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Exploration 9-3a: Integration by Parts Practice
Date:
Objective: Integrate by parts to find antiderivatives quickly.
Find the integral. Show the steps you take.
1.
x
2
2.
x
5
3.
e
sin 3x dx
ln 4x dx
5x
4.
x(ln x)
5.
sin
x cos x dx (Be clever!)
6.
sin
x dx, in terms of
10
10
3
dx
sin
8
x dx
cos 6x dx
7. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-4a: Reduction Formulas
Date:
Objective: Integrate by parts to find a reduction formula, and use that formula
in a volume problem.
1. Integrate by parts to find a formula for sin10 x dx in
terms of sin8 x dx.
4. Problem 3 involves sin6 x dx. Use the reduction
formula to evaluate this indefinite integral. Use the
result to find the volume of the solid algebraically,
using the fundamental theorem. Does the result
agree with the answer to Problem 3?
2. Based on the pattern you observe in the answer to
Problem 1, write a formula for sinn x dx in terms
of sinnD2 x dx. Check with Section 9-4 to make
sure your formula is correct.
3. The graph shows y H sin3 x. Find numerically the
volume of the solid generated by rotating about the
x-axis the region under this graph from x H 0 to
x H π.
y
1
x
5. What did you learn as a result of doing this
Exploration that you did not know before?
3
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Exploration 9-5a: Integrals of Special Powers
of Trig Functions
Date:
Objective: Integrate odd powers of sine and cosine, and even powers of secant
and cosecant.
1. Integrate cos7x sin x dx as a power.
5. Integrate tan5 x sec2 x dx as a power.
2. Explain why cos7x dx cannot be integrated as a
power.
6. Explain why sec8 x dx cannot be integrated as a
power.
3. Associate all but one of the cosines in the integral
in Problem 2, change them to sines using the
Pythagorean properties, then evaluate the integral
as a sum of powers.
7. Associate all but two of the secants in the integral
in Problem 6, change them to tangents using the
Pythagorean properties, then evaluate the integral
as a sum of powers.
8. Explain why sec7 x dx cannot be integrated by the
method of Problem 7.
4. Explain why cos6 x dx cannot be integrated by the
method of Problem 3.
9. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-5b: Other Special
Trigonometric Integrals
Date:
Objective: Find the antiderivatives sin2 x dx and cos2 x dx and cos ax sin bx dx.
1. You recall that
cos (A C B) H cos A cos B D sin A sin B
Use this property to write the double-argument
property for cos 2x.
6. Products of sine and cosine with two different
arguments can be integrated by parts. Integrate
cos 7x sin 5x dx.
2. Transform the double-argument property so that
cos 2x is expressed in terms of cos x alone.
3. Transform the double-argument property so that
cos 2x is expressed in terms of sin x alone.
4. Transform the properties in Problems 2 and 3 so
that cos2 x is expressed in terms of cos 2x and
so that sin2 x is expressed in terms of cos 2x.
7. Use the result of Problem 6 to write a formula for
cos ax sin bx dx.
5. Use the results of Problem 4 to find the integrals
cos2 x dx and sin2 x dx.
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-6a: Introduction to Integration
by Trig Substitution
Date:
Objective: Integrate a square root of a quadratic by making a rationalizing substitution.
Write the integral in Problem 2 in terms of trig
functions of θ.
8 y
x
–2
7
4. Using the techniques you have learned for
integrating powers of trig functions, evaluate
the integral in Problem 3 algebraically.
1. The diagram above shows the circle for which
y H J
64 D x 2
Use the most time-efficient method to find the area
of the zone of this circle from x H D2 to x H 7.
2. In Problem 1, you evaluated the definite integral
7
D2
5. Do the reverse substitution to get the answer in
Problem 4 into terms of x.
64 D x 2 dx
numerically. Although the radical can be written as
(64 D x 2)1/2, explain why the indefinite integral
cannot be found using the power rule.
3. The radical in Problems 1 and 2 looks like the third
side of a right triangle (diagram below) with one
leg x, hypotenuse 8, and θ in standard position.
6. Use the answer to Problem 5 to find exactly the area
of the zone in Problem 1. Show that a decimal
approximation of this exact answer agrees with the
answer found numerically.
v
8
x
u
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Exploration 9-7a: Integrals of Rational Functions
by Partial Fractions
Date:
Objective: Integrate algebraically a function of the form (polynomial)/(polynomial).
1. The integral
10x D 32
dx
x D 4x D 5
2
has an integrand that is a rational algebraic
function. It can be transformed to
10x D 32
(x D 5)(x C 1) dx
The integrand can then be broken up into two
partial fractions,
A
B
C xD5 xC1
Find the constants A and B.
3. Ask your instructor to show you the way to find the
partial fractions in one step, in your head.
4. Study Heaviside’s method in Section 9-7 to see why
the shortcut works. Check here when you feel you
understand it well enough to explain it to other
members of your study group.
5. Heaviside’s method and its shortcut can be extended
to integrals with any number of distinct linear
factors in the denominator, as long as the numerator
is of degree at least one lower than the denominator.
Do this integration:
11x D 22x D 13
dx
x D 2x D 5x C 6
2
3
2
2. The integral in Problem 1 can be evaluated as a sum
of two terms, each of which has the form of the
reciprocal function. Find the indefinite integral.
6. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-8a: Integrals of Inverse
Trigonometric Functions
Date:
Objective: Use integration by parts to learn how to integrate an inverse
trigonometric function.
For Problems 1 and 2, let I H sinD1 x dx.
1. The integral I can be written as a product in two
different ways:
(1)(sin
D1
x dx)
or
(sin
D1
3. Find I H secD1 x dx. Assume that x > 0. To do this,
you will have to recall sec x dx and integration by
trigonometric substitution, as well as integration
by parts.
x)(1 dx)
Why is the first form not useful for integration by
parts, while the second form is?
2. Evaluate the integral to find I. Check your answer
with the answer in the text. If it does not agree, go
back and find the source of your differences.
4. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-9a: Introduction to Hyperbolic Functions
Date:
Objective: Starting with two combinations of exponential functions, show that they
have properties similar to those of trigonometric functions.
For this exploration, let
1
u H 2(e x C eDx )
1
and v H 2(ex D eDx)
1. On the same screen, plot the graphs of y1 H u, y2 H v,
and y3 H (1/2)e x using a window with [D2, 2] for x
and [D4, 4] for y. Sketch the results. Tell how the
graphs of u and v are related to the graph of (1/2)e x.
4. The equation u2 D v 2 H 1 that you found in
Problem 3 plots as a unit equilateral hyperbola in
a uv-coordinate system, as shown here. Use the two
parametric equations to find u(1.3) and v(1.3). Show
that the point (u, v) is on the hyperbola.
u
1
v
1
2. Show that uQ H v, uP H u, uQP H v, and u(4) H v, where
uQP and u (4) are the third derivative and the fourth
derivative of u, respectively. How do these
derivatives compare to the first four derivatives
of y H cos x?
5. Tell why the branch of the hyperbola for negative
values of u is extraneous.
6. The figure shows the unit circle in a uv-coordinate
system. Show that the point (u, v) H (cos 1.3, sin 1.3)
is on this unit circle.
1
v
u
3. Functions u and v are parametric functions of x.
Eliminate the parameter x by finding u 2 D v 2.
−1
1
−1
7. Why do you suppose that cosine and sine are called
circular functions and u and v at the beginning of
this Exploration are called hyperbolic functions?
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-9b: Chain Experiment
Date:
Objective: Find the particular equation for a hanging chain, and verify it by actual
measurement.
4. Calculate y for each 20 cm from the vertex, from
the left end to the right end of the chain. Use a
time-efficient method. Round to one decimal place.
y
x
y
x
y
x
1. Hang a chain from the frame around the chalkboard,
as shown in the diagram. Let x be the horizontal
distance (cm) from the vertex to a point on the chain,
and let y be the vertical distance (cm) from the chalk
tray to the point. Measure x and y for the vertex and
for the two endpoints.
Vertex:
Ends: Left:
Right:
2. From Problem Set 9-9, you found that the general
equation for a hanging chain is
h
w
y H wcosh hx C C
where h is the horizontal tension in the chain and
w is the weight of the chain per unit length. Let
k H h/w. Calculate the constants k and C. Show your
work. Store the answers in your grapher.
5. Remove the chain from the board. Plot the data on
the board as accurately as you can. Then hang the
chain again. How closely do the calculated data fit
the shape of the actual chain?
6. Use the equation to calculate the length of the chain
between the two endpoints. Then measure the chain
to see how close your calculated value is to the
actual value.
7. On the back of this sheet, derive the equation
h
w
y H wcosh hx C C
3. Weigh and measure the chain. Use the results to find
values of h and w.
using the fact that the tension in the chain is
equal to the vector sum of the horizontal tension
(of magnitude h) and the vertical tension, and is
always directed along the chain.
8. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-10a: Introduction to Improper Integrals
Date:
Objective: Find the limit of a definite integral as the upper limit of integration
becomes infinite.
Calvin is driving at 80 ft/s (about 55 mi/h). At time t H 0 s,
he lifts his foot from the accelerator and coasts. His
velocity decreases as shown in the graph.
3. Phoebe figures that her velocity is given by
v(t) H 80eD0.1t
How far has she gone after 10 s, 20 s, and 50 s?
v (t)
80
Area = distance
traveled.
t
0
10
b
20
1. Calvin figures that his velocity is given by
4. Find algebraically the distance Phoebe has gone after
t H b s. Use the result to find out how long it will
take her to go a distance of 1000 ft.
v(t) H 320(t C 4)D1
How far has he gone after 10 s, 20 s, and 50 s?
2. Find algebraically the distance Calvin has gone after
t H b s. Use the result to find out how long it will
take him to go a distance of 1000 ft.
Phoebe does the same thing Calvin did. Her velocity-time
graph is shown below.
v (t)
80
Area = distance
traveled.
5. Show that Phoebe’s distance approaches a limit as b
approaches infinity, but Calvin’s distance does not.
6. How do you reconcile the fact that Phoebe’s distance
approaches a limit, and the fact that her velocity
never reaches zero?
t
0
10
b
20
7. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 9-11a: Miscellaneous
Integration Practice!
Date:
Objective: Evaluate an antiderivative when the particular technique is not specified.
1.
2.
tan
5
4x dx
x 3eDx dx
1
1 C t dt
2
5.
3.
∞
4.
x
(x D 2)(x D 3)(x D 4) dx
tanh x dx
(Over)
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Exploration 9-11a: Miscellaneous Integration
Practice! continued
6.
sin
7.
(x
5
4
x dx
9.
x e
ax
cos bx dx
C 2)3 dx
10.
8.
e
Date:
2 x3
sin
D1
ax dx
dx
11. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration 8-7a
1. The graph agrees with the given figure.
10
2. r(0.3) H 3 D 2 c os 0.3 H 9.1799...
x(0.3) H r(0.3) • cos 0.3 H 8.7699...
y(0.3) H r(0.3) • sin 0.3 H 2.71286...
The sample point shown is at about (8.8, 2.7). The angle
measures about 17−, which is 0.296... radian. So all three
values agree with the graph.
7. The graph shows that at an apparent intersection point in the
second quadrant, r1, is positive, but r2 is negative, meaning
that the rose is being traced in the fourth quadrant when the
circle is being traced in the second quadrant. By the time the
rose gets to the second quadrant points, the circle is being
traced in the fourth quadrant.
Apparent
intersection
r1 is positive.
3. At θ H 0.3, the radius (of the sector of circle that the wedge
approximates) is r(0.3), and the central angle is dθ H 0.1, so
0.1
2
the approximate area is 2π • π • 9.1799... H 0.4589... unit.
r2 is negative.
dq
of a circle of radius r and area
4. The sector is approximately 2π
dq
1
• πr 2 H r 2 dθ.
πr 2, so dA H 2π
2
5. A H
θH2π
θH0
2
10
1
2 3 D 2 cos θ
dθ
H 84.2977... (numerically)
6. The y-radius is approximately 4.5, so the area formula
predicts that the area will be π • 6 • 4.5 H 84.8230.... Close!
(y-radius is exactly 6 sin (cosD1 23) H 4.4721..., making the area
π • 6 • 4.4721... H 84.2977..., which is precisely the numerical
answer.)
8. Answers will vary.
Exploration 8-7c
1. The graph agrees with the given figure.
2. r(0.8) H 35.9554... M 36.0 thousand miles
1
3. dA H 2r 2 dθ H 3528(10 D 11 cos θ)D2 dθ
7. Answers will vary.
Exploration 8-7b
1. The graph agrees with the given figure.
5.5
AH
0.8
3528(10 D 11 cos θ)D2 dθ H 308.9953...
M 309 million square miles
2
4. dL H (dr/dθ
C r 2 dθ
)
2
H [D84(1
0 D 11
cos θ
)211 sin θ]
C [84
(10 D 1
1 cos θ
)1]2 dθ
LH
5.5
θH0.8
dL H 82.4852... M 82.5 thousand miles
5. Check: r(0.8) M 36 as the comet approaches Earth, and
r(5.5) H 38.1015... M 38 as the comet recedes from Earth. So
the distance traveled on the curved path should be a bit
greater than 36 C 38 H 72 thousand miles. Thus the 82.5
thousand miles is reasonable.
6. Answers will vary.
2. 6 sin 2θ H 4
θ H 0.5 sinD1 (4/6)
H 0.5(0.7297... C 2πn) or 0.5(π D 0.7297... C 2πn)
H 0.3648... or 1.2059... in Quadrant I (Store as a and b.)
3. dA H 0.5r 22 D r 21 dθ
A H 0.5
b
a
(36 sin2 20 D 16) dθ
H 5.3132... square units
4. Counting squares in the region gives about 5.3, thus
confirming the answer.
5. A H 0.5
π/2
36 sin2 θ dθ
0
H 14.1371... H 4.5π, an interesting multiple of π.
Chapter 9
Exploration 9-3a
1.
x
2
sin 3x dx
u
dv
x2
+
sin 3x
–
– 3 cos 3x
2x
2
0
+
–
1
1
– 9 sin 3x
1
27 cos 3x
1
2
2
H D3x 2 cos 3x C 9x sin 3x C 27 cos 3x C C
6. dL H (dr)2 C
(r dθ)2 H 144 co
s2 (2θ
) C 36
sin2 (2θ)
dθ
LH
π/2
θH0
dL H 14.5326… units
14.5326.../π H 4.6258..., so the answer is not an interesting
multiple of π.
200 / Solutions for the Explorations
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2.
x
5
Exploration 9-4a
ln 4x dx
u
ln 4x
dv
x5
+
1.
sin
10
x dx
1
u
x6
x –1
6
------- ---------1
1
6
–
–9 sin8 x cos x
x5
1
+
0
sin 9 x
36
1 6
1
x 6 ln 4x D H6
36 x C C
3.
e
5x
–cos x
sin
x cos x C 9 sin
x cos x C 9 sin
8
x cos2 x dx
H Dsin9
8
x (1 D sin2 x) dx
sin x dx
10 sin x dx H Dsin x cos x C 9 sin x dx
sin x dx H D110 sin x cos x C 190 sin x dx
nD1
1
2. sin x dx H Dn sin
x cos x C n sin x dx
H Dsin9
cos 6x dx
8
10
u
e5x
5e
5x
25e
5x
dv
cos 6x
+
1
sin 6x
–
6
+
– 36 cos 6x
e
cos 6x dx
5
1
e5x sin 6x C e5x cos 6x C C
e cos 6x dx H 6
36
6 5x
5
e sin 6x C e5x cos 6x C C1
e5x cos 6x dx H 61
61
3
3. V H
dx
0
π(sin3 x)2 dx
π
4. V H
π
0
π (sin3 x)2 dx
π
Hπ
6
x cos2 x dx
H Dsin
8
x (1 D sin x) dx
10
10
9
9

Exploration 9-5a
2
sin x dx
10 sin x dx H Dsin x cos x C 9 sin x dx
sin x dx H D110 sin x cos x C 190 sin x dx
8
x dx
5. Answers will vary.
–cos x
8
H Dsin9
2
sin x
H Dsin9 x cos x C 9
9
3
dv
sin
x cos x C 9 sin
x cos x C 9 sin
4
1
5
3
H D6 sin5 x cos x D 24 sin x cos x
15 1
1
x
C
24 D2 sin x cos x C 2 C C
1
5
3
H D6 sin5 x cos x D 24 sin x cos x
5
5
x
D
16 sin x cos x C 16 C C
1
5
3
V H π D6 sin5 x cos x D 24 sin x cos x
π
5
5
x
D
16 sin x cos x C 16
0
5
π
Hπ 0D0D0C
16 C 0 C 0 C 0 C 0
5 2
π H 3.0842..., which agrees with Problem 3.
H
16
x dx
–9 sin8 x cos x
sin x dx
5 1
3
x cos x C 6 D4 sin x cos x C 4 sin
1
5
x dx H D6 sin5 x cos x C 6
1
H D6 sin5
16
+
–
sin6 x dx
0
sin
1
3
3
3
H 2x2 (ln x)3 D 4x 2 (ln x)2 C 4x2 ln x D 8x 2 C C
1
11
5. sin10 x (cos x dx) H 11 sin x C C
u
sin6 x dx
0
1
sin9 x
nD2
H 3.0842... (numerically)
2
6(ln x) x –1
4x
------------------------------------1
x
6 ln x
4
+
1 2
–1
x
6x
8
------------------------------------1
x
8
6
–
1 2
+
x
0
10
8
nD1
π
Hπ
u
dv
(ln x)3 + x
1 2
x
3(ln x)2 x –1
2
------------------------------------1
x
3(ln x)2
2
–
6. sin
8
(n ≠ 0)
5x
5x
4. x (ln x)
10
9
n
1
x dx D 9
9
10
5
1
25
e5x sin 6x C e5x cos 6x D H6
36
36
61
36
sin x
H Dsin x cos x C 9
9
x6
dv
+
–
x dx D9
1.
cos
7
1
x(sin x dx) H 8 cos8 x C C
10
8
8
7. Answers will vary.
2. The differential sin x dx of cos x does not appear in the
integrand.
3.
cos
7
(cos x) cos x dx
H (1 D sin x) cos x dx
H (1 D 3 sin x C 3 sin x D sin x) cos x dx
H cos x dx D 3 sin x cos x dx
C 3 sin x cos x dx D sin x cos x dx
x dx H
2
3
2
3
2
4
6
2
4
6
3
1
H sin x D sin3 x C 5 sin5 x D 7 sin7 x C C
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4. The power of cos x is even, so converting to powers of
(1 D sin2 x) leaves no cos x dx to be the differential of sin x.
5.
1
tan6 x C C
tan x (sec x dx) H 6
5
2
6. The differential sec x tan x dx of sec x does not appear in the
integrand.
7.
sec
8
(sec x) sec x dx
H (tan x C 1) sec x dx
H (tan x C 3 tan x C 3 tan x C 1) sec x dx
H tan x sec x dx C 3 tan x sec x dx
C 3 tan x sec x dx C sec x dx
x dx H
2
3
2
2
3
2
6
4
6
2
2
2
4
2
2
2
2
1
3
H 7 tan7 x C 5 tan5 x C tan3 x C tan x C C
Exploration 9-6a
1. A H 2
7
D2
64 D x 2 dx H 126.9622... (numerically)
2. The differential 2x dx of the “inside function” 64 D x 2 does
not appear in the integrand, so the integral cannot be done
as the antiderivative of a power.
x
3. 8 H sin θ ⇒ x H 8 sin θ
dx H 8 cos θ dθ
By the Pythagorean theorem, the horizontal leg of the right
triangle is 64 D x 2.
Thus, 8 cos θ H 64 D x 2.
64 D x dx H (8 cos θ) • 8 cos θ dθ
H 64 cos θ dθ
4. 64 cos θ dθ H 32 (1 C cos 2θ) dθ
N
2
2
2
8. The power of sec x is odd, so converting to powers of
(tan2 x C 1) does not leave sec2 x dx to be the differential
of tan x.
H 32θ C 16 sin 2θ C C
5. 64
9. Answers will vary.
cos
2
Exploration 9-5b
1. cos 2x H cos (x C x) H cos2 x D sin2 x
6. A H 2
2. cos 2x H cos x D sin x
H cos2 x D (1 D cos2 x)
H 2 cos2 x D 1
2
2
7
D2
θ dθ H 32θ C 16(2 sin θ cos θ) C C
64 D x x
x H 32 sinD1 8 C 32 • 8 • CC
8
1
D1 x
2
H 32 sin 8 C 2x 64 D x C C
2
64 D x 2 dx H 64 sinD1 8 C x 64 D x 2
D2
x
7
7
D1
H 64 sinD1 8 C 715
D 64 sinD1 4 C 260
H 126.9622...,
which agrees with the answer found numerically in
Problem 1.
3. cos 2x H cos2 x D sin2 x
H (1 D sin2 x) D sin2 x
H 1 D 2 sin2 x
7. Answers will vary.
4. cos 2x H 2 cos x D 1
1
cos2 x H 2(1 C cos 2x)
cos 2x H 1 D 2 sin2 x
1
sin2 x H 2(1 D cos 2x)
2
5.
6.
Exploration 9-7a
10x D 32
A
B
1. x D 5 C x C 1 H (x D 5)(x C 1)
⇒ A(x C 1) C B(x D 5) H 10x D 32
⇒ A C B H 10, A D 5B H D32
⇒ A H 3, B H 7
1 1
C cos 2x dx
cos x dx H
2 2
1
1
sin 2x C C
H 2x C 4
1 1
D cos 2x dx
sin2 x dx H
2 2
1
1
sin 2x C C
H 2x D 4
2
2.
H 3 ln x D 5 C 7 lnx C 1 C C
3. (Instructor input on Heaviside method.)
cos 7x sin 5x dx
u
cos 7x
–7 sin 7x
–49 cos 7x
+
–
+
4. See Section 9-7 for reasons behind the Heaviside method.
5. x 3 D 2x 2 D 5x C 6 H (x D 1)(x C 2)(x D 3)
11x 2 D 22x D 13
dx
(x D 1)(x C 2)(x D 3)
4
5
2
H x D 1 dx C x C 2 dx C x D 3 dx
dv
sin 5x
1
– 5 cos 5x
–
1
25
sin 5x
1
49
7
H D5 cos 7x cos 5x D 25 sin 7x sin 5x C 25
cos 7x sin 5x dx
24
D
25 cos 7x sin 5x dx
1
7
H D5 cos 7x cos 5x D 25 sin 7x sin 5x C C cos 7x sin 5x dx
5
7
H
24 cos 7x cos 5x C 24 sin 7x sin 5x C C1
7.
10x D 32
3
7
(x D 5)(x C 1) dx H x D 5 dx C x C 1 dx
cos ax sin bx dx
H 4 lnx D 1 C 5 lnx C 2 C 2 lnx D 3 C C
6. Answers will vary.
Exploration 9-8a
1. In the first form, dv would equal sinD1 x dx, and you don’t
know how to do this integration. In the second form, dv H 1
dx, which you can integrate.
a
a
H
sin ax sin bx C cos ax cos bx C C
a2 D b2
a2 D b2
8. Answers will vary.
202 / Solutions for the Explorations
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Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 203
1
(e2x C 2 C eD2x)
3. u 2 H 4
2.
u
dv
sinD1 x
+
dx
(1 D x 2)D1/2
–
x
(1 D x )
1
x C 2 (1 D x )
I H x sinD1 x D
H x sinD1
2 D1/2
1
(e2x D 2 C eD2x)
u2 H 4
1
(4)
u2 D v 2 H 4
2
2
u Dv H1
4. u(1.3) H 1.9709..., v(1.3) H 1.6903...
x dx
2 D1/2
v
(D2x dx)
1 2
H x sinD1 x C 2 • 1(1 D x2)1/2 C C
H x sinD1 x C 1 D x2 H C
which agrees with the text.
Here
1
u
1
3.
u
sin
D1
dv
x
+
1
2
1/2
x(x D 1)
–
D1
I = x sec
xD
For x I 0,
I H x secD1 x D
Let I1 H
dx
x
5. u is always positive. The extraneous negative branch arises
from squaring in Problem 3.
x
2
1/2 dx
x(x D 1)
6. cos 1.3 H 0.2674..., sin 1.3 H 0.9635...
1
dx
(x D 1)
2
1
1/2
v
1
dx.
(x D 1)
2
Here
1/2
Let x H sec θ.
N dx H sec θ tan θ dθ and
(x 2 D 1)1/2 H tan θ.
sec q tan q dq
I1 H H sec θ dθ
tan q
u
–1
1
–1
H lnsec θ C tan θ C C
= ln x C (x 2 D 1)1/2 C C
N I H x secD1 x D ln x C (x 2 D 1)1/2 C C,
which agrees, for x I 0, with the text’s
x secD1 x D sgn x lnx C x 2 D 1  C C.
7. The coordinates u and v defined here have the same
relationship to a unit hyperbola as u = cos x and v H sin x
have to a unit circle, hence the names circular functions and
hyperbolic functions.
4. Answers will vary.
8. Answers will vary.
Exploration 9-9a
Exploration 9-9b
1. u and v are asymptotic to y3.
1. (The following is simulated data, agreeing with the figure
shown in the Exploration. Actual data will depend on
the chain used.)
Vertex: (0, 20)
Left end: (D90, 120), Right end: (90, 120)
4 y
u
v
–2
x
2
–4
1
2. uQ = 2(e x D eDx ) H v
cosQ x = Dsin x
1 x
Dx
uP = 2(e C e ) H u
cosP x = Dcos x
1 x
uPQ = 2(e D eDx ) H v
cosPQ x = sin x
1 x
(4)
Dx
u = 2(e C e ) H u
cos(4) x = cos x
The u and v derivatives follow the same pattern, but without
any minus signs.
Calculus: Concepts and Applications Instructor’s Resource Book
©2005 Key Curriculum Press
1
2. y H k cosh kx C C
20 H k cosh 0 C C ⇒ 20 H k C C
90
120 H k cosh k C C
Substitute 20 D k for C.
90
120 H k cosh k C 20 D k
90
0 H k cosh k D 100 D k
Solving numerically gives k H 51.7801....
C H 20 D 51.7801... H D31.7801...
Equation is
1
x
y H 51.7801... cosh 51.7801... D 31.7801...
3. A typical medium-weight tow chain weighs about
0.015 lb/cm, which is the value of w.
h H (51.7801...)(0.015) M 0.8 lb
Solutions for the Explorations / 203
Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 204
4. x
y
0
20
20
23.9
40
36.2
6. It is a seeming paradox that the velocity remains positive but
the distance approaches a limit! The Greeks, including Zeno
of Elea (ca. 490–430 B.C.E.), wrestled with such paradoxes. It
was the invention of calculus that allowed people to find out
what happens in such situations.
60
58.8
7. Answers will vary.
80
95.1
Exploration 9-11a
5. The chain fits closely the data points.
y
1.
tan
5
4x dx H
(tan
3
4x)(sec2 4x D 1) dx
(tan 4x) dx
4x D (tan 4x)(sec 4x D 1) dx
1
4x D 8 tan 4x C tan 4x dx
1
4
H
16 tan 4x D
1
4
H
16 tan
1
4
H
16 tan
3
2
2
1
1
1
4
2
H
16 tan 4x D 8 tan 4x C 4 lnsec 4x C C
2.
1 C t dt
2
Let t/1 H tan θ.
Thus dt H sec θ dθ and 1 C t 2 H sec θ.
x
v
dx 2 C dy 2
6. dL H 1
H 1 C sinh2 kx dx
L H dL H 285.3490... M 285 cm
√1 + t 2
90
θ
D90
u
1
The measured length should be close to this.
1 C t dt
H sec θ sec θ dθ H sec
7. See the text derivation.
2
8. Answers will vary.
2
Exploration 9-10a
1. For Calvin, v(t) H 320(t C 4)D1, so distance d(t) is
d(t) H 320(t C 4)D1 dt H 320 ln (t C 4) C C for t L 0.
d(0) H 0 ⇒ C H D320 ln 4
d(t) H 320(ln (t C 4) D ln 4)
d(10) H 320(ln 14 D ln 4) H 400.8841... M 401 ft
d(20) H 320(ln 24 D ln 4) H 573.3630... M 573 ft
d(50) H 320(ln 54 D ln 4) H 832.8606... M 833 ft
2. d(b) H 320(ln (b C 4) D ln 4) for b L 0
1000 H 320(ln (b C 4) D ln 4)
b H 87.0395... M 87 s
(Exactly 4(e3.125 D 1))
3. For Phoebe, v(t) H 80eD0.1t, so distance d(t) is
d(t) H 80eD0.1t dt H D800eD0.1t C C.
d(0) H 0 ⇒ C H 800
d(t) H D800eD0.1t C 800 H 800(1 D eD0.1t )
d(10) H 800(1 D eD1) H 505.6964... M 506 ft
d(20) H 800(1 D eD2) H 691.7317... M 692 ft
d(50) H 800(1 D eD3) H 794.6096... M 795 ft
4. d(b) H 800(1 D eD0.1b )
1000 H 800(1 D eD0.1b )
1.25 H 1 D eD0.1b
eD0.1b H D0.25, which is impossible.
Phoebe never reaches 1000 ft!
5. lim 800(1 D eD0.1b) H 800 D 800 lim eD0.1b H 800
b→∞
t
b→∞
lim 320(ln (b C 4) D ln 4) H ∞, since ln x is unbounded.
b→∞
204 / Solutions for the Explorations
3
θ dθ
1
1
H 2 sec θ tan θ C 2 ln sec θ C tan θ C C
1
1
H 2 t 1 C t 2 C 2 ln
1 C t 2 C t C C
tanh x dx H ln (cosh x) C C
4. x e dx
3.
3
u
x3
3x 2
6x
6
0
Dx
+
–
+
–
+
dv
e –x
–e –x
e –x
–e –x
e –x
H DeDx(x 3 C 3x 2 C 6x C 6) C C
∞
1
x 3 eDx dx H lim
b
b→∞ 1
3
2
x 3 eDx dx
H lim (DeDb(b C 3b C 6b C 6) C 6)
b→∞
b3
∞
lim eDb b3 H lim b → ∞
b→∞
b→∞ e
2
3b
∞
H lim b → ∞
b→∞ e
6b
∞
H lim b → ∞
b→∞ e
6
6
H lim b → ∞
b→∞ e
H0
Similarly, each power of b multiplied by eDb approaches zero
as b approaches infinity.
N
∞
1
x 3 eDx dx H 0 C 0 C 0 C 0 C 6 H 6
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Calc2_IR_656_SE_09-12.qxd 10/26/06 5:52 PM Page 205
5.
x
(x D 2)(x D 3)(x D 4) dx
1
D3
2
H
x D 2 dx C x D 3 dx C x D 4 dx
4. t
H ln x D 2 D 3 lnx D 3 C 2 lnx D 4 C C
6.
sin
5
(1 D cos x) sin x dx
H (1 D 2 cos x C cos x) sin x dx
x dx H
2
2
2
7.
(x
4
2
1
H Dcos x C 3 cos3 x D 5 cos5 x C C
C 2) dx H
4
3
(x
12
4
1 13 2 9 12 5
x C x C x C 8x C C
H
13
3
5
x
9. e
8.
2
3
1
e x dx H 3
ax
x3
ae
ax
2
ax
1
–
+
b
sin bx
1
– b2
e
ax
cos bx dx
11
203
1910
30
D4
238
4115
40
D13
153
6070
50
D20
D12
6775
60
0
D112
6155
ax dx H x sinD1 ax D
ax dx H 0 C C
17
1
(6t D 0.3t 2) dt H (3t 2 D 0.1x 3)
17
1
1
4. No. 2[v(1) C v(17)] H 10.5, not 23.3.
5.
v(t )
ax
dx
1 D (a
x)2
1
H x sinD1 ax C a1
D (ax
)2 C C
37 2.8 ft
3. Average velocity H 17 D 1 s H 23.3 ft/s
cos bx dx
b
a
eax sin bx C eax cos bx C C1
H
a2 C b2
a2 C b2
sin
20
H 372.8 ft
ax
D1
455
2. Displacement H
e
D1
88
1. v(1) H 6 D 0.3 H 5.7
v(17) H 6 • 17 D 0.3 • 289 H 15.3
Both (1, 5.7) and (17, 15.3) appear to lie on the graph.
cos bx
sin
12
Exploration 10-3a
a ax
a2
1
H beax sin bx C e
cos bx D 2
2
b
b
a2 C b2
eax cos bx dx
2
b
1 ax
a ax
H e sin bx C e
cos bx C C
b
b2
10.
0
10
7. Answers will vary.
dv
cos bx
+
d
3
6. The object did not go back beyond its starting point since the
displacement at t H 60 is 6155 m, which is still positive.
1 3
(3x 2 dx) H 3e x C C
cos bx dx
u
e ax
a e
e
v
5
5. The fact that v changes sign somewhere between t H 0 and
t H 60 shows that the object stops and begins going
backward.
C 6x C 12x C 8) dx
8
a
0
30
(a ≠ 0)
(a H 0)
20
11. Answers will vary.
Chapter 10
10
Exploration 10-2a
1. Assume the average acceleration for the first time interval is
1
a M 2(5 C 12) H 8.5.
If the initial velocity is v H 3, then the velocity at the end of
the first time interval is
v H 3 C 8.5(10) H 88 m/s at t H 10.
2. t
a
v
0
5
3
10
12
88
20
11
203
30
D4
238
40
D13
153
50
D20
D12
60
0
D112
t
0
10
20
6. See the graph in Problem 5, showing that the area of the
rectangle equals the area of the shaded region.
7. The shaded area above the rectangle equals the unshaded
region within the rectangle.
8. There are about 26 squares under the curve for 0 K t K 10,
so the displacement is about 26 units, and the average
total disp.
26
H H 2.6.
velocity is 10
time
(The equation is v(t) H 6eD0.2t, so the area is 25.9399..., and
the average is 2.5939....)
9. Answers will vary.
3. Assume the average velocity for the first time interval is
1
v H 2(3 C 88) H 45.5.
If the initial displacement is d H 0, then the displacement at
the end of the first time interval is d H 0 C 45.5(10) H 455 m
at t H 10.
Calculus: Concepts and Applications Instructor’s Resource Book
©2005 Key Curriculum Press
Solutions for the Explorations / 205