<<

. 9
( 11)



>>

e’x dx
2
k=6
(a)
0
2
k = 10
sin(ex ) dx
(b)
’2
0
k=5
cos x 2 dx
(c)
’2
4 ex
k = 12
(d) dx
2 + sin x
0
12. Do each of the problems in Exercise 11 with “method of rectangles” replaced
by “trapezoid rule.”
13. Do each of the problems in Exercise 11 with “method of rectangles” replaced
by “Simpson™s Rule.”
This page intentionally left blank.
BIBLIOGRAPHY




[CRC] Zwillinger et al., CRC Press Handbook of Tables and Formulas,
34th ed., CRC Press, Boca Raton, Florida, 1997.
[SCH1] Robert E. Moyer and Frank Ayres, Jr., Schaum™s Outline of
Trigonometry, McGraw-Hill, New York, 1999.
[SCH2] Fred Sa¬er, Schaum™s Outline of Precalculus, McGraw-Hill,
New York, 1997.
[SAH] S. L. Salas and E. Hille, Calculus, John Wiley and Sons, New York,
1982.




263
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
This page intentionally left blank.
SOLUTIONS
TO EXERCISES




This book has a great many exercises. For some we provide
sketches of solutions and for others we provide just the answers.
For some, where there is repetition, we provide no answer. For the
sake of mastery, we encourage the student to write out complete
solutions to all problems.



Chapter 1
’5
1. (a)
24
43219445
(b)
1000000
’148
(c)
3198
19800
(d)
34251
’73162442
(e)
999000
’108
(f)
705

265
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
266 Solutions to Exercises

14
(g)
885
32115422
(h)
9990000
√ √
2. In Fig. S1.2, set A = 3.4, B = ’π/2, C = 2π , D = ’ 2+1, E = 3·4,
F = 9/2, G = ’29/10.

G B D A F CE

x
0

Fig. S1.2


3.



Fig. S1.3(a)





Fig. S1.3(b)





Fig. S1.3(c)





Fig. S1.3(d)





Fig. S1.3(e)




Fig. S1.3(f)
267
Chapter 1

√√ √
4. Let A = (2, ’4), B = (’6, 3), C = (π, π 2 ), D = (’ 5, 8), E = ( 2π, ’3),
F = (1/3, ’19/4).

y


C




B D


x

E
A
F



Fig. S1.4


5.




Fig. S1.5(a) Fig. S1.5(b)
268 Solutions to Exercises




Fig. S1.5(d)



Y
Fig. S1.5(c)
FL
AM
TE




Fig. S1.5(f)
Fig. S1.5(e)
269
Chapter 1

6.




Fig. S1.6(a) Fig. S1.6(b)




Fig. S1.6(c)
270 Solutions to Exercises




Fig. S1.6(d)




Fig. S1.6(f)
Fig. S1.6(e)
271
Chapter 1

4’6 ’2
slope = =
7. (a)
2 ’ (’5) 7
4’2
= 1 hence requested line has slope ’1
(b) Given line has slope
3’1
Write y = ’(3/2)x + 3 hence slope is ’3/2
(c)
Write x ’ 4y = 6x + 6y or y = (’1/2)x hence slope is ’1/2
(d)
9’1 ’8
slope = =
(e)
(’8) ’ 1 9
Write y = x ’ 4 hence slope is 1
(f)
Slope is ’3/8 hence line is y ’ (’9) = (’3/8) · (x ’ 4)
8. (a)
Slope is 1 hence line is y ’ (’8) = 1 · (x ’ (’4))
(b)
y ’ 6 = (’8)(x ’ 4)
(c)
3’4 1
= ’ hence line is y ’ 3 = (’1/8)(x ’ 2)
(d) Slope is
2 ’ (’6) 8
y = 6x
(e)
Slope is ’3 hence line is y ’ 7 = (’3)(x ’ (’4))
(f)
9.
y y

x

x




(a) (b)


y


y




x x



(c) (d)
272 Solutions to Exercises

y


y




x x



(e) (f)




10. (a) Each person has one and only one father. This is a function.
(b) Some men have more than one dog, others have none. This is not a
function.
(c) Some real numbers have two square roots while others have none.
This is not a function.
(d) Each positive integer has one and only one cube. This is a function.
(e) Some cars have several drivers. In a one-car family, everyone drives
the same car. So this is not a function.
(f) Each toe is attached to one and only one foot. This is a function.
(g) Each rational number succeeds one and only one integer. This is a
function.
(h) Each integer has one and only one successor. This is a function.
(i) Each real number has a well de¬ned square, and adding six is a well
de¬ned operation. This is a function.
11.




(a) (b)
273
Chapter 1




(c) (d)




(e) (f)



sin(8π/3) = sin(2π/3) = 3/2
12. (a)
√ √
tan(’5π/6) = [’1/2]/[’ 3/2] = 1/ 3
(b)

sec(7π/4) = 1/ cos(7π/4) = 2
(c)
√ √
csc(13π/4) = csc(5π/4) = 1/ sin(5π/4) = 1/[’ 2/2] = ’ 2
(d)
cot(’15π/4) = cot(’7π/4) = cot(π/4) = cos(π/4)/ sin(π/4) =
(e) √ √
[ 2/2]/[ 2/2] = 1

cos(’3π/4) = ’ 2/2
(f)
13. We check the ¬rst six identities.

cos π/3 = 1/2, sin π/3 = 3/2, cos2 π/3 + sin2 π/3 = [1/2]2 +
(a) √
[ 3/2]2 = 1/4 + 3/4 = 1.
√ √
cos π/3 = 1/2, sin π/3 = 3/2, ’1 ¤ 1/2 ¤ 1, ’1 ¤ 3/2 ¤ 1.
(b)
√ √2
tan π/3 = 3, sec π/3 = 2, tan π/3 + 1 = [ 3] + 1 = 3 + 1 =
2
(c)
22 = sec2 π/3.
√ √ √2
cot π/3 =√ 3, csc π/3 = 2/ 3, cot π/3 + 1 = [1/ 3] + 1 =
2
(d) 1/
4/3 = [2/ 3]2 = csc2 π/3.
274 Solutions to Exercises

sin(π/3 + (’π/6)) = sin(π/6) = 1/2, sin π/3 cos(’π/6) +
(e) √ √
cos π/3 sin(’π/6) = [ 3/2][ 3/2] + [1/2][’1/2] = 1/2.

cos(π/3 + (’π/6)) = cos(π/6) = √ cos π/3 cos(’π/6) ’
(f) 3/2,
√ √
sin π/3 sin(’π/6) = [1/2][ 3/2] ’ [ 3/2][’1/2] = 3/2.
14. We shall do (a), (c), (e).




F/2

Fig. S1.14(a)


Fig. S1.14(c)




Fig. S1.14(e)

= (15/2)—¦
15. (a) θ
= ’60—¦
(b) θ
= 405—¦
(c) θ
= (405/4)—¦
(d) θ
= (540/π )—¦
(e) θ
= (’900/π )—¦
(f) θ
θ = 13π/36 radians
16. (a)
θ = π/18 radians
(b)
θ = ’5π/12 radians
(c)
275
Chapter 1

θ = ’2π/3 radians
(d)
θ = π 2 /180 radians
(e)
θ = 157π/9000 radians
(f)
f —¦ g(x) = [(x ’ 1)2 ]2 + 2[(x ’ 1)2 ] + 3; g —¦ f (x) = ([x 2 + 2x +
17. (a)
3] ’ 1)2 .
3√
32
f —¦ g(x) = x ’ 2 + 1; g —¦ f (x) = [ x + 1]2 ’ 2.
(b)
f —¦ g(x) = sin([cos(x 2 ’ x)] + 3[cos(x 2 ’ x)]2 ); g —¦ f (x) =
(c)
cos([sin(x + 3x 2 )]2 ’ [sin(x + 3x 2 )]).
f —¦ g(x) = eln(x’5)+2 ; g —¦ f (x) = ln(ex+2 ’ 5).
(d)
f —¦g(x) = sin([ln(x 2 ’x)]2 +[ln(x 2 ’x)]); g—¦f (x) = ln([sin(x 2 +
(e)
x)]2 ’ [sin(x 2 + x)]).
’x 2 2 x2 2
f —¦ g(x) = e[e ] ; g —¦ f (x) = e’[e ] .
(f)
f —¦ g(x) = [(2x ’ 3)(x + 4)] · [(2x ’ 3)(x + 4) + 1] · [(2x ’ 3)(x +
(g)
4)+2]; g—¦f (x) = (2[(x(x +1)(x +2)]’3)([(x(x +1)(x +2)]+4).
f is invertible, with f ’1 (t) = (t ’ 5)1/3 .
18. (a)
g is not invertible since g(0) = g(1) = 0.
(b)
h is invertible, with h’1 (t) = sgn t · t 2 .
(c)
f is invertible, with f ’1 (t) = (t ’ 8)1/5 .
(d)
g is invertible, with g ’1 (t) = ’[ln t]/3.
(e) √
h is not invertible, since sin π/4 = sin 9π/4 = 2/2.
(f)
f is not invertible, since tan π/4 = tan 9π/4 = 1.
(g) √
g is invertible, with g ’1 (x) = sgn x · |x|.
(h)
19. We will do (a), (c), (e), and (g).




Fig. S1.19(a)
276 Solutions to Exercises




Fig. S1.19(c)
Fig. S1.19(e)


(g) Not invertible.

’1
Invertible, f (t) = t.
20. (a)
Invertible, g ’1 (t) = et .
(b)
Invertible, h’1 (t) = Sin’1 t.
(c)
Invertible, f ’1 (t) = Cos’1 t.
(d)
Invertible, g ’1 (t) = Tan’1 t.
(e)
Not invertible because h(’1) = h(1) = 1.
(f)

Invertible, f ’1 (t) = [3 + 9 + 4t]/2.
(g)


Chapter 2
lim x · ex = 0 because x tends to 0 and ex tends to 1.
1. (a)
x’0
x2 ’ 1
= lim x + 1 = 2.
(b) lim
x’1 x ’ 1 x’1
lim (x ’ 2) · cot(x ’ 2) = lim [(x ’ 2)/ sin(x ’ 2)] · cos(x ’ 2) =
(c)
x’2 x’2
1 · 1 = 1. [Here we use the non-trivial fact, explored in Chapter 5,
that lim (sin h/ h) = 1.]
h’0
lim x · ln x = lim ln x x = ln 1 = 0. [Here we use the non-trivial
(d)
x’0 x’0
fact, explored in Chapter 5, that lim x x = 1.]
x’0
’ 7t + 12
t2
= lim (t ’ 4) = ’1.
(e) lim
t ’3
t’3 t’3
277
Chapter 2

s 2 ’ 3s ’ 4
= lim (s + 1) = 5.
(f) lim
s’4
s’4 s’4
ln x
= lim ln[x 1/(x’1) ] = lim ln(1 + h)1/ h = ln e = 1.
(g) lim
x’1 x ’ 1 x’1 h’0
[Here we use the non-trivial fact, explored in Chapters 5 and 6, that
lim (1 + h)1/ h = e, where e is Euler™s number.]
h’0
x2 ’ 9
= lim x ’ 3 = ’6.
(h) lim
x’’3 x + 3 x’’3
2. (a) lim f (x) does not exist, so f is not continuous.
x’’1
lim f (x) = 1/2 and f (3) = 1/2 so f is continuous at c = 3.
(b)
x’3
lim f (x) = 0. If we de¬ne f (0) = 0, which is plausible from the
(c)
x’0
graph, then f is continuous at 0.
lim f (x) = 0. If we de¬ne f (0) = 0, which is plausible from the
(d)
x’0
graph, then f is continuous at 0.
lim f (x) = 1 and f (1) = 1 so f is continuous at c = 1.
(e)
x’1
lim f (x) does not exist so f is not continuous at c = 1.
(f)
x’1
lim f (x) = 0 and f (0) = 0 so f is continuous at c = π .
(g)
x’π
lim f (x) = 2 · e2 and f (2) = 2 · e2 so f is continuous at c = 2.
(h)
x’2
3. (a) We calculate
f (2 + h) ’ f (2)
f (2) = lim
h
h’0
[(2 + h)2 + 4(2 + h)] ’ [22 + 4 · 2]
= lim
h
h’0
[4 + 4h + h2 + 8 + 4h] ’ [4 + 8]
= lim
h
h’0
h2 + 8h
= lim
h
h’0
= lim h + 8
h’0
= 8.
The derivative is therefore equal to 8.
(b) We calculate
f (1 + h) ’ f (h)
f (1) = lim
h
h’0
278 Solutions to Exercises

[’1/(1 + h)2 ] ’ [’1/12 ]
= lim
h
h’0
’1 ’ [’(1 + h)2 ]
= lim
h(1 + h)2
h’0
2h + h2
= lim
h’0 h + 2h2 + h3
2+h
= lim
h’0 1 + 2h + h2
=2
The derivative is therefore equal to 2.
(x 2 + 1) · 1 ’ x · 2x 1 ’ x2
d x
= =2
4. (a) .
dx x 2 + 1 (x 2 + 1)2 (x + 1)2
d d d2
sin(x 2 ) = sin (x 2 ) · x = [cos(x 2 )] · 2x.
(b)


Y
dx dx dx
d d d
FL
tan(t 3 ’t 2 ) = tan (t 3 ’t 2 )· (t 3 ’t 2 ) = sec2 (t 3 ’ t 2 ) ·
(c)
dt dt dt
(3t ’ 2t).
2
AM

d x2 ’ 1 (x 2 + 1) · (2x) ’ (x 2 ’ 1) · (2x) 4x
= =2
(d) .
dx x 2 + 1 (x 2 + 1)2 (x + 1)2
x · cos x
cos x
d
[x · ln(sin x)] = 1·ln(sin x)+x · = ln(sin x)+
(e) .
TE



sin x sin x
dx
d s(s+2)
= es(s+2) · [1 · (s + 2) + s · 1] = es(s+2) · [2s + 2].
(f) e
ds
d sin(x 2 ) d
2 2
= esin(x ) · [sin(x 2 )] = esin(x ) · cos(x 2 ) · 2x.
(g) e
dx dx
ex + 1
1
ln(e + x) = x · (e + 1) = x
x x
(h) .
e +x e +x
Since the ball is dropped, v0 = 0. The initial height is h0 = 100.
5. (a)
Therefore the position of the body at time t is given by
p(t) = ’16t 2 + 0 · t + 100.
The body hits the ground when
0 = p(t) = ’16t 2 + 100
or t = 2.5 seconds.
(b) Since the ball has initial velocity 10 feet/second straight down, we
know that v0 = ’10. The initial height is h0 = 100. Therefore the
279
Chapter 2

position of the body at time t is given by
p(t) = ’16t 2 ’ 10 · t + 100.
The body hits the ground when
0 = p(t) = ’16t 2 ’ 10t + 100
or t ≈ 2.207 seconds.
(c) Since the ball has initial velocity 10 feet/second straight up, we
know that v0 = 10. The initial height is h0 = 100. Therefore the
position of the body at time t is given by
p(t) = ’16t 2 + 10 · t + 100.
The body hits the ground when
0 = p(t) = ’16t 2 + 10t + 100
or t ≈ 2.832 seconds.
1
d
sin(ln(cos x)) = cos(ln(cos x)) · · (’ sin x)
6. (a)
cos x
dx
’ sin x
= cos(ln(cos x)) · .
cos x
d sin(cos x)
= esin(cos x) · cos(cos x) · (’ sin x).
(b) e
dx
1
d
ln(esin x + x) = sin x · (cos x + 1).
(c)
+x
dx e
1
d
arcsin(x 2 + tan x) = · [2x + sec2 x].
(d)
1 ’ [x 2 + tan x]2
dx
’1 1 ex
d
arccos(ln x ’ e /5) = · ’
x
(e) .
1 ’ [ln x ’ e 5
dx x
x /5]2
1
d
arctan(x 2 + ex ) = · [2x + ex ].
(f)
1 + (x 2 + ex )2
dx
7. Of course v(t) = p (t) = 12t ’ 5 so v(4) = 43 feet/second. The average
velocity from t = 2 to t = 8 is
p(8) ’ p(2) 364 ’ 34
vav = = = 55.
6 6
The derivative of the velocity function is (v ) (t) = 12. This derivative never
vanishes, so the extrema of the velocity function on the interval [5, 10] occur
at t = 5 and t = 10. Since v(5) = 55 and v(10) = 115, we see that the
maximum velocity on this time interval is 115 feet per second at t = 10.
280 Solutions to Exercises

8. (a) We know that
1 1
[f ’1 ] (1) = =.
3
f (0)
(b) We know that
1 1
[f ’1 ] (1) = =.
8
f (3)
(c) We know that
1 1
[f ’1 ] (1) = = 2.
f (2) π
(d) We know that
1 1
[f ’1 ] (1) = =.
40
f (1)

Chapter 3
1.




Fig. S3.1

2. Figure S3.2 shows a schematic of the imbedded cylinder. We see that the
volume of the imbedded cylinder, as a function of height h, is
V (h) = π · h · (3 ’ h/2)2 .
Then we solve
0 = V (h) = π · 9 ’ 6h + 3h2 /4 .
281
Chapter 3




Fig. S3.2


The roots of this equation are h = 2, 6. Of course height 6 gives a trivial
cylinder, as does height 0. We ¬nd that the solution of our problem is height
2, radius 2.
3. We know that
V = ·w·h
hence
dV d dw dh
= ·w·h+ · ·h+ ·w·
dt dt dt dt
= 1 · 60 · 15 + 100 · (’0.5) · 15 + 100 · 60 · 0.3
= 900 ’ 750 + 1800 = 1950 in/min.
4. We know that v0 = ’15. Therefore the position of the body is given by
p(t) = ’16t 2 ’ 15t + h0 .
Since
0 = p(5) = ’16 · 52 ’ 15 · 5 + h0 ,
we ¬nd that h0 = 475. The body has initial height 475 feet.
5. We know that
1
V= · π r 2 · h.
3
Therefore
1 dh 1
d dr
0= V = · π · r2 · + · π · 2r · · h.
3 3
dt dt dt
At the moment of the problem, dh/dt = 3, r = 5, h = 12/(5π ). Hence
dr 12
0 = π · 52 · 32 + π · (2 · 5) · ·
dt 5π
282 Solutions to Exercises

or
dr
0 = 225π + 24 ·.
dt
We conclude that dr/dt = ’75π/8 microns per minute.
6. Of course
10000 = V = π · r 2 · h.
We conclude that
10000
h= .
π · r2
We wish to minimize
A = (area of top) + (area of sides)
10000
= π · r 2 + 2π · r · h = π · r 2 + 2π · r · .
πr2
Thus the function to minimize is
20000
A(r) = π · r 2 + .
r
Thus
20000
0 = A (r) = 2π r ’ .
r2
We ¬nd therefore that
10000
r3 =
π

or r = 3 10000/π . Since the problem makes sense for 0√ r < ∞, and
<
since it clearly has no maximum, we conclude that r = 10000/π, h =
3
√3
10000/π .
7. We calculate that g (x) = sin x +x cos x and g (x) = 2 cos x ’x sin x. The
roots of these transcendental functions are best estimated with a calculator
or computer. Figure S3.7 gives an idea of where the extrema and in¬‚ection
points are located.
8. We know that v0 = ’5 and h0 = 400. Hence
p(t) = ’16t 2 ’ 5t + 400.
The body hits the ground when
0 = p(t) = ’16t 2 ’ 5t + 400.
Solving, we ¬nd that t ≈ 4.85 seconds.
283
Chapter 3




Fig. S3.7

9. We see that
x
h(x) =
x2 ’ 1
x2 + 1
h (x) = ’ 2
(x ’ 1)2
2x(x 2 + 3)
h (x) =
(x 2 ’ 1)3
We see that the function is unde¬ned at ±1, decreasing everywhere, and
has an in¬‚ection point only at 0. The sketch is shown in Fig. S3.9.
10. We know that
4π 3
V= r.
3
Therefore

dV dr
= · 3r 2 .
3
dt dt
Using the values V = 36π, r = 3, dV /dt = ’2, we ¬nd that
dr
’2 = 4π · 32 ·
dt
hence
1
dr
=’ in. per sec.
18π
dt
284 Solutions to Exercises




Fig. S3.9

11. The acceleration due to gravity, near the surface of the earth, is about
’32 ft/sec2 regardless of the mass of the object being dropped. The two
stones will strike the ground at the same time.
12. He can drop a rock into the well and time how long it takes the rock to strike
the water. Then he can use the equation
p(t) = ’16t 2 + 0t + h0
to solve for the depth. If the well is very deep, then he will have to know
the speed of sound and compensate for how long it takes the splash to reach
his ears.
13. Refer to Fig. S3.13 to see the geometry of the situation.
Let (x, y) be the point where the rectangle touches the line. Then the area
of the rectangle is
A = x · y.
But of course 3x + 5y = 15 or y = 3 ’ (3/5)x. Hence
A = x · [3 ’ (3/5)x].
We may differentiate and set equal to zero to ¬nd that x = 5/2 and y = 3/2
is the solution to our problem.
14. Let s be a side of the base and let h be the height. The area of the base is s 2
and the same for the top. The area of each side is s · h. Thus the cost of the
base and top is
C1 = [s 2 + s 2 ] · 10 cents
285
Chapter 3

y




x




Fig. S3.13

while the cost of the sides is
C2 = 4(s · h) · 20 cents.
We ¬nd that the total cost is
C = C1 + C2 = 20s 2 + 80sh. (—)
But
100 = volume = s 2 · h
hence
h = 100/s 2 .
Substituting this last formula into (—) gives
8000
C(s) = 20s 2 + 80s · [100/s 2 ] = 20s 2 + .
s
We may calculate that
8000
0 = C (s) = 40s ’ .
s2
Solving for s gives the solution s ≈ 5.8479 and then h ≈ 2.9241.
15. We see that
x2 ’ 1
f (x) = 2
x +1
4x
f (x) = 2
(x + 1)2
’12x 2 + 4
f (x) =
(x 2 + 1)3
286 Solutions to Exercises


Thus there are a critical point at x = 0 and in¬‚ection points at x = ±1/ 3.
Figure S3.15 exhibits the complete graph.




Fig. S3.15

16. We see that the equation for the position of a falling body will now be
20 2
p(t) = ’ t + v0 t + h 0 .
2
It is given that v0 = 0 and h0 = 100. Hence
p(t) = ’10t 2 + 0t + 100.
The body hits the surface when
0 = p(t) = ’10t 2 + 100.

This occurs at time t = 10.

Chapter 4
F (x) = x 3 /3 + cos x + C
1. (a)
F (x) = e3x /3 + x 5 /5 ’ 2x + C
(b)
F (t) = t 3 + [ln t]2 /2
(c)
F (x) = ’ ln(cos x) ’ sin x ’ [cos 3x]/3 + C
(d)
F (x) = [sin 3x]/3 ’ [cos 4x]/4 + x + C
(e)
F (x) = esin x + C
(f)
’ cos x 2
x sin x dx = +C
2
2. (a)
2
287
Chapter 4

3 3
ln x 2 dx = ln2 x 2 + C
(b)
4
x
1
sin x · cos x dx = sin2 x + C
(c)
2
1
tan x · ln cos x dx = ’ ln2 cos x + C
(d)
2

sec2 x · etan x dx = etan x + C
(e)

12
(2x + 1) · (x 2 + x + 7)43 dx = (x + x + 7)44 + C
(f)
44
3. (a) We have
k 2
2 1
j j
x + x dx = lim 1+ + 1+ ·
2
k k k
k’∞
1 j =1
k
j2
2j j1
= lim 1+ + 2 +1+
k kk
k
k’∞
j =1
k
j2
2 3j
= lim + 2+ 3
k k k
k’∞
j =1

2 k2 + k 3 2k 3 + 3k 2 + k 1
= lim k· + · 2+ ·3
2 6
k k k
k’∞

3 3 1 1 1
= lim 2+ + ++ +2
2 2k 3 2k 6k
k’∞
31
=2+ +
23
23
= .
6
(b) We have
2
2j
’1 +
k
1 x2 2
k
’ dx = lim ’ ·
3 3 k
k’∞
’1 j =1
k
’2 4j 2
4j
= lim 1’ +2
3k k k
k’∞
j =1
288 Solutions to Exercises

k
8j 2
2 8j
= lim ’ + 2’ 3
3k 3k 3k
k’∞
j =1
’2 k 2 + k 8 2k 3 + 3k 2 + k ’8
= lim k · + · 2+ ·3
3k 2 6
3k 3k
k’∞
2 4 16
=’ + ’
3 3 18
2
=’ .
9
3
3 x3
x ’ 4x + 7 dx = ’ x + 7x
2 3 4
4. (a)
3
1 1
33 13
= ’ 34 + 7 · 3 ’ ’ 14 + 7 · 1
3 3
172
= (9 ’ 81 + 21) ’ (1/3 ’ 1 + 7) = ’ .


Y
3
6
2
sin2 x
6 ex
FL
x2
’ sin x cos x dx = ’
(b) xe
2 2
2
2
AM

e36 sin2 6 e4 sin2 2
= ’ ’ ’ .
2 2 2 2
4
’ cos x 2
ln2 x
4
TE



ln x
+ x sin x 2 dx = +
(c)
2 2
x
1
0

ln2 4 ’ cos 42 ln2 1 ’ cos 12
= + ’ +
2 2 2 2

ln2 4 cos 16 cos 1
= ’ + .
2 2 2
2
2 sin x 3
tan x ’ x cos x dx = ’ ln | cos x| ’
2 3
(d)
3
1 1
sin 23 sin 13
= ’ ln | cos 2| ’ ’ ’ ln | cos 1| ’
3 3
sin 8 sin 1
= ’ ln | cos 2| ’ + ln | cos 1| + .
3 3
e
ln2 x 2 ln2 e2 ln2 12
ln x 2
e
dx = = ’ = 1 ’ 0 = 1.
(e)
4 4 4
x
1
1
289
Chapter 4

8
sin2 x 3
8
x · cos x sin x dx =
2 3 3
(f)
6
4
4

sin2 83 sin2 43 sin2 512 sin2 64
= ’ = ’ .
6 6 6 6
5
5 x3 x2
Area = x + x + 6 dx = + + 6x
2
5. (a)
3 2
2 2
53 52 2 3 22 405
= + +6·5 ’ + +6·2 = .
3 2 3 2 6
π/4
sin2 x
π/4
Area = sin x cos x dx =
(b)
2
0
0

sin2 π/4 sin2 0 1
= ’ =.
2 2 4
2
2 2 2
2 e2 e1 e4
ex e
x2
Area = dx = = ’ = ’.
(c) xe
2 2 2 2 2
1
1
e
ln2 x ln2 e ln2 1
e ln x 10 1
Area = dx = = ’ =’=.
(d)
2 2 2 22 2
x
1
1
6.


y




1
2

π 2π x

positive for x µ(kπ/6, (k + 1) π/6), k even
negative for x µ(kπ/6, (k + 1) π/6), k odd




Fig. S4.6(b)
Fig. S4.6(a)
290 Solutions to Exercises




Fig. S4.6(c)

Fig. S4.6(d)


0 2
Area = ’(x + 3x) dx + x 3 + 3x dx
3
7. (a)
’2 0
0 2
x4 3x 2 x 4 3x 2
=’ + + +
4 2 4 2
’2 0

0 0 (’2)4 3 · (’2)2
=’’+ +
42 4 2

24 3 · 2 2 0 0
+ + ’’ = 20.
4 2 42

11 (j +1)π/6
Area = (’1)j sin 3x cos 3x dx
(b)
j =’12 j π/6
11 (j +1)π/6 1
= (’1)j sin 6x dx
2
j =’12 j π/6
(j +1)π/6
11
(’1)j cos 6x
= ’
2 6 j π/6
j =’12
291
Chapter 4

11
1 1
= ’’
6 6
j =’12
= 8.
1 e
ln x ln x
Area = ’ dx +
(c) dx
x x
1/2 1
1 e
ln2 x ln2 x
=’ +
2 2
1/2 1

ln2 (1/2) 12 02
0
=’’’ + ’
2 2 2 2
≈ 0.7404.
0 3
3 x4 4
Area = ’x e dx + x 3 ex dx
(d)
’3 0
0 3
4 4
ex ex
=’ +
4 4
’3 0

e81 e81 1 e81 1
1
=’’’ + ’ = ’.
4 4 4 4 2 2
2
2 x 4 3x 2
Signed Area = x + 3x dx = +3
8. (a)
4 2
’2 ’2

24 3 · 22 (’2)4 3 · (’2)2
= + ’ + = 0.
4 2 4 2
2π 2π 1
Signed Area = sin 3x cos 3x dx =
(b) sin 6x dx
2
’2π ’2π

1 cos 6x 1 1
= ’ =’ ’ ’ = 0.
2 6 6 6
’2π
e
ln2 x
e ln x
Signed Area = dx =
(c)
2
x
1/2
1/2
12 ln2 (1/2)
= ’
2 2
≈ 0.2598.
292 Solutions to Exercises

3
4
3 e81 e81
ex
3 x4
Signed Area = dx = = ’ = 0.
(d) xe
4 4 4
’3
’3
1 1
Area = [’3x + 10] ’ [2x ’ 4] dx = ’5x 2 + 14 dx
2 2
9. (a)
’1 ’1
1
’5x 3 ’5 5 74
= + 14x = + 14 ’ ’ 14 = .
3 3 3 3
’1
1
1 x3 x4
Area = x ’ x dx = ’
2 3
(b)
3 40
0
11 00 1
= ’ ’ ’ =.
34 34 12
1
1 x3
Area = [’x + 3] ’ 2x dx = ’ + 3x ’ x 2
2
(c)
3
’3 ’3
’27
1
= ’ + 3 · 1 ’ 12 ’ ’ + 3 · (’3) ’ (’3)2
3 3
32
=.
3
e
x2
e
Area = x ’ ln x dx = ’ [x ln x ’ x]
(d)
2
1 1
e2 ’ 3
e2 12
= ’ [e · 1 ’ e] ’ ’ [1 · 0 ’ 1] = .
2 2 2
π/4
x2
π/4
Area = x ’ sin x dx = + cos x
(e)
2
0 0
√ √
π2 π2
2 0 2
= + ’ +1 = + ’ 1.
32 2 2 32 2
3
3 x2
Area = e ’ x dx = e ’
x x
(f)
20
0
9 0 11
= e3 ’ ’ e0 ’ = e3 ’ .
2 2 2
1
1 x2 x3 11 00
Area = x ’ x dx = ’ = ’ ’ ’
2
10. (a)
2 3 23 23
0 0
1
=.
6
293
Chapter 5

1
1√ x 3/2 x 3
Area = x ’ x dx = ’
2
(b)
3/2 3
0 0
21 00 1
= ’ ’ ’ =.
33 33 3



√ 3 ( 3)5
3
3 x5
Area = √ 3x 2 ’x 4 dx = x 3 ’ = ( 3) ’
(c)
5 ’√3 5
’3


√ ( ’3)5 12 3
’ ( ’3)3 ’ = .
5 5
1
’2x 3
1 x5
Area = [’2x + 3] ’ x dx = + 3x ’
2 4
(d)
3 5
’1 ’1

<<

. 9
( 11)



>>