Calculus Third Edition: Chapter 8 Solutions
1.
2.
a.
cylinder
b.
Answers vary.
c.
An example is shown at right.
a.
The thickness times the area gives the volume of each slice if they are modeled as
cylinders, and the summation notations add up the volumes of n slices.
b.
By integration:
R
∫L A(x)dx , from the left side of the line to the right, where A(x) is the
area of each cross-section, which takes the limit of estimates with smaller and smaller
slices.
3.
a.
π
((
) + ( 14.32 )2 + ( 16.52 )2 + ( 182 )2 + ( 192 )2 + ( 18.12 )2 + ( 15.32 )2 + ( 112 )2 + ( 8.42 )2 + ( 27 )2 ) ⋅ 5
11.8 2
2
≈ 8259.404 cm3
4.
5.
1
b.
It would be more accurate.
c.
Integrate the areas, which takes the limit of using thinner and thinner sections of the
vase.
a.
dx represents an infinitesimal change in x.
b.
It represents the area of an infinitesimally thin rectangle.
y
See graph at right.
a.
height = x , width = Δx = ni (9 – 5)
in a Riemann sum or dx in an integral.
b.
∫4
9
x dx =
x3 2 9
3 4
2
=
2
3
(27 − 8) =
38
3
un2
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x
Calculus
6.
7.
a.
= 13 (5x 2 − 17x)−2 3 (10x − 17)
b.
= (12x2 + 2)(8 – 7x3) + (4x3 + 2x2 – 5)(–21x2)
= 96x2 – 84x5 + 16 – 14x3 – 84x5 – 42x3 + 105x2
= –168x5 – 56x3 + 201x2 + 16
c.
= cos[sin(t)] ⋅ cos(t)
d.
ln[cos2(x)][–sin(x)] – 7ln(49x2)
1
1
−1
−1
= π ∫ (4x 4 − 8x 2 + 4 − (x 4 − 2x 2 + 1))dx = π ∫ (3x 4 − 6x 2 + 3)dx
= π ( 53 x 5 − 2x 3 + 3x)
8.
10.
11.
2
(
)
= π 1 53 − (−1 53 ) = 165π
a.
= e3x sin(5x) + C
b.
=
c.
= ∫ sec(x) tan(x)dx −1 = sec(x) + C
d.
u = sin2(x), du = 2 sin(x) cos(x) dx = sin(2x) dx;
5
e4 x+2 =
1
4
3
= ∫ 2u du =
9.
1
−1
1
4
1
ln(2)
e22 − 14 e14
1 ⋅ 2 sin
2u + C = ln(2)
2(x)
∫ sin(2x)2sin
2(x)
dx
+C
2
2
∫0 (16x 3 − 16x 2 + 4x)dx = 4x 4 − 163 x 3 + 2x 2 0 = 4 ⋅16 − 163 ⋅ 8 + 8 = 72 − 1283 = 29 13
e.
=
a.
250 ⋅ 1.1211 ≈ $869.64
b.
11
1
x
11 0 250 ⋅1.12 dx
a.
At x = 2 there is a hole.
b.
At x = –4 and 1 there is no limit, and at x = 2 g is undefined.
c.
At x = –4, 1, and 2, because it is not continuous, and at x = –2 there is a cusp.
a.
3
∫
1 ⋅ 250 ⋅
1
x
= 11
ln(1.12) ⋅1.12
b.
–2
11
0
=
c.
250(1.1211 −1)
11 ln(1.12)
≈ $497.06
2
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d.
Does not exist.
Calculus
12.
height = f (x) or
b.
The slice is a thin cylinder with depth Δx and radius f (x), and its volume is
V = Δx ⋅ π f (x)2 = Δx ⋅ π ( x )2 .
c.
Cylindrical disks represent the area of the circle (a circle is just the set of points that
make up the circumference). Cylindrical disks are cylinders whose thickness is very
small.
8
∫0
d.
13.
x , width = Δx or dx
a.
8
8
2 8
0
0
0
dx ⋅ π f (x)2 = ∫ π ( x )2 dx = ∫ π ⋅ x dx = π 2x
= 32π un3
a.
Again, each slice is a thin cylinder (or disk), with thickness Δx (or dx).
b.
Δx(π)( f (c))2
y
b
∫a
c.
π ( f (x))2 dx
y
14.
See graph at right.
a.
x
x
See diagram at far right.
2
3
∫0 π (x 2 )2 dx = ∫0 π x 4 dx = π
b.
15.
3
3
x5
5
0
radius = x
=π
35
5
=
243 π
5
un3
5
8
5
8
2
5
2
5
width = Δx
π ∫ ( 6 − x )2 dx + π ∫ (x − 4)2 dx = π ∫ (6 − x)dx + π ∫ (x 2 − 8x + 16)dx =
(
π 6x − 12 x 2
) 2 + π ( 13 x 3 − 4x 2 + 16x ) 5 = 28.5π ≈ 89.54 units3
5
8
y
16.
17.
3
See graph at right.
4
a.
2
See sketch at far right.
Radius = x; x2 = y or x = y
Thickness: dy
4
b.
π∫
a.
π∫
0
2
4
radius = x
x
thickness = dy
( y ) dy = π ∫0 ydy = 12 π y2 04 = π (8 − 0) = 8π un 3
2
(
2 1
0 2
4
)
2
x 3 + 3 dx
b.
( 3 2y − 6 )
3
π∫
7
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2
dy
Calculus
18.
a.
u = 6x4 – 3x, du = 24x3 – 3:
−1 dx = 1
∫ 68xx4 −3x
∫3
(24 x 3 −3)dx
6 x 4 −3x
3
= π ⎛ (x+1)
− 9x ⎞
⎝ 3
⎠
5
3
b.
=π
2
=∫
1 du
3 u
= 13 ln u + C = 13 ln 6x 4 − 3x + C
( 2163 − 45 ) − π ( 273 − 18 ) = 27π − (−9)π = 36π
2
c.
= 5 ⋅ x2 − 2x + 5 ln x + 3 + C
d.
1 sin(11x 2 − 3) − 22x dx
u = 11x 2 − 3, du = 22xdx : ∫ 2x sin(11x 2 − 3)dx = ∫ 11
1 sin(u) du = − 1 cos(u) + C = − 1 cos(11x 2 − 3) + C
= 11
∫
11
11
e.
u = x + 2, x = u – 2, du = dx:
∫
0
2 x dx
−1 x+2
0 2(u−2)
du
−1 u
=∫
= 2(x + 2) − 4 ln x + 2
f.
u = 3x, du = 3dx:
∫
0
−1
0
( 2 − u4 ) du = 2u − 4 ln u −1
−1
=∫
0
= ( 4 − 4 ln(2) ) − ( 2 − 4 ln(1) ) = 2 − 4 ln(2)
3
1−9 x 2
dx = ∫
du
1−u 2
= sin–1(u) + C = sin–1(3x) + C
19.
It will look like a cone with a ridge. See diagram at right.
20.
a.
= lim ( 2−1 x +
b.
0
= 18−18+1
=
c.
5
= 2(−∞) − 3 + (−∞)+6
= −∞ + 0 = –∞ or DNE
d.
(2− x )
x→4 2− x
e.
1− 5 + 8
2
3
lim 24x 54x
+
x→
x2 x 3
f.
= lim
x→2
lim
0
1
x−1 ) = lim 1+(x−1)
2− x
x→2 2− x
= lim
x
2−
x
x→2
Does not exist.
=0
= lim 2 + x = 2 + 2 = 4
x→4
= ∞ or DNE
(x−2)(2 x−1)
(x−2)(5
x+ 3)
x→2
= lim
2 x−1
x→2 5 x+ 3
3
= 13
21.
πR2 – πr2
22.
B: The middle integral computes a different area, but I and III compute the area of the same
region.
4
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Calculus
(ln(x))3
x
23.
C. If you take the derivatives of I and II you get
24.
a.
10 inches
b.
0.5 inches
c.
π(radius)2(thickness) – π(radius of hole)2(thickness)
d.
π(10 – 0.5i)2(2) – π(0.5)2(2)
9
e.
.
9
∑ ( π (10 − 0.5i)2 − π (0.5)2 ) ⋅ 2 = ∑ 2π ( (10 − 0.5)2 − 0.25 )
i=0
i=0
10
(
or ∑ 2π (5 + 0.5i)2 − 0.25
i=1
25.
)
f.
1237.5π = 3887.721 in3
a.
A cylinder with a cone cut out of it: V = 4 ⋅ π ⋅ 4 2 − 13 ⋅ 4 ⋅ π ⋅ 4 2 = 64π −
b.
Δx ⋅ (πR2 – πr2) where R is the bigger radius and r is the smaller radius:
Δx ⋅ (π ⋅ 42 – π ⋅ x2) = Δx(16π – x2π)
c.
4
∫0
(16π − x 2π )dx = 16π x −
x3
3
π
4
0
64
3
π = 1283 π un 3
128π
3
= 64π − 64
3 π = 3 un
y
26.
a.
See graph above right.
4
9
3
4
π ∫ (y − 3)2 dy + π ∫ (− y + 3)2 dy ≈ 5.760 un 3
b.
2
See graph below right.
∫0 π ( (x − 3)4 − (x + 3)2 ) dx ≈ 93.829 un 3
1
–3
3
x
3
x
–2
y
5
–3
–5
5
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Calculus
27.
a.
See diagram at right. The washers are
perpendicular to the x-axis.
b.
Δx ⋅ (π(–2x2 + 2)2 – π(–x2 + 1)2)
c.
∫−1 (π (−2x 2 + 2)2 − π (−x 2 + 1)2 )dx
d.
y
2
1
x
1
2 ∫ (π (−2x 2 + 2)2 − π (−x 2 + 1)2 )dx
0
–2
28.
a.
Vertical rectangles:
A=
b.
4
∫1 (
x3 2
3
2
+
x2 4
2 1
=
(
8
3 2+8
)−(
2
3
+
1
2
) = 163 + 243 − 67 = 806 − 67 = 736 un2
Horizontal rectangles. Intersections are (0, 2) and (3, –1):
A=∫
2
−1
=
29.
x − (−x))dx =
−8
3
y = ±3 1 −
((−y 2
+ 4) − (2 − y))dy = ∫
−1
+2+4−
x2
4
2
(
1+1
3 2
)
−2 =
10
3
(−y 2
+ y + 2)dy = −
( )=
− −
7
6
and x goes from –2 to 2: A =
20
6
2
+ −
7
6
∫−2 2 ⋅ 3
27
6
1−
=
y3
3
9
2
+
=
x 2 dx
4
y2
2
+ 2y
2
−1
4.5 un 2
=
2
∫−2 3
4 − x 2 dx ≈ 18.850 un 2
or 6π using a calculator or noticing that it is a semicircle.
30.
y = ±3 1 −
x2
4
: the horizontal cross-sections are disks with radius 3 1 −
2
⎛
V = ∫ π ⎜ 3 1−
−2 ⎝
x2
4
2
x2
4
so
2
2
⎞
9π x 3 3
⎟⎠ dx = ∫−2 9π x − 4 ⋅ 4 x
−2
= (18π – 6π) – (–18π + 6π) = 12π – (–12π) = 24π un3
6
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Calculus
31.
a.
x 2 − x−6
x→3 x− 3
lim
b.
(x− 3)(x+2)
x− 3
x→3
= lim
= lim (x + 2) = 5 , so f (3) must be 5, h = 5.
x→3
Except h = 1 does not work, notice the original equation is
sqaured, which left room for –3 + h and 3 + h to be
opposites, so h = 6.
−3 + h = 3 + h
(−3 + h)2 = 3 + h
9 − 6h + h 2 = 3 + h
h 2 − 7h + 6 = 0
(h − 6)(h − 1) = 0 → h = 1 or 6
c.
a: When graphed, the function looks like the line x + 2. h = 5, fills in the hole at
(3, 5). Therefore the function is differentiable for all x.
b: f ′( x + 6 ) =
=
1
2 x+6
=
1
2 3+6
1
6
at x = 3.
f ′(–x + 6) = –1 at x = 3.
Therefore the function is not differentiable at x = 3.
32.
a.
Use the reverse Chain Rule. u = e x , du = e x dx : ∫
ex
0 (2−e x )2
=
b.
=
1
2−e
1 =
− 2−1
1
2−e
1
du
x=0 (2−u)2 x=0
1
− 1 = 1−(2−e)
2−e =
e−1
2−e
x
4
∫
x2
f (x)dx = h1 (x)⋅ g1 (h(x)) = 2x ⋅ f (x 2 )
u = sec(x), du = sec(x) tan(x)dx:
∫
sec(x) tan(x)
dx
1+sec2 (x)
=∫
d.
u = x2, du = 2xdx:
e.
=∫2
du
1+u 2
∫
= tan −1 (u) + C = tan −1 ( sec(x) ) + C
x
1−x 4
π
= ∫ 12 ⋅
( )
π /2
= −2 ln u x=0 = −2 ln ( cos ( 2x ) )
sin(x)/2
dx, u − cos 2x
0 cos(x)/2
= −2 ln
33.
1
1
2−e x 0
dx = ∫
Use the Chain Rule with g(x) = ∫ f (x)dx and h(x) = x2:
d
dx 4
c.
1
f ′(x) =
( ) + 0 = −2 ln(2
2
2
(x−3)(−1)−(1)(1−x)
(x−3)2
=
2 xdx
1−x 4
=
, du = − 12 sin
−1 2 ) =
3−x−1+x
(x−3)2
=
π /2
x=0
1
2
∫
( )
x
2
du
1−u 2
= 12 sin −1 (u) + C = 12 sin −1 (x 2 ) + C
π
dx : ∫ 2
sin(x)/2
dx
0 cos(x)/2
=∫
π /2
x=0
−2du
u
( ( ) ) + 2 ln ( cos(0) )
= −2 ln cos
π
4
( − 12 ) ⋅(−2) ln(2) = ln(2) ≈ 0.693
2
(x−3)2
, f 11 (x) = −2 ⋅
2
(x−3)3
=
−4
(x−3)3
, which is positive
when x < 3: in (–∞, 3).
7
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Calculus
34.
Intersections are when
3 = x = 25 − y 2 , 9 = 25 − y 2 , y 2 = 16, y = ±4; (3, ±4)
V=∫
4
−4
π
(
25 − y 2
) − π (3) dy = ∫
2
2
4
−4
4
π (25 − y 2 − 9)dy = ∫ π (16 − y 2 )dy
−4
( 12 ) = 0
35.
Yes (Rolle’s Theorem) or f′ (x) = 2x – 1; f ′
36.
a.
Horizontally, washers.
b.
Horizontally, prisms.
c.
Vertically, disks.
a.
b.
A disk (thin cylinder), with thickness Δy or dy.
y
37.
c.
d.
8
10
∫0
10
2 10
π ( y )2 dy = ∫ π y dy = π ⋅ y2
0
The inverse functions is y = x :
= 50π un 3
0
10
∫0
2 10
π ( x )2 dx = π ⋅ x2
0
= 50π un 3
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Calculus
38.
a.
Since most of the region is closer to the y-axis than the x-axis, one
should suspect the volume of the rotation about the x-axis to have
greater volume.
b.
Rotation about the x-axis produces a cone with a chocolate
chip shape of equal height and radius removed. Rotation
about the y-axis produces a bell shape with a cone of equal
height and radius removed. For rotation about the x-axis,
the inner and outer radii are x2 and 2x and the width is Δx:
y
x
2
U = ∫ (π (2x)2 − π (x 2 )2 )dx
0
2
= ∫ π (4x 2 − x 4 )dx = π ⋅ 43 x 3 − π
0
x5
5
32π
3
32π
5
64 π
15
2
0
=
−
=
For rotation about the y-axis, the inner and outer
radii are
y
2
and
Δy = V = ∫
4
0
( (π (
y and the width is
c.
39.
)
y
4
= 8π –
y
y )2 − π ( 2 )2 dy
= ∫ π ⎛ y−
⎝
0
16
3
≈ 13.404 un 3
y2
4
π=
⎞ dy = π
⎠
8
3
y2
2
y3
12
−π
4
0
x
π ≈ 8.378 un
3
Most of the region is closer to the y-axis and farther from the x-axis. So even though
the second integral is twice as long (0 to 4), the radii are about 12 the size of the radii
of the first integral, and within the integral the radii are squared. The height in the xaxis revolution is greater. Since the height determines the radius of the washers, and
since the radii are squared, the volume will be greater.
These are washers with inner and outer radii of 2 – x and 4 – x2
and width Δx: V = ∫
2
−1
2
=∫
−1
2
( π (4 − x 2 )2 − π (2 − x)2 ) dx
π (16 − 8x 2
+
x4
− 4 − 4x −
y
x
x 2 )dx
= ∫ π (x 4 − 9x 2 + 4x + 12)dx
−1
=π
(
x5
5
3
2
− 9 2x + 4 2x + 12x
)
2
−1
= π (14.4 − (−7.2)) = 21.6π ≈ 67.858 un 3
9
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Calculus
40.
a.
1
(
x7
7
23 + π ⋅ 9
⋅ 14
14
=
1
π (x 3 + 1)2 dx = ∫ π (x 6 + 2x 3 + 1)dx = π
∫−1
−1
=π
(
1
7
) (
)
+ +1 − π − + −1 = π
1
2
1
7
1
2
4
+ 2 ⋅ x4 + x
32
14
π=
16
7
)
1
−1
π ≈ 7.181 un 3
b.
A washer.
c.
The inner radius is 1 (the distance between y = –1 and the x-axis) and the outer radius
is f (x) + 1 = x3 + 1 – (–1) = x3 + 2 (the distance between y = –1 and f (x)).
d.
1
1
π (x 3 + 2)2 − π ⋅12 dx = π ∫ x 6 + 4x 3 + 4 − 1dx = π
∫−1
−1
=π
e.
(
x7
7
+ x 4 + 3x
( 17 + 1+ 3) − π ( − 17 + 1− 3) = 297 π + 157 π = 447 π ≈ 19.747 un 3
)
1
−1
Basically the integral is the same and gives the same answer:
1
∫−1 π (x 3 + 2)2 − π ⋅12 dx = 447π ≈ 19.747 un 3
41.
For a disk, it is as if it were a washer with an inner radius of 0.
42.
a.
4
∫2
u = 9 –2x, du = –2dx:
1
9−2 x dx
=∫
4
x=2
− 12 u1 du = − 12 ln u
= − 12 ln 9 − 2x
6
3
3
= 2 ⋅ x6 + 3⋅ x3 + x
c.
=∫
d.
=
e.
u = sin(x), du = cos(x) dx:
2 (x−1)(x+1)
dx
x−1
3
4
∫1
−3
=
x6
3
b.
+ x3 + x
−3
3
2
52 4
1
π 2
= 85 ⋅ x 5 2
x=2
x=2
= − 12 ⋅(0) + 12 ⋅ ln(5) = 12 ln(5)
3
= − ∫ (x + 1)dx = −
4x 3 2 dx = 4 ⋅ x5/2
4
4
= 243 + 27 + 3 – (243 – 27 – 3) = 60
(
4
1
x2
2
+x
)
=−
3
2
= 85 ⋅ (32 − 1) =
( 92 + 3) + (2 + 2) = − 27
8⋅31
5
=
248
5
π 2
∫π 4 sin 3 (x) cos(x)dx = ∫x=π 4 u 3 du
=
10
u4
4
π /2
x=π /4
=
sin 4 (x)
4
π /2
π /4
=
1
4
−
( 2 2)4
4
=
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1
4
1 = 3
− 16
16
Calculus
43.
=
b.
= sec[5 ln(x)] tan[5 ln(x)] ⋅
45.
= 18 x+6−182x+6 =
x 3 2 cos(2 x)− 3x 2 sin(2 x)
x6
c.
44.
(3x+1)⋅6− 3⋅(6 x−2)
(3x+1)2
a.
(3x+1)
=
12
(3x+1)2
5
x
2 x cos(2 x)− 3 sin(2 x)
x4
d.
= 2k ⋅ 93k–1 + k2 ⋅ (ln(9)) ⋅ 93k–1 ⋅ 3 = 2k ⋅ 93k–1 + (3 ln(9))k293k–1
e.
= [2 sin(x) cos(x)]cos(x) + sin2(x)[– sin(x)] = 2 sin(x) cos2(x) – sin3(x) or
2 sin(x)[1 – sin2(x)] – sin3(x) = 2 sin(x) – 3 sin3(x)
a.
s(0) = 136 ft
b.
s′(t) = –32t + 120, so initial velocity is s′(0) = 120 ft/s, which is just the t term.
c.
s′(t) = –32t + 120 = 0, 120 = 32t, t =
d.
s
e.
0 = s(t) = –16t2 + 120t + 136, 0 = –2t2 + 15t + 17 = (–2t + 17)(t + 1),
17
17
t = 17
2 : s ′ 2 = −32 2 + 120 = –272 + 120 = –152 ft/s: it will explode
120
32
=
30
8
3
= 15
4 = 3 4 sec
( 154 ) = −16 ( 154 )2 + 120 ( 154 ) + 136 = –225 + 450 + 136 = 361 ft
( )
( )
Let x be the total price deduction, so the charge is 159 – x, the occupancy is 265 + 2x, and
the revenue is (159 – x)(265 + 2x).
d
d
0 = dx
((159 − x)265 + 2x)) = dx
(159 ⋅ 265 − 265x + 318x − 2x 2 )
=
d
dx
(159 ⋅ 265 + 53x − 2x 2 ) = 53 − 4x, x =
53
4
= $13 14
So they should charge 159 − 13 14 = $145.75 .
46.
11
1
2
a.
k=
c.
k = –1
(by symmetry)
b.
k=1
d.
k=1
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Calculus
47.
1− 3x 2
2y
y = ± x − x 3 , dx = ± 12 (x − x 3 )−1 2 ⋅(1− 3x 2 ) = ±
b.
There is a vertical tangent when dx is undefined. This occurs when the denominator
from part (a) is 0. 2y = 0, so y = 0 at each of these points.
c.
x = 0, 1, –1
dy
or
dy
0 = 2 x − x3
−6 x⋅2 y−2(1− 3x 2 )
d.
48.
1−3x 2
2 x−x 3
a.
dy
dx
4 y2
=
0 = x − x3
2
−12 xy2 −(1− 3x 2)
4 y3
0 = x(1 − x 2 )
0 = x(1 − x)(1 + x)
x = 0, 1, −1
The vertex is (0, 0): the distance is d(t) = (y − 0)2 + (x − 0)2 = (x 2 )2 + x 2 = x 4 + x 2 ,
so
d
dt
d(t) =
d
dt
x 4 + x 2 , d ′(t) =
1
2
(x 4 + x 2 )−1 2 ⋅ (4x 3 + 2x) ⋅ x1 (by the extended Chain Rule.
d ′(−3) = 12 (81+ 9)−1 2 ⋅(−4 ⋅ 27 − 6)⋅(0.5) =
1
2 90
⋅(−114)⋅ 12 =
−57
2 90
=
−57
6 10
≈ −3.004 units/s
or 3.004 units is also appropriate.
49.
a.
The shape is a cylinder with a whirlwind cut out of it. A slice is a washer with inner
and outer radii of x2 and 4 and width Δx.
2
2
0
0
(
V = ∫ (π ⋅ 4 2 − π (x 2 )2 )dx = ∫ (16 − x 4 )dx = π 16x −
(
)
x5
5
)
2
0
128π
3
= π 32 − 32
5 = 5 ≈ 80.425 un
b.
V=
c.
y and width Δy .
The shape is like a bowl. A slice is a disk with radius
4
∫0 π (
4
y )2 dy = π ∫ ydy = π ⋅
0
y2 4
2 0
= 8π ≈ 25.133 un 3
The shape is almost like a quarter of a donut and will be hard to draw. A slice is a
washer with inner and outer radii of 1 + x2 and 5 and width Δx.
2
2
0
0
(
V = ∫ (π ⋅ 5 2 − π (1+ x 2 )2 )dx = π ∫ (25 − 1− 2x 2 − x 4 )dx = π 24x − 2 ⋅ x3 −
=π
d.
(
48 − 163
−
32
5
)=π ( )=
544
15
x5
5
)
2
0
≈ 113.935 un 3
Again this is like a quarter donut. A slice is a washer with inner radius r − y and
outer radius 3 and width Δy.
( π ⋅ 32 − π (3 −
0
V=∫
)
4
y3 2
= π ⎛ 6⋅ 3 2 −
⎝
12
544 π
15
3
y2
2
⎞
⎠
(9 − 9 + 6
0
y )2 dy = π ∫
4
0
4
= π ⎛ 4 ⋅ y3 2 −
⎝
y2
2
⎞
⎠
4
)
y − y dy = π ∫
4
0
(6
)
y − y dy
= π (32 − 8) − 0 = 24π ≈ 75.398 un 3
0
© 2017 CPM Educational Program. All rights reserved.
Calculus
50.
a.
The radius of the outside cylinder is 5. The radii of the inside disks are y = 2x + 1.
2
2
0
0
2
2
0
0
π ∫ 5 2 dx − π ∫ (2 x + 1)2 dx = π ∫ 25 − 2 2 x − 2 ⋅ 2 x − 1dx = π ∫ (24 − 2 2 x − 2 x+1 )dx
(
2x
x+1
= π 24x − 22ln(2) − 2ln(2)
(
)
)
2
=π
0
(( 48 −
16
2 ln(2)
) (
8
1 − 2
− ln(2)
− 0 − 2 ln(2)
ln(2)
))
27
= π 48 − 2 ln(2)
≈ 28.524 un 3
b.
The radius is x, but y = 2x + 1 or x =
5
(
2
c.
)
5 ln(y−1) 2
dy
ln 2
2
π ∫ x 2 dy = π ∫
2
2
0
0
≈ 16.981 un 3
(
x
2x
)
2
2
2
0
0
≈ 8.196π ≈ 25.747 un 3
0
The radius is x + 1, but y = 2x + 1 or x =
( (x + 1)2 − 12 ) dy = π ∫2 ( ln(y−1)
ln(2) + 1 )
2
π∫
e.
.
π ∫ (5 − y)2 dx = π ∫ (5 − (2 x + 1))2 dx = π ∫ (4 − 2 x )2 dx = π ∫ 16 − 8 ⋅ 2 x + 2 2 x dx
8⋅2 + 2
= π 16x − ln(2)
2 ln(2)
d.
ln(y−1)
ln 2
5
5
2
ln(y−1)
ln(2)
.
− 12 dy ≈ 12.749π ≈ 40.052 un 3
The radius of the outside cylinder is 7. The radii of the inside disks are y + 2 = 2x + 3.
2
2
0
0
π ∫ 7 2 dx − π ∫ (2 x + 3)2 dx ≈ 135.752 un 3
13
© 2017 CPM Educational Program. All rights reserved.
Calculus
51.
a.
This is a sideways bowl. A slice is a disk with radius
V=
b.
4
∫0 π (
x )2 dx =
4
2 4
0
∫0 π xdx = π ⋅ x2
x and width Δx.
= 8π ≈ 25.133 un 3 .
This is a cylinder with a whirlwind cut out. A slice is a washer with inner and outer
radii of y2 and 4 and width dy.
V = ∫ (π ⋅ r 2 − π (y 2 )2 )dy = π ∫ 16 − y 4 dy = π ⎛ 16y −
⎝
0
0
2
2
y5
5
⎞
⎠
2
0
128π
= π (32 − 32
5 )= 5
≈ 80.425 un3
c.
It is like a quarter donut. A slice is a washer with inner and outer radii of 2 + y2 and 6
and width Δy.
V=∫
2
0
(
(
)
π ⋅ 6 2 − π (2 + y 2 )2 dy = π ∫
)
32
= π 64 − 32
3 − 5 =
d.
704 π
15
2
0
(
)
36 − 4 − 4y 2 − y 4 dy = π ⎛ 32y − 4 −
⎝
y3
3
−
y5
5
≈ 147.445 un 3
⎞
⎠
2
0
It is like a quarter donut. A slice is a washer with inner and outer radii of 5 − x and
5 and width Δx.
4
π ⋅ 5 2 − π (5 − x )2 ) dx = π ∫ ( 25 − 25 + 10 x − x ) dx
(
0
0
4
4
3/2
2
136π
3
= π ∫ (10 x − x ) dx = π ⋅ (10 ⋅ x3 2 − x2 ) = π ( 20
3 ⋅ 8 − 8 ) = 3 ≈ 142.419 un
0
0
V=∫
52.
4
2
a.
π ∫ (3x + 1)2 dx ≈ 166.421 un 3
b.
The outer radius is 2, the inner radius is x. Since y = 3x + 1, x = ln(y−1)
. The outside
ln(3)
cylinder is solid from y = 0 to y = 2, so subtract out the “inside” from y = 2 to y = 10.
0
0
c.
(
)
10 ln(y−1) 2
ln(3)
2
10
π ∫ 2 2 dx − π ∫
dy ≈ 40π – 16.5π = 23.5π ≈ 73.827 un3
The outer radius is (3x + 1) + 1 = 3x + 2, the inner radius is 1.
2
2
0
0
π ∫ (3x + 2)2 dx − π ∫ (1)2 dx ≈ 73.54π – 2π ≈ 224.74 un3
53.
Answers vary. Students might show that a2 + b2 ≠ (a + b)2.
54.
This is the inside half of a donut (half torus). It will have a cylindrical exterior and a
donut’s interior. A slice is a washer with inner and outer radii of 2 − 1 − x 2 and 2 and
1
1
−1
1
−1
width Δx. V = ∫ π ⋅ 2 2 − π (2 − 1− x 2 )2 dx = π ∫ 4 − 4 + 4 1− x 2 − (1− x 2 )dx
= π ∫ 4 1− x 2 − 1+ x 2 dx ≈ π ⋅ 4.950 ≈ 15.550 un 3
−1
14
© 2017 CPM Educational Program. All rights reserved.
Calculus
55.
Vertical slices: disks with radius f (x) and width dx.
b.
Vertical slices: washers with inner and outer radii of 1 and 1 + f (x) and width dx.
2
V = ∫ π (1+ f (x))2 − π (1)2 dx or
−1
2
∫−1 π (2 f (x) + f (x)2 )dx .
3
∫1 π (g(y))2 dy .
V=
c.
Horizontal slices: disks with radius g(y) and width dy.
d.
Horizontal slices: washers with inner and outer radii of 5 – g(y) and 5 and width dy.
3
V = ∫ (π ⋅ 5 2 − π (5 − g(y))2 )dy or
1
56.
2
∫−1 π ( f (x))2 dx .
V=
a.
a.
3
∫1 π (10g(y) − g(y)2 )dy .
A typical slice is a horizontal disk with radius
6
y
−
5
y2
y
and width dy. See diagram at right.
V=∫
13
1
π ⎛ 6y −
⎝
2
5
y2
⎞ dy
⎠
13
= ∫ π (36y −2 − 60y −3 + 25y − 4 )dy
1
= π (36 ⋅
y−1
−1
= π ⎛ −36
+
⎝ y
57.
y−2
−2
+ 25 ⋅
y−3
−3
13
1
x
13
30 − 25 ⎞
≈ 36.875 in.3
y2
3y 3 ⎠
1
b.
Height is 12: V = πr2h = πr2 ⋅ 12 = 36.875, r2 =
a.
y=h–
b.
Using x = r −
c.
Yes: V = ∫ π r − hr y
h
r
x (slope is
h
0
= π r2
15
− 60 ⋅
r
h
0−h
r−h
=
−h
r
36.875
12π
) or hx + ry = rh or x = r –
h
(
r
h
y as the radius of a disk, V = ∫ π r − hr y
0
(
⎛ y−
⎝
y2
h
36.875
12π
,r=
≈ 0.989 in.
y.
)2 dy .
)2 dy = π ∫0 r 2 − 2 rh2 y + hr22 y2 dy = π r 2 ∫0 ⎛⎝ 1− 2 hy + hy2 ⎞⎠ dy
+
h
y3
3h 2
⎞
⎠
h
0
h
(
2
= π r 2 h − hh +
h3
3h 2
) = πr (
2
h
3
2
) = π r32h
© 2017 CPM Educational Program. All rights reserved.
Calculus
58.
Now these are washers with inner and outer radii of 1 and 1 + 6y −
13
V=∫
1
=π∫
13
1
π ⎛ 1+ 6y −
⎝
2
5
y2
⎞ − π ⋅12 dy = π 13 1+
∫1
⎠
12y −1 + 26y −2
− 60y −3 + 25y − 4 dy = π
(
= π 12 ln(y) − 26y −1 + 30y −2 − 25
3
59.
)1
13
+
36
y2
25
y4
5
y2
+ 12y − 102 − 603 − 1dy
y
y
13
⎛ 12 ln(y) + 26 y −1 − 60 ⋅ y−2 + 25 ⋅ y−3 ⎞
−1
−2
−3 ⎠
⎝
1
≈ 104.572 in 3
The graph of x − 4 is symmetric about x = 4, so
a.
8
8
x − 4 dx = 2 ∫ (x − 4)dx = 2
∫0
4
(
x2
2
− 4x
)
8
= 2 ⋅ (32 – 32 – (8 – 16)) = 2 ⋅ 8 = 16
4
(This is calculated as the area of two triangles.)
w4
4
w2
2
b.
= ∫ (w 3 − w)dw =
c.
= −∫
d.
When graphed, this is a quarter circle:
−1
−5
−7dx =
−
+C
−1
∫−5 7dx = 4 ⋅ 7 = 28
(This region is the area of a rectangle.)
π ⋅5 2
4
=
25π
4
60.
Each doll is like a shell of the largest doll, so that when you add up all their volumes,
ideally you will get the full volume of the largest doll.
61.
sum = ∫ 2π x dx = 2π ⋅ x2
62.
r
2 r
0
0
r
= π x 2 = π r 2 = area of a circle
0
Rotating y = r 2 − x 2 about the x-axis yields disk slices with radius
r
r
−r
−r
(
V = ∫ π ( r 2 − x 2 )2 dx = ∫ (r 2 − x 2 )dx = π r 2 ⋅ x −
(
)
x3
3
)
r
−r
=π
((
r 2 − x 2 and width dx:
3
)(
3
r 3 − r3 − −r 3 + r3
))
= π 2r 3 − 23 r 3 = π ⋅ 43 r 3 = 43 π r 3
16
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Calculus
63.
a.
Use washers, but with 2 sections (2 integrals).
b.
y = 0.5x 2 , 2y = x
V=∫
0.5
0
(π (3)2 − π (1)2 )dy + ∫
4.5
0.5
0.5
4.5
(
0.5
= 8π y 0 + π ∫
9 − 2y dy
= 4π + π (9y − y 2 )
= 4π + π
(π (3)2 − π ( 2y )2 )dy
4.5
0.5
)
( 812 − 814 ) − π ( 92 − 14 )
64 π
17
3
= 4π + 81
4 π − 4 π = 4π + 4 = 20π un
64.
65.
66.
a.
Positive slope, concave up: B
b.
Negative slope, concave down: D
c.
Positive slope, concave down: C
d.
Negative slope, concave up: A, E
a.
shells of the inner dolls
b.
The area of small rectangles formed by consecutive circumferences.
c.
He used the circumference of a strip and dx as the rectangle’s length and width.
Note: These are the actual volumes using the shell method.
1A.
b
∫a 2π x ⋅ f (x)dx
∫0 2π x(x − 2)2 dx = 2π ∫0 (x 3 − 4x 2 + 4x)dx = 2π ( 14 x 4 − 43 x 3 + 2x 2 ) 0 = 83 π
2
2
2
un 3
1B. In this case the radius is 2 – x.
2
2
∫0 2π (2 − x)(x − 2)2 dx = 2π ∫0 (−x 3 + 6x 2 − 12x + 8)dx =
(
2π −
1
4
x 4 + 2x 3 − 6x 2 + 8x
) 20 = 8π un 3
1C. In this case the radius is x + 1.
∫0 2π (x + 1)(x − 2)2 dx = 2π ∫0 (x 3 − 3x 2 + 4)dx = 2π ( 14 x 4 − x 3 + 4x ) 0 =
2
2
2
2π(4 – 8 + 8) = 8π un3
2A.
∫0 2π x(4 − x)dx = 2π ∫0 (4x − x 2 )dx = 2π ( 2x 2 − 13 x 3 ) 0 = 643 π
4
4
4
un 3
Solution continues on next page. →
17
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Calculus
66.
Solution continued from previous page.
2B. In this case the radius is 4 – x.
4
∫0
(
4
2π (4 − x)(4 − x)dx = 2π ∫ (16 − 8x + x 2 )dx = 2π 16x − 4x 2 + 13 x 3
0
(
2π 64 − 64 +
64
3
) = 1283 π un 3
) 40 =
2C. In this case the radius is 2 + x.
∫0 2π (2 + x)(4 − x)dx = 2π ∫0 (8 + 2x − x 2 )dx = 2π ( 8x + x 2 − 13 x 3 ) 0 = 1603 π
4
3A.
4
∫0 2π x 2 dx = 2π ( 13 x 3 ) 0 = 1283 π
4
4
4
un 3
un 3
3B. In this case the radius is 1 + x.
∫0 2π (1+ x)(x)dx = 2π ∫0 x + x 2 dx = 2π ( 12 x 2 + 13 x 3 ) 0 = 1763 π
4
4
4
3C. In this case the radius is 4 – x.
4
∫0
4A.
4
(
4
2π (4 − x)(x)dx = 2π ∫ (4x − x 2 )dx = 2π 2x 2 − 13 x 3
0
∫0 2π x
4
x dx = 2π ∫ x 3/2 dx = 2π
0
4
2π
4
x dx = 2π ∫ 4x1/2 − x 3/2 dx = 2π
0
3
( 643 − 645 ) = 256
15 π un
4C. In this case the radius is 5 – x.
4
∫0 2π (5 − x)
2π
67.
4
x dx = 2π ∫ 5x1/2 − x 3/2 dx = 2π
0
3
( 803 − 645 ) = 236
15 π un
( 83 x 3/2 − 25 x 5/2 ) 0 =
4
( 103 x 3/2 − 25 x 5/2 ) 0 =
a.
A thin box with length 2πx, width f (x), and thickness dx.
b.
2πx ⋅ f (x)dx
c.
∫a 2π x ⋅ f (x)dx
d.
18
) 40 = 643 π un 3
( 25 x 5/2 ) 40 = 1285 π un 3
4B. In this case the radius is 4 – x.
∫0 2π (4 − x)
un 3
4
b
3
∫1
3
2π x(0.5x 2 )dx = ∫ π x 3 dx = π
1
x4
4
3
1
=π
( 814 − 14 ) = 20π ≈ 62.832 cm3
© 2017 CPM Educational Program. All rights reserved.
Calculus
68.
2 = 1+ 1
x
y
2 −1= 1
x
y
2− x
1
=y
x
x = y
2− x
a.
y′ =
1⋅(2− x)− x(−1)
(2− x)2
=
2
(2− x)2
At (1, 1) it is 2.
x2 y − y3 = 8
b.
At (–3, 1) it is 1.
2xy + x 2 y′ − 3y 2 y′ = 0
2xy = 3y 2 y′ − x 2 y′
2xy = y′(3y 2 − x 2 )
2 xy
3y2 − x 2
69.
= 2⋅ t
a.
3/2
3
2
3 4
+ t3
1
= y′
= 43 ⋅ 8 + 64
3 −
u = x2 + 1, du = 2xdx:
b.
( 43 + 13 ) = 963 − 53 = 913
∫ x2x+1 dx = ∫ 12 ⋅ 2x2xdx+1 = 12 ∫ duu = 12 ln u + C = 12 ln(x 2 + 1) + C
(using substitution)
c.
u = cos(u), du = – sin(u) du:
π /4
π /4 − sin(u)du
π /4
du
cos(u) = − − π /4 u
/4 π
π /4
=
−
ln
cos(u)
= − ln 22 + ln
− π /4
− π /4
∫−π /4 tan(u)du = − ∫−π /4
= − ln u
∫
( ) ( )=0
2
2
The best way to realize it is 0 is by symmetry (tangent is odd).
∫ ln(e2 x ) dx = ∫ 2x dx = x 2 + C
d.
70.
E:
2
∫1
−1 2
2x 2 dx = 2 ⋅ x−1
1
=
−2 2
x 1
= −1+ 2 = 1
∫0 ( π (2x)2 − π (2x 2 )2 ) dx = ∫0 (4x 2 − 4x 4 )dx
1
1
71.
B: This is
72.
D: At x = 0, lim f (x) = 1 (by graphing of l’Hôpital’s rule), so I is false.
x→0
As x → ∞, f (x) → 0, so II and III are true.
73.
19
D: I is false because f ′(b) does not exist. II and III are true.
© 2017 CPM Educational Program. All rights reserved.
Calculus
74.
E: The average is
∫ ( ln(x 2 ) ) dx ≈ 1.164
e
1
e−1 1
Without a calculator if one notes that f is concave down in this interval, with endpoints at
(1, 0) and (e, 2), then it is apparent that its average value must be above the average of 0
and 2, i.e., greater than 1.
75.
D:
c
2
∫0 (x 2 + 1)dx = ∫c (x 2 + 1)dx
x3
3
c
+x =
0
x3
3
+x
2
c
c3 + c = 8 + 2 − c3 − c
3
3
3
3
c + 3c = 8 + 6 − c 3 − 3c
2c 3 + 6c = 14
c 3 + 3c − 7 = 0
76.
→
c ≈ 1.406 by using a calculator
3
∫1 ( −(x − 3)2 + 4 ) dx = ∫1 (−x 2 + 6x − 5)dx = − x3 + 3x 2 − 5x 1
1
2
= ( − 64
3 + 48 − 20 ) − ( − 3 + 3 − 5 ) = 9 un
4
a.
b.
4
∫1
4
4
(
1
(
= 2π (−64 + 128 − 40) − − 14 + 2 − 52
77.
))
(
( ))
= π 48 − − 23
=
x4
4
99π
2
3
+ 6 x3 − 5 2x
It would have two different parts of f (x) as its bounds (radii).
b.
Shells would only requite one integral, instead of 2 with washers/discs.
c.
x ranges from a = 0 to b = 3.
d.
radius = 1 + x, height = f (x) = –x2 + 2x + 3
3
∫0
)
4
1
3
2π (1+ x)(−x 2 + 2x + 3)dx = 2π ∫ (−x 3 + x 2 + 5x + 3)dx
(
= 2π −
2
2
≈ 155.509 un 3
a.
e.
78.
(
4
2π x(−(x − 3)2 + 4)dx = 2π ∫ (−x 3 + 6x 2 − 5x)dx = 2π −
x4
4
+
x3
3
2
+ 5 2x + 3x
)
0
3
0
(
)
27 45
= 2π − 81
4 + 3 + 2 +9 =
81π
2
≈ 127.235 un 3
2
∫−1 2π (x + 1)(−x 2 + 2x + 3 − (x + 1))dx = 2π ∫−1 (x + 1)(−x 2 + x + 2)dx
Note: The height is the difference of the two functions.
20
© 2017 CPM Educational Program. All rights reserved.
Calculus
79.
a.
See graph at right.
5x 0.5 − 0.1x1.5 = 0
y
5x 0.5 = 0.1x1.5
0.5
1.5
5 ⋅ x 0.5 = 0.1⋅ x0.5
x
x
x
5 = 0.1x
x = 50 ft
b.
V=∫
50
0
π (5x 0.5 − 0.1x1.5 )2 dx = π ∫
0
= π (31250 − 125000
3
c.
50
(
2
(25x − x 2 + .01x 3 )dx = π 25 ⋅ x2 −
+ 15625) = π ⋅ 5208 ≈ 16362.462
1
3
x3
3
4
+ .01⋅ x4
)
50
0
ft 3
The 120 aliens take up 10% ⋅ 16362.462 ft3 ≈ 1636.246 ft3 and the average weight is
13.635 ft3 ⋅ 3.5lbs
≈ 47.724 lbs.
3
ft
80.
total displacement
total time
m
= 1200
96 sec = 12.5 m/sec
a.
Vavg =
b.
d(t) ≈ 0.000145x4 – 0.0246x3 + 1.315x2 – 15.808x + 5.582,
d′(t) ≈ 0.000579x3 – 0.0734x2 + 2.629x – 15.808
d′(96) ≈ 69.952 m/s ≈ 0.069952 km/s ⋅ 602 s/hr ≈ 251.829 km/hr
Which might be too fast.
c.
See graph at right. It is a cubic. Velocity speeds up and
slows down due to curves in the track.
v(t)
40
20
20
81.
82.
a.
40
60
80
t
c
b
c
f (x)dx = − ∫ f (x)dx = − ⎛ ∫ f (x)dx + ∫ f (x)dx ⎞ =
⎝
⎠
a
a
b
–(2.5 – 5) = 2.5
a
∫c
⎛
c
b
f (x)dx + ∫ f (x)dx ⎞ = 2(2.5 – 5) = –5
⎠
b
c
b.
∫a 2 f (x)dx = 2 ⎝ ∫a
c.
d
b
c
d
= − ∫ f (x)dx = − ⎛ ∫ f (x)dx + ∫ f (x)dx + ∫ f (x)dx ⎞ = –(2.5 – 5 + 1.5) = –(–1) = 1
⎝ a
⎠
a
b
c
d.
0
B: f ′(x) =
d
dx
(x 3 − 10x 2 + 21x) = 3x 2 − 20x + 21 = 0, x =
20± 400−4(3)⋅21
6
=
20± 148
6
,
max is at x = 20− 6 148 = 10−3 37 and is about 12.597. (Note: Remember to check the
endpoints, f (0) = 0 and f (7) = 0.)
21
© 2017 CPM Educational Program. All rights reserved.
Calculus
83.
84.
C: x 2 = 8 + 2xy
y′ =
1
2
− 4 ⋅ (−1)x −2 =
x 2 −8 = y
2x
y = 2x − 4x
y′ =
1
2
+
1
∫0 x
A:
4
4
=
3
2
1
2
+
4
x2
at x = 2
1− x 2 dx → u = 1− x 2 , du = −2xdx
1
= − 12
∫0 (−2x)
= − 12
u 3/2
3
2
1− x 2 dx = − 12
1
∫x=0
= − 13 u 3/2 = − 13 (1− x 2 )3 2
u du
1
0
= − 13 (0) + 13 ⋅1 =
1
3
85.
D: f ′(x) = –e–x – 3x2 + 6x – 10
f ′′(x) = e–x – 6x + 6
0 = e–x – 6x + 6 ⇒ using a calculator f ′′(x) = 0 and changes sign at ≈ 1.058.
86.
B (At horizontal tangents, the derivative is 0.)
y
87.
a.
or
b.
(
6
)
Shells: V = ∫ 2π x x + 1− (x 2 − 6x + 7) dx
1
6
∫1 2π x(−x 2 + 7x − 6)dx
x
6
Washers: V = ∫ (π (3 + x + 1)2 − π (3 + x 2 − 6x + 7)2 )dx
1
6
or π ∫ ((x + 4)2 − (x 2 − 6x + 10)2 )dx
1
y
x
Solution continues on next page →
22
© 2017 CPM Educational Program. All rights reserved.
Calculus
87.
Solution continued from previous page.
c.
1
Washers: V = ∫ (π (−y 2 + 3)2 − π (y 2 + 1)2 )dy
y
−1
1
or π ∫ (−8y 2 + 8)dy
−1
x
y
d.
1
Shells: V = ∫ 2π (2 − y)(−y 2 + 3 − (y 2 + 1))dy
−1
1
or 2π ∫ (2 − y)(−2y 2 + 2)dy
x
−1
e.
Disks: V = ∫
2π
0
π ( cos(x) + 2 ) dx
2
y
x
f.
Shells: V = ∫
2π
0
2π x ( cos(x) + 2 ) dx
y
x
y
g.
2
Washers: V = ∫ (π (6 −
0
x)2
− π (2 x )2 )dx
2
= π ∫ ((6 − x)2 − 4 x )dx
x
0
h.
Shells: V = ∫
2
0
y
2π x(−x + 6 − 2 x )dx
x
23
© 2017 CPM Educational Program. All rights reserved.
Calculus
88.
a.
2
f (t)dt = 12 (2)(2) = 2 (triangle),
4
f (t)dt = ∫ f (t)dt + ∫ f (t)dt = 2 − π ⋅24
g(2) =
∫0
g(4) =
∫0
2
4
0
2
b.
No, it is the opposite. g(–2) = –2,
c.
∫−2 f (x)dx = ∫−2 f (t)dt + ∫0
d.
Yes,
2
0
x
d
dx 0
∫
2
2
= 2 – π (triangle – quarter circle)
0
∫−2 f (t)dt = 2
f (t)dt = –g(–2) + g(2)
f (t)dt = f (x) in this interval. Since f is continuous on the interval, g is
differentiable.
89.
e.
This is where its derivative changes from positive to negative: x = 2.
f.
g(4) = 2 – π and g′(4) = f (4) = –2, so the tangent line is
y = –2(x – 4) + 2 – π = –2x + 10 – π.
g.
This is where the derivative of f changes sign: x = 0 and 4.
a.
=
ye y −e y
y2
b.
=
1
2
c.
=
d
dθ
, using the Quotient Rule.
( sin(x) )−1/2 ⋅ cos(x) , using the Chain Rule.
( 5 ( cos (θ ) + sin (θ ) ) ) =
2
d
dx
2
( x) =
1
2
d
dθ
(5) = 0 , by simplifying first.
x −1 2 .
90.
D: This is
91.
A: There are twenty terms, all multiplied by
92.
C: The intersection is at ex = –x + 3 or ex + x – 3 = 0, x ≈ 0.792:
A≈∫
.792
0
93.
94.
24
(−x + 3 − e x )dx = −
x2
2
+ 3x − e x
0.792
0
1
20
, which is the width of a rectangle.
≈ –0.145 – (–1) ≈ 0.855
C: It is flat at x = 0 and 2.
E: =
(
)
x 1 x −1 2 e x − 1 x −1 2 e x
2
2
x
=
e x (1− x −1 2 )
2x
=
e x (x− x )
.
2 x2
© 2017 CPM Educational Program. All rights reserved.
Calculus
b
b
a
a
95.
C: = ∫ f (x)dx + ∫ (−3)dx = 2b – a + (–3)(b – a) = 2b – a – 3b + 3a = –b + 2a
96.
D: I uses cylindrical shells and II uses disks.
97.
a.
Slice horizontally
b.
Slice horizontally, but you would get three different similar shapes.
c.
Slice vertically.
d.
No convenient way to slice.
n
98.
b.
∑ A(i)Δx
c=1
c.
With an integral:
R
∫L A(x)dx
The limit of the sums of volumes as the number of slices → ∞.
101. It does not matter when some cards in a deck protrude out of it; the deck maintains its
volume, as each card has the same height. Since the volume is the sum of the slices, it does
not matter if the slices of both solids have the same orientation. All that matters is that each
corresponding slice has the same area.
102. a.
c.
103. D: =
∫0 π ( 12 x cos(x) + 4 )
5
5
∫0
2π (6 − x)
−1
1−(3x)2
104. B: Vavg = π1
∫0
−3
1−9 x 2
∫0 2π x ( 12 x cos(x) + 4 ) dx
5
b.
dx
( 12 x cos(x) + 4 ) dx
⋅3=
π
2
5
∫0
d.
(
π
( 12 x cos(x) + 11)2 − π ⋅ 49 ) dx
by the Chain Rule
(t + sin(t))dt = π1
(
t2
2
− cos(t)
)
π
0
= π1
(
π2
2
)
− (−1) − (−1) = π2 + π2 .
105. C: F ′(5) = 9 − 5 = 2 , but F(–7) is negative on the graph, and F ′′(x) =
25
© 2017 CPM Educational Program. All rights reserved.
−1
2 9− x
is negative.
Calculus
106. D:
dP
dt
= 0.02P + 357
dP
0.02P+357
= dt
dP
= ∫ dt
∫ 0.02P+357
–17850 + C = 18000
1
0.02
C = 35850
ln(0.02P + 357) = t + C
P(6) = –17850 + 35850e0.02⋅6
≈ 22571
ln(0.02P + 357) = 0.02t + C
0.02P + 357 = e0.02t+C
0.02P = Ce0.02t
0.02P = –357 + Ce0.02t
P = –17850 + Ce0.02t
d x(3x − 12) =
107. D: xy = x(3x – 12): dx
(
)
minimum product = 2(6 – 12) = –12.
d
dx
(3x2 – 12x) = 6x – 12 = 0, x = 2,
108. D: The derivative of g is f: D starts out in a positive direction and is flat at a and b.
3
109. 1(a): 2 ∫ ( 9 − x 2 )2 dx = 36 un 3
0
3
1(b): 2 ∫
1
0 2
( 9 − x 2 )2 dx = 18 un3
3
1(c): 2 ∫ 2( 9 − x 2 )2 dx = 72 un 3
0
1(d): 2π
( 12 ) ∫0 ( 12
3
1(e): 2 ∫
3
1(f): 2 ∫
3
1
0 2
9 − x2
) dx =
2
9
2
( 9 − x 2 )2 dx = 18 un3
3
0 2
( 9 − x 2 )2 dx = 18 3 un 3
31
0 2
(
3π
0 4
( 9 − x 2 )2 dx = 9π un3
1(g): 2 ∫
1(h): 2 ∫
π un 3
9 − x2 +
9 − x2
1
2
)(
1
2
)
9 − x 2 dx =
3 3 (9 −
4 0
∫
x 2 )dx =
27
2
un 3
3
1(i): 2π ∫ ( 12 9 − x 2 )2 dx = 9π un3
0
Note: For number 2, x 2 = 9 − y 2 .
2(a):
3
3
∫0 (2x)(2x)dy = 4 ∫0 (9 − y2 )dy = 72 un 3
Solution continues on next page →
26
© 2017 CPM Educational Program. All rights reserved.
Calculus
109. Solution continued from previous page.
3
3
2(b):
∫0 (2x)(x)dy = 2 ∫0 (9 − y2 )dy = 36 un 3
2(c):
∫0 (2x)(4x)dy = 8 ∫0 (9 − y2 )dy = 144 un 3
2(d):
π 3 (x)(x)dy = π 3 (9 − y 2 )dy =
2 0
2 0
2(e):
1 3
(2x)(2x)dy
2 0
2(f):
∫0 12 (2x)(x
2(g):
∫0 12 (2x + x)(x)dy = 23 ∫0 (9 − y2 )dy = 27 un 3
2(h):
π 3 (2x)2 dy = π 3 (9 − y 2 )dy =
4 0
0
3
3
∫
∫
∫
3
9π un3
3
= 2 ∫ (9 − y 2 )dy = 36 un 3
0
3
3)dy = 3 ∫ (9 − y 2 )dy = 18 3 un 3
0
3
3
∫
∫
3
3
0
0
18π un3
2(i): π ∫ (x)(x)dy = π ∫ (9 − y 2 )dy = 18π un3
3
3
0
0
3(a): 2 ∫ (3 − y)2 dx = 2 ∫ (6 − 2x)2 dx = 72 un3
3
3(b): 2 ∫
1
0 2
3
(3 − y)2 dx = ∫ (6 − 2x)2 dx = 36 un3
0
3
3
0
0
3(c): 2 ∫ 2(3 − y)2 dx = 4 ∫ (6 − 2x)2 dx = 144 un3
∫0 ( 12 (3 − y) ) dx = π8 ∫0 (6 − 2x)2 dx = 92 π
3
3
2
3(d):
π
2
3(e):
3
1
2 0 (3 −
3(f):
1 3 3
2 0 2
3(g):
∫0 12 ( (3 − y) + 12 (3 − y) ) ( 12 (3 − y) ) dx = 83 ∫0 (6 − 2x)2 dx = 272
3(h):
π 3 (3 −
4 0
∫
∫
y)2 dx =
3
1
2
2 0 (6 − 2x) dx
∫
(3 − y)2 dx =
3
4
un 3
= 18 un3
3
∫0 (6 − 2x)2 dx = 9
3 un 3
3
3
∫
y)2 dx = π4
3
un 3
∫0 (6 − 2x)2dx = 9π un
3
1 (3 − y) 2 dx = π
(
) 4 ∫0 (6 − 2x)2dx = 9π un3
0 2
3(i): π ∫
4(a):
3
3
3
3
∫−3 (x − (−3))2 dx = ∫−3 (x + 3)2 dx = 72 un 3
Solution continues on next page →
27
© 2017 CPM Educational Program. All rights reserved.
Calculus
109. Solution continued from previous page.
4(b):
1 3 (x +
2 −3
∫
3)2 dx = 36 un 3
3
4(c): 2 ∫ (x + 3)2 dx = 144 un 3
−3
∫ (
)2
4(d):
π 3 1
2 −3 2
4(e):
1 3
(x +
2 −3
4(f):
1 3 3
2 −3 2
4(g):
∫−3 12 ( (x + 3) + 12 (x + 3) ) ( 12 (x + 3) ) dx = 83 ∫−3 (x + 3)2 dx = 27 un 3
4(h):
π 3 (x + 3)2 dx
4 −3
∫
∫
3
∫−3 (x + 3)2 dx = 9π un
(x + 3) dx = π8
3
3)2 dx = 36 un 3
(x + 3)2 dx =
3
∫−3 (x + 3)2 dx = 18
3
4
3
3 un 3
3
∫
= 18π un3
3
1 (x + 3) 2 dx = π
2
(
)
4 ∫−3 (x + 3) dx = 18π un
−3 2
3
4(i): π ∫
3
3
5(a):
∫−1 (
5(b):
∫−1 12 (
5(c):
∫−1 2(
5(d):
∫−1 π2 ( 12
5(e):
∫−1 12 (
5(f):
x + 1)2 dx =
3
3
3
3
∫−1 (x + 1)dx = 8 un 3
x + 1)( x + 1)dx =
= 4 un 3
3
−1
x +1
)
2
3
∫−1 (x + 1)dx = π un
dx = π8
x + 1)( x + 1)dx =
∫−1 12 (
3
x + 1)(
5(g):
∫−1 12 (
3
x +1 +
5(h):
∫−1 π4 (
3
x + 1)2 dx = π4
5(i):
∫−1 π ( 12
3
∫
x + 1)( x + 1)dx = 2 ∫ (x + 1)dx = 16 un 3
3
3
1 3 (x + 1)dx
2 −1
x +1
3
2
)
1
2
2
3
1 3
(x + 1)dx
2 −1
∫
x + 1)dx =
x +1
) ( 12
3
3
4
3
3
∫−1 (x + 1)dx = 2
)
x + 1 dx =
∫−1 (x + 1)dx
dx = π4
= 4 un 3
3 un 3
3 3 (x + 1)dx
8 −1
∫
= 8 un 3
= 2π un3
∫−1 (x + 1)dx = 2π un
3
3
( 3 sin(x) )2 dx = 18 ∫0 ( sin(x) )2 dx ≈ 28.26 un 3
0
6(a): 2 ∫
Solution continues on next page →
28
© 2017 CPM Educational Program. All rights reserved.
Calculus
109. Solution continued from previous page.
2
3 sin(x) ) ( 23 sin(x) ) dx = 9 ∫ ( sin(x) ) dx ≈ 14.13 un 3
(
0
0
3
6(b): 2 ∫
3
3
3
( 3 sin(x) ) ( 6 sin(x) ) dx = 36 ∫0 ( sin(x) )2 dx ≈ 56.52 un 3
0
6(c): 2 ∫
3π
0 2
( 3 sin(x) )2 dx = 9π ∫0 ( sin(x) )2 dx ≈ 14.13π
6(e): 2 ∫
3
1
0 2
( 3 sin(x) ) ( 3 sin(x) ) dx = 9 ∫0 ( sin(x) )2 dx ≈ 14.13 un 3
6(f): 2 ∫
3
( 3 sin(x) )
3
1
0 2
( 3 sin(x) + 23 sin(x) ) ( 23 sin(x) ) dx = 274 ∫0 ( sin(x) )2dx ≈ 10.60 un 3
3π
0 4
( 3 sin(x) )2 dx = 92 π ∫0 ( sin(x) )2 dx ≈ 7.065π
6(d): 2 ∫
1
0 2
6(g): 2 ∫
6(h): 2 ∫
3
un 3
3
(
3 3
2
)
9 3 3
2
0
2
∫ ( sin(x) ) dx ≈ 12.24 un 3
sin(x) dx =
3
3
3
6(i): 2 ∫ π ( 3 sin(x) ) dx = 18π ∫
2
0
3
0
( sin(x) )2 dx ≈ 28.26π
un 3
un 3
Note: For number 7 the radius/length is (3 – x), where x = y2 – 1. Thus r = (4 – y2).
2
2
7(a):
∫0 (3 − x)2 dy = ∫0 (4 − y2 )2 dy = 17 151 = 256
15
7(b):
∫0 (4 − y2 ) ( 12 (4 − y2 ) ) dy = 12 ∫0 (4 − y2 )2 dy = 256
30
7(c):
∫0 (4 − y2 ) ( 2(4 − y2 ) ) dy = 2 ∫0 (4 − y2 )2 dy = 512
15
7(d):
∫0 π2 ( 12 (4 − y2 ) )
2
2
2
2
2
2 1
0 2
un 3
2
dy =
π
8
2
∫0 (4 − y2 )2 dy = 1532 π
1 2 (4
2 0
un 3
un 3
7(e):
∫
7(f):
∫0 12 (4 − y2 ) (
7(g):
∫0 12 ( (4 − y2 ) + 12 (4 − y2 ) ) ( 12 (4 − y2 ) ) dy = 83 ∫0 (4 − y2 )2 dy = 325
7(h):
∫0 π4 (4 − y2 )2 dy = π4 ∫0 (4 − y2 )2 dy = 1564 π
7(i):
∫0 π ( 12 (4 − y2 ) )
2
(4 − y 2 )(4 − y 2 )dy =
3
2
∫
un 3
)
(4 − y 2 ) dy =
− y 2 )2 dy =
3
4
256
30
2
∫0 (4 − y2 )2 dy =
2
64 3
15
un 3
2
2
2
un 3
2
2
dy =
π
4
2
un 3
un 3
∫0 (4 − y2 )2 dy = 1564 π
un 3
Solution continues on next page →
29
© 2017 CPM Educational Program. All rights reserved.
Calculus
109. Solution continued from previous page.
3
3
8(a):
∫−3 (3 − x)2 dy = ∫−3 (3 − y)2 dy = 72 un 3
8(b):
1 3
(3 −
2 −3
∫
y)(3 − y)dy = 36 un 3
3
8(c): 2 ∫ (3 − y)2 dy = 144 un 3
−3
∫ (
)2
8(d):
π 3 1
2 −3 2
8(e):
1 3
(3 −
2 −3
8(f):
1 3 3
2 −3 2
8(g):
∫−3 12 ( (3 − y) + 12 (3 − y) ) ( 12 (3 − y) ) dy = 83 ∫−3 (3 − y)2 dy = 27 un 3
8(h):
π 3 (3 −
4 −3
(3 − y) dy = π8
∫
∫
3
∫−3 (3 − y)2 dy = 9π un
3
y)2 dy = 36 un 3
(3 − y)2 dy =
3
4
3
∫−3 (3 − y)2 dy = 18
3
3 un 3
3
∫
y)2 dy = 18π un3
3
1 (3 − y) 2 dy = π
2
(
)
4 ∫−3 (3 − y) dy = 18π un
−3 2
8(i): π ∫
3
3
Note: For number 9 the radius/length is 2x = y + 3.
∫−3 (y + 3)2 dy = 72 un 3
9(b):
∫−3 12 (y + 3)(y + 3)dy = 36 un 3
9(c):
∫−3 2(y + 3)(y + 3)dy = 144 un 3
3
3
3 π
−3 2
3
∫−3 ( 12 (y + 3) ) dy = π8 ∫−3 (y + 3)2dy = 9π un
2
3
3
9(d):
∫
9(e):
∫−3 12 (y + 3)(y + 3)dy = 36 un 3
x 2 dy = π2
3
3 1
−3 2
(
)
3
9(f):
∫
9(g):
∫−3 12 ( (y + 3) + 12 (y + 3) ) ( 12 (y + 3) ) dy = 83 ∫−3 (y + 3)2 dy = 27 un 3
9(h):
∫−3 π4 (y + 3)2 dy
9(i):
30
3
9(a):
(y + 3)
3
2
(y + 3) dy =
3
4
∫−3 (y + 3)2 = 18
3
3
3
3
∫−3
3 un 3
π x 2 dy = π ∫
3
= 18π un3
(
1
−3 2
)
2
(y + 3) dy = π4
3
∫−3 (y + 3)2 dy = 18π un
3
© 2017 CPM Educational Program. All rights reserved.
Calculus
110.
b
∫a ln(x)dx ;
No, Cavalieri’s principle.
111. The radius is x. x =
112.
4
∫0 (
9
∫0 (
y
4
x )2 dx = ∫ x dx =
0
1
2
x2
4
0
y )2 dy
2
= 16
2 − 0 = 8 un
2
2(4 − x 2 ) ) dx
(
0
113. 2 ∫
2
114. dl = 6 cm/sec, dw = 2 cm/sec, l = 30 cm, w = 15 cm, A = lw
dA = dl (w) + (l) dw = (6 cm/sec)(15 cm) + (30 cm)(2 cm/sec) = 90 + 60 = 150 cm2/sec
dt
dt
dt
115. a.
Let u = x2 – 1 ⇒ du = 2x dx. If x = 1, u = 0. If x = 5, u = 24.
24
∫0
b.
1
2 u
(
0
5
1
2u
= u1/2
24
0
= 24 − 0 = 24
(
1
) 12 ( 23 u 3/2 ) 10 = 13
u − 12 du =
0
5
du = 12 ln(u) =
1
1
2
( ln(5) − ln(1) ) = 12 ln(5)
Let u = –x2 ⇒ du = –2xdx. If x = 2, u = –4. If x = 0, u = 0.
∫0
b.
)
u − 12 du = − ∫
−4
116. a.
∫
Let u = x2 + 1 ⇒ du = 2xdx. If x = 2, u = 5. If x = 0, u = 1.
∫1
d.
24
1
−1/2 du
2 0 u
Let u = 1 – x2 ⇒ du = –2xdx. If x = 1, u = 0. If x = 0, u = 1.
∫1
c.
du =
x
15
=
− 23 eu du = ∫
0
3 eu du
−4 2
42
60
= 23 eu
0
−4
= 23 − 23 e− 4
⇒ x = 10.5 cm3
V = 13 π r 3 ⇒ dV = π r 2 dr
15 = π (1.7)2 dr ⇒ dr =
15
π (1.7)2
= −1.652 cm/min
Note: The radius is decreasing, so dr must be negative.
c.
Only if it is chocolate.
117. g′(x) =
31
1
f ′ (g(x))
⇒ g′(3) =
1
f ′ (g(3))
=
1
f ′ (1)
=
1
2
© 2017 CPM Educational Program. All rights reserved.
Calculus
118. The average rate is the average slope. y(8) =
=6
y(0) = 12 (0)2 +2(0) + 1 = 1. m = 49−1
8−0
(8)2 + 2(8) + 1 = 49,
1
2
The slope is y′(t) = t + 2. t + 2 = 6 ⇒ t = 4
119. a.
= 00 , so use l’Hôpital’s Rule ⇒ lim
b.
= 00 , so use l’Hôpital’s Rule ⇒ lim
c.
lim
d.
120. a.
x→25
sin 2 (x)
x
x→∞
sin(x)
x
= lim
x→∞
x
3
x→∞
1
2
4−
x 2 dx
3
π 4
2 0
( )
b.
∫0 (
⎛
9 − x2 ⎜
⎝
c.
5
5
1
1
2
1
(2y)(4y)dy
=
2 0
2 0 8y dy = 2
d.
π
2
e.
∫0 (
∫0 (
⋅ dx = 4 −
12
∫
3
x
2
)
)
2
=
∫
(−3) (4−x/3)
3
4
= π2
9−x 2
2
⎞
⎟⎠ dx =
)
x +1 −2 x
2
)
y 2
2
dy + ∫
2
⎞
⎟⎠ dx =
(
8 y
4 2
π
8
3 12
1
7
=0
→ ∞.
(
= − 12 4 −
0
( )
x2
8
4
12
x 3
3
0
)
= −0 + 12 4 3 = 32 cm3
= π un 3
0
3
∫ ( 9 − x 2 ) dx = 12 ( 9x − x3 ) 0 = 12 (27 − 9) = 9 un 3
3
3
1
2 0
∫
∫0 ⎜⎝
4
1
−2
x dx
0 4
∫
(
sin(x)
x
=
dx
dx = π2
)
1⎛
1
2(25)−1
4 =−1)
(the slope is − 12
3
x 2
3
c.
=
x→∞
(
y2
2
1
2
⋅ lim ( sin(x) ) = 0 because lim
3x + 7 x
x→∞ e x +10 x
y=4 –
=
(1/2)(2 x−1)−1/2 (2)
1
3x > ex since 3 > e. Therefore lim
b.
121. a.
32
x→6
1/(x−4)
1
∫0 ( (
( 83 y 3 ) 0 = 12 ( 83 ) (125) = 5003
5
)
1
2
x + 1) − 2 x dx ≈ 0.0582412
)
2
− 4y − 16 dy =
y3
12
4
0
+ 16
=
15
64
12
+ 16
=
15
( π8 ) ≈ 0.023 un 3
32
5
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un 3
un 3
Calculus
20y = x + 50, x = 20y − 50 , so
For a given y, y = 0.05(x + 50)2, 20y = (x + 50)2,
122. a.
the length is 2 (50 − 20y ) = 100 − 4 5y , so a slice has volume (100 − 4 5y )2 dy .
12
∫0
b.
12
(100 − 4 5y )2 dy = ∫ (10000 − 800 5y + 80y)dy
0
y3 2
32
= 10000y – 800 5
c.
Full volume is
= 109333
123. b.
h
b2
y=h–
2h
b
h
∫0
12
0
(100 − 4 5y )2 dy = 10000y − 16003
yards , so there are about
109333 13
5
y 3 2 + 40y 2
20
0
– 76186 ≈ 33147.520 yards3 left.
≈ 552.5 days or ≈ 1.514 years left, so no delay.
y
or
x = h − 2h
b x
b y=
x = h − y, x = b2 − 2h
b
2h
(h − y)
2
∫0 ( 2x ) dy = ∫0 ( 2 ( 2hb (h − y) ) )
c.
≈ 76185.813 yards3
3
1
3
33147.5 yrds 3
60 yrds 3 /day
d.
20
+ 40y 2
h
(
= − bh ⋅ 13 b − bh y
2
h
2
b − bh y ) dy
(
0
dy = ∫
h
h
)3 0 = −0 + bh ⋅ 13 ⋅b 3 = 13 b2h
x
b
124. b.
The length of the side of a square is 2x. x = –2y + 20
10
∫0
10
∫0
(2(20 − 2y))2 dy =
10
(1600 − 320y + 16y 2 )dy =
5
3
1600y − 160y 2 + 16
3 y
125. y′ =
−1
(1+ x 2 )2
⋅ 2x =
At x = –1, y =
At x = 2, y =
Therefore
y=
33
1
2
y
1
2
1
5
1
2
−2 x
(1+ x 2 )2
10
0
3
= 16000
3 in
20
x
by the Chain Rule.
and y′ = 12 , so the tangent line is y =
and y′ =
–20
−4
25
1
2
(x + 1) + 12 =
1
2
x +1.
4 (x − 2) + 1 .
, so the tangent line is y = − 25
3
4 (x – 2) +
x + 1 = – 25
1
5
, 25x + 50 = –8x + 16 + 10, 33x = –24, x =
21 −24 21
( −24
33 ) + 1 = 33 : ( 33 , 33 ) ≈ (–0.727, 0.636)
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−24
33
,
Calculus
(x− 4)3 2 8
32
4
8
x − 4dx =
c
x − 4 dx = ∫
126. a.
∫4
b.
∫4
8
=
43 2
32
y
= 16
un 2
3
x
x − 4 dx ,
c
c
8
(x−4)3 2
(x−4)3 2
=
32
32
4
c
3
2
32
(c−4)
− 0 = 163 − (c−4)
32
32
4 (c − 4)3/2 = 16
3
3
(c – 4)3/2 = 4,
c – 4 = 43/2
c = 4 + 42/3 ≈ 6.520
8
∫4 π (
c.
8
= π ∫ (x − 4)dx = π
x − 4 )2 dx
4
(
x2
2
− 4x
)
8
=
4
π(32 ⋅ 32 – (8 – 16)) = 8π un3 using disks.
d
∫4 π (
d.
(
π(
x2
2
π
d2
2
d2
128. C:
d
dx
)
d
( )
− 4d − (−8) ) = π ( 0 − (
− 4x
d2
2
d
=π
4
− 4x
8
d
d2
2
− 4d
))
2
− 8d + 8 = 0
8+ 64−4(8
2
4
∫0
= 4+
32
2
= 4 + 4 2 2 = 4 + 2 2 ≈ 6.828
(
4
( 4 − x )2 dx = ∫ (4 − x)dx = 4x −
0
(x 2 + y 2 ) =
d
dx
(25)
2x + 2yy′ = 0
2yy′ = −2x
y'=
34
x2
2
− 4d + 8 = − d2 + 4d
d=
127. A: V =
8
x − 4 )2 dx = ∫ π ( x − 4 )2 dx
−x
y
d
dx
( y′ ) =
y′′ =
=
d
dx
x2
2
)
4
= 16 – 8 = 8
0
(− )
x
y
y(−1)− y′ (− x)
y2
− y2 −x 2
y3
=
=
− y+xy′
y2
−(x 2 +y2 )
y3
=
=
− y+x(− x y)
y2
−25
y3
© 2017 CPM Educational Program. All rights reserved.
Calculus
1 ⋅ (−2x) = −2 x .
4− x 2
4− x 2
d f (x) = d ln(x 2 − 4) = 1
dx
dx
x2 − 4
129. D: For x in (–2, 2), f ′(x) =
For x > 2 or x < –2,
⋅ 2x =
−2 x
4− x 2
.
Both yield the same answer: D
130. E: f ′(x) = −
2
x3
= 8, 28 =
1 − 8x + 7, f (x) = −(−2) ⋅ 1 − 8 = 2 − 8 , which is zero when
′′
x2
x3
x3
1
3
−2
1
3
−2
3
, and undefined at x = 0. It is positive for
x = 4 , x = (2 ) = 2
x in the
interval (0, 2–2/3).
2=±
131. C: 4ydy = (2x − 3)dx
∫ 4ydy = ∫ (2x − 3)dx
2y 2 = x 2 − 3x + C
y2 =
x2
2
y=±
−
x+C
3
x
x2
2
−
3
2
1
2
− 23 + C
4 = −1 + C
C=5
y=±
x2
2
−
3
2
x + 5, f (2) = ± 2 − 3 + 5 = ±2
x+C
132. E: y′ = 4x3 + 9x2, y′(0) = 0, so the normal line is vertical; x = 0.
133. The real length is
4
∫− 4
1+ (2x)2 dx ≈ 33.637 , so answers should between 30 and 35 by
using secant lines.
134. The more secant lines that are used, the better they approximate the curve, so that the
estimated arc length becomes more accurate. That is, the shorter the line segments, the
better the segment follows the curve.
m=
f (a+h)− f (a)
h
b.
l=
((a + h) − a)2 + ( f (a + h) − f (a) ) = h 2 + ( f (a + h) − f (a) )
c.
Answers vary. Decreasing the value of h will improve the accuracy of both m and l.
135. a.
35
2
© 2017 CPM Educational Program. All rights reserved.
2
Calculus
136. a.
m = lim
h→0
f (a+h)− f (a)
h
= f ′(a) . It is the slope of the tangent line at x = a.
b.
dl = lim h 2 + ( f (a + h) − f (a) )
c.
f(a + h) – f(a) = h f ′(a) ⇒ dl = lim h 2 + ( hf ′(a) )
d.
lim (dx)2 + ( f ′(a)dx )
2
h→0
2
h→0
2
h→0
e.
lim (dx)2 + ( f ′(b)dx )
i.
2
h→0
lim (dx)2 + ( f ′(10)dx )
ii.
2
h→0
lim (dx)2 + ( f ′(x)dx )
iii.
2
h→0
lim (dx)2 + ( f ′(x)dx ) = lim (dx)2 (1+ f ′(x)2 ) = lim dx 1+ f ′(x)2
2
137. a.
h→0
b
=∫
a
h→0
1+ ( f ′(x) ) dx
2
f ′(x) = 2x ⇒
b.
h→0
4
∫− 4
1+ (2x)2 dx ≈ 33.637 in
“Smooth and continuous” implies that f ′ exists.
c.
π
138. a.
∫0
b.
∫1
9
1+ cos2 (x) dx ≈ 3.820 units (by calculator)
1+ ((x − 1)1/2 )2 dx = ∫
1
2
∫−2
f ′(x) = e x :
139. a.
e
∫1
f ′(x) = 1x :
b.
( )
140. ln(y) = ln e x
(2
1
V=∫
e
9
2
1+ (x − 1) dx = ∫
9
1
x dx =
2
3
2
1
54
3
− 23 =
52
3
units
1+ e2 x dx ≈ 9.010 units (by calculator)
x 2 +1
x2
≈ 2.0034 units (by calculator)
y
= x 2 → x = ln(y),
)
9
x 3/2 =
e
ln(y) dy = ∫ 4 ( ln(y) ) dy = 4 ( y ln(y) − y )
1
e
1
x
3
= 4(e – e) – 4(0 – 1) = 4 un
36
© 2017 CPM Educational Program. All rights reserved.
Calculus
141. xn+1 = xn −
f ′(x) = −
f (xn )
f ′ (xn )
(xn )
1 ⋅ 2x
(1+x 2 )
=
−2 x
(1+x 2 )2
12
x2 = 1− 1 2 = 2
x3 = 3.25, x4 = 5.023, …
The values keep getting larger, but the function has no root for Newton’s Method to
approach; only an asymptote of y = 0.
142. a.
∫0 2π x(2x + 1− (x 2 + 1))dx
Use shells:
c.
Use washers:
d.
Use shells:
2
∫0
x
(π (4 − x 2 )2 − π (4 − 2x)2 )dx
2
∫0 2π (4 − x)(2x + 1− (x 2 + 1))dx
f = x, dg = 5xdx, df = dx, g =
1
ln 5
⋅ 5x:
x 5x −
5 dx = x ⋅ 5 x − 5
5
+ C = ln(5)
( x − ln(5)1 ) + C
∫ x5 x dx = ln(5)
∫ ln(5)
ln(5)
(ln(5))2
x
dx
x
e ln(x)+1
dx
x
1
:∫
x
=∫
e
x
2 e
u du = u2
=
e
(ln(x)+1)2
2
1
b.
u = ln x + 1, du =
c.
u = x2 – 4, du = 2xdx:
∫ duu = 12 ln u + C = 12 ln x 2 − 4 + C
d.
u = x2 + 4, du = 2xdx:
∫ x2x+4 dx = 12 ln u + C = 12 ln(x 2 + 4) + C
144. a.
b.
37
y
2
b.
143. a.
2
∫0 (π (2x + 1)2 − π (x 2 + 1)2 )dx
Use washers:
x=1
x=1
= 2 − 12 =
3
2
Between 3 p.m. and 8 p.m., 1.25 feet of (accumulated) snow melted.
–0.25; The snow melted at an average rate of 0.25 ft/hour between 3 p.m. and 8 p.m.
© 2017 CPM Educational Program. All rights reserved.
Calculus
145. a.
Vavg =
1
20
=
1
20
⎛ 10 (10000 + 100t)dt + 20 10t dt ⎞
∫10
⎝ ∫0
⎠
(
10000t + 50t 2
10
0
)
20 ⎞
⎛ 1
+ ⎜ 20 ln(10)
10t ⎟
⎝
10 ⎠
1
= 100250 + 20 ln(10)
(10 20 − 1010 ) ≈ 2.1710 ⋅1018 km/hr
38
b.
The Mean Value Theorem does not apply because v is not continuous, but v does
equal Vavg at 10t = 2.171 ⋅ 1018, t = log(2.171 ⋅ 1018) ≈ 18.337 sec.
c.
aavg =
d.
The Mean Value Theorem does not apply, but a = aavg at
1
(ln(10))10t = 5(1018), t = log( ln(10)
⋅ 5(1018)) ≈ 18.337 sec.
Vfinal −V0
20seconds
= 10
20 −10000
20
≈ 5(1018) km/hr/sec
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Calculus