y)dy.
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-
Example 49.1Find the volume of a cone of radius r and height
h.
Solution.Assume that the cone is placed with its vertex in the
origin and its axis onthe x-axis as shown in Figure 125.
Figure 125
The cross section of the cone at each point x is a circular disk
of radius y.Using similar triangles, we find y = xr
h. Hence its area is A(x) = π(xr
h)2. The
volume of the cone can now be computed by Cavalieri’s
formula:
V =∫ h0
πr2
h2x2dx =
πr2
h2[x3
3]h0 =
1
3πr2h.
Example 49.2There is a solid whose bottom face is the disk x2 +
y2 ≤ 1 and every cross-section of the solid perpendicular to x-axis
is a square. Find the volume ofthe solid.
Solution.We view the solid a cardboard model shown in Figure
126.
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-
Figure 126
A typical cross-section is a square of length side s as shown in
Figure 127.
Figure 127
The length s is given by the expression s = 2√
1− y2. Thus, the area of across section is A(y) = s2 = 4(1− y)2.
By Cavalieri’s formula the volume is
V =∫ 1−1
4(1− y2)dy = 4 y − 13y3∣∣∣∣1−1
=4
3.
• Volume of Solids of RevolutionBy a solid of revolution we mean
a solid obtained by revolving a regionaround a line. Consider the
solid of revolution obtained by revolving a planeregion under the
graph of f(x) around the x-axis. See Figure 128.
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-
Figure 128
Each cross section is a circular disk of radius y, so its area
is A(x) = πy2 =π[f(x)]2. Hence, by Cavalieri’s principle, the
volume of the solid is
V =∫ b
aπ[f(x)]2dx.
Example 49.3The region bounded by the curve y =
√x + 1 and the x-axis between x = 0
and x = 9 is revolved around the x-axis. Find the volume of this
solid ofrevolution.
Solution.
The solid of revolution is given in Figure 129.
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-
Figure 129
A cross-section is a disk of area A(x) = π(√
x + 1)2. Thus, the total volumeis given by
V =∫ 90 π(
√x + 1)2dx =
∫ 90 π(x + 2
√x + 1)dx
= π[x2
2+ 4
3x
32 + x]90
= π 1712≈ 268.61 cubic units
If the revolution is performed around the y-axis, the roles of x
and y areinterchanged so in that case the formula is∫ b
aπx2dy,
where x must be written as a function of y, i.e. x = f−1(y).
Example 49.4The curve y = x2, 0 ≤ x ≤ 1 is rotated about the
y-axis. Find the volume ofthe resulting solid of revolution.
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-
Solution.The solid of revolution is shown in Figure 130.
Figure 130
A cross-section is a disk of area A(y) = πy. Thus, by
Vavalieri’s principle thevolume is
V =∫ 10
πydy = πy2
2
∣∣∣∣∣1
0
≈ 1.571. cubic units
If the region being revolved is the area between two curves y =
f(x) andy = g(x), then each cross section is an annular ring (or
washer) with outerradius f(x) and inner radius g(x) (assuming f(x)
≥ g(x) ≥ 0.) See Figure131.
Figure 131
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-
The area of the annular ring is A(x) = π[(f(x))2−(g(x))2], hence
the volumeof the solid will be
V =∫ b
aπ[(ytop)
2 − (ybottom)2]dx =∫ b
aπ[f(x)2 − g(x)2]dx.
If the revolution is performed around the y-axis, then
V =∫ b
aπ[(xright)
2 − (xleft)2]dy.
Example 49.5Find the volume of the solid obtained by revolving
the area between y = x2
and y =√
x around the x-axis.
Solution.First we need to find the intersection points of these
curves in order to findthe interval of integration. Setting x2
=
√x and solving for x we find (0, 0)
and (1, 1). Hence, we must integrate from x = 0 to x = 1.
V =∫ 10 π[(
√x)2 − (x2)2]dx =
∫ 10 π(x− x4)dx
= π[
12x2 − 1
5x5]10
= 3π10
.
• Arc LengthThe definite integral can also be used to compute
the length of a smoothcurve (i.e. a curve with no corner points).
Recall that when using the inte-gral to find the area of a region
one approximates the region by rectanglesthe sum of whose areas
approximates the area of the region. In finding thelength of an arc
one approximates the arc by a finite set of straight line
seg-ments. An approximation of the length of the arc is made by
using the wellknown formula for the length of a line segment and
taking a sum . A limitingprocess then yields the definite integral
which is equal to the length of thearc.To elaborate the above
statement, if an arc is just a line segment with end-points (x1,
y1) and (x2, y2) then its length can be found by the
Pythagoreantheorem:
s =√
(x2 − x1)2 + (y2 − y1)2 =√
∆x + ∆y.
Now, if the arc is the graph of a function f(x) defined on an
interval [a, b],then we divide the interval into n equal
subintervals. See Figure 132. The
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corresponding points in the arc have coordinates (xi, f(xi)), so
two consecu-tive points are seperated by a distance equal to
si =√
(xi − xi−1)2 + (f(xi)− f(xi−1))2.
But by the Mean Value Theorem there is a point x∗i in the
interval [xi−1, xi]such that
f(xi)− f(xi−1) = f ′(x∗i )(xi − xi−1) = f ′(x∗i )∆x.
Hence,
si =√
(∆x)2 + [f ′(x∗i )∆x]2 =
√1 + [f ′(x∗i )]
2∆x.
The total length of the arc is
s ≈n∑
i=1
si =n∑
i=1
√1 + [f ′(x∗i )]
2∆x.
Again, we recognize the sum on the right-hand side as a Riemann
sum whichconverges to the following integral
s =∫ b
a
√1 + [f ′(x)]2dx =
∫ ba
√1 + (
dy
dx)2dx.
Figure 132
Example 49.6Find the length of the arc defined by the curve y =
x
32 between the points
(0, 0) and (1, 1).
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Solution.Using the arc length formula we have
s =∫ 10
√1 + ( dy
dx)2dx =
∫ 10
√1 + [(x
32 )′]2dx
=∫ 10
√1 + (3
2x
12 )2dx =
∫ 10
√1 + 9x
4dx
= [ 127
(4 + 9x)32 ]10 =
127
(1332 − 8) unit length
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Large Practice Problems
In Exercises 1 - 3, the region is rotated around the x-axis.
Find the vol-ume.
Exercise 49.1Bounded by y = ex, y = 0, x = −1, x = 1.
Exercise 49.2Bounded by y = 4− x2, y = 0, x = −2, x = 1.
Exercise 49.3Bounded by y = cos x, y = 0, x = 0, x = π
2.
In Exercises 4 - 5, find the arc length of the given function
from x = 0 tox = 2.
Exercise 49.4f(x) =
√4− x2.
Exercise 49.5f(x) =
√x3.
Exercise 49.6Find the length of the parametric curve x = cos
(et), y = sin (et) for 0 ≤ t ≤ 1.
Exercise 49.7Consider the hyperbola x2 − y2 = 1 in Figure
133.
(a) The shaded region 2 ≤ x ≤ 3 is rotated around the x-axis.
What isthe volume generated?(b) What is the arc length with y ≥ 0
from x = 2 to x = 3?
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Figure 133
In Exercises 8 - 9, sketch the solid obtained by rotating each
region aroundthe indicated axis. Using the sketch, show how to
approximate the volumeof the solid by a Riemann sum, and hence find
the volume.
Exercise 49.8Bounded by y = x3, x = 1, y = −1. Axis: y =
−1.Exercise 49.9Bounded by the first arch of y = sin x, y = 0.
Axis: x-axis.
In Exercises 10 - 13 consider the region bounded by y = ex, teh
x-axis, andthe lines x = 0 and x = 1. Find the volume of the
following solids.
Exercise 49.10The solid obtained by rotating the region about
the x-axis.
Exercise 49.11The solid obtained by rotating the region about
the horizontal line y = −3.Exercise 49.12The solid whose base is
the given region and whose cross-sections perpendic-ular to the
x-axis are squares.
Exercise 49.13(a) Write an integral which represents the
circumference of a circle of radiusr.(b) Evaluate the integral, and
show that you get the answer you expect.
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50 Density and Center of Mass
In this section we discuss two of the applications of definite
integrals, namely, the concepts of density and the center of
mass.
DensityDensity is used in different situations to describe
similar concepts. We presentsome of the situations.
• There is the density of a substance, which indicates how much
mass pervolume unit (i.e. grams per cm3)the substance has.• There’s
population density (i.e. people per mile).• Density of typed words
on a page.• There’s density of fog, referring to the amount of
water vapor in a volumeof unit of air ( i.e. kg/m3).
In all these cases we can use density to compute total mass,
people, words,amount of water vapor, etc. If density is uniform,
simply multiply the entirearea/volume/etc by the density. If
density is not uniform, then we dividethe region /solid/etc. into
small pieces so that the density is approximatelyuniform on each
piece, then add all the pieces together to obtain a Riemannsum.
Making the pieces smaller and smaller, i.e. letting n →∞ we obtain
adefinite integral.
Example 50.1Find the total mass of a rod of length l and
(line-)density δ(x) where x isthe distance a length element from
the left end.
Solution.If the density δ was uniform then the mass ∆m of a
length element ∆x would(by definition of δ) be simply
∆m = δ∆x.
However, the density is not uniform. So we can approximate the
total massof the rod by slicing it into thin segments xi ≤ x ≤ xi+1
each of length ∆xwhere the density is constant there, say δ(x) ≈
δ(xi) for all x in the interval[xi, xi+1]. See Figure 134.
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-
Figure 134
Then the mass of the ith piece is
∆mi ≈ δ(xi)∆x
and
M =n−1∑i=0
∆mi ≈n−1∑i=0
δ(xi)∆x.
Taking the limit n →∞ (infinitely many thin segments) we
obtain
M =∫ l0
δ(x)dx.
Example 50.2Find the total mass of a circular object of radius R
and (area-)density δ(r)where r is the radius of an area element
from the center.
Solution.If the density δ was uniform then the mass ∆m of an
area element ∆A would(by definition of δ) be simply
∆m = δ∆A.
Since the density is not uniform, we approximate the total mass
of the circleby slicing it into thin concentric, circular rings, ri
≤ r ≤ ri+1 each of area∆A(ri) where the density is constant there,
say δ(r) ≈ δ(ri) for all r in theinterval [ri, ri+1]. See Figure
135.
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-
Figure 135
Then the mass of the ith ring is
∆mi ≈ δ(ri)∆A(ri).
But∆A(ri) = π(ri + ∆r)
2 − πr2i= 2πri∆r + π(∆r)
2.
Since each slice is assumed to be very thin then we can ignore
(∆r)2. Thus,obtaining
∆mi2πriδ(ri)∆r.
It follows that the total mass is approximated by the sum
M =n−1∑i=0
∆mi ≈n−1∑i=0
2πriδ(ri)∆r.
Taking the limit n →∞ (infinitely many thin rings) we obtain
M =∫ R0
2πrδ(r)dr.
Example 50.3Suppose we know the (volume-)density δ(r), i.e. the
mass per volume ele-ment, of a spherical object of radius R as a
function of the radius r from thecenter. Estimate the total mass of
the object.
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-
Solution.If the density δ was uniform then the mass ∆m of a
volume element ∆Vwould (by definition of δ) be simply
∆m = δ∆V.
Since the density is not uniform, then we approximate the total
mass of thesphere by slicing it into thin concentric, spherical
shells, ri ≤ r ≤ ri+1 eachof volume ∆V (ri) where the density is
constant there, say δ(r) ≈ δ(ri) forall r in the interval [ri,
ri+1].
Then the mass of the ith spherical shell is
∆mi ≈ δ(ri)∆V (ri).
But the volume of a thin spherical shell with inner radius ri
and thickness∆r is
∆Vi(ri) =43π(ri + ∆r)
3 − 43πr3i
= 43π[r3i + 3r
2i ∆r + 3ri(∆r)
2 + (∆r)3 − r3i ]≈ 4
3π3r2i ∆r = 4πr
2i ∆r
since (∆r)2 ≈ 0 and (∆r)3 ≈ 0. Thus,
∆mi ≈ 4πr2i δ(ri)∆r
and
M =n−1∑i=0
∆mi ≈n−1∑i=0
4πr2i δ(ri)∆r.
Taking the limit n →∞ (infinitely many thin shells) we
obtain
M =∫ R0
4πr2δ(r)dr.
Center of MassThe center of mass is the so-called ”balancing
point” of an object (or sys-tem.) For example, when two children
are sitting on a s
