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MATHEMATICS 121, Worked Problems
1) Evaluate
a) 1 + 2x + 3x2 + 4x3 + · · ·+ 100x99 b)n∑
j=1
j2j
Solution. a)
1 + 2x + 3x2 + 4x3 + · · ·+ 100x99 =100∑
i=1
ixi−1 = ddx
100∑
i=0
xi = ddx
1−x101
1−x
= −101x100(1−x)+(1−x101)(1−x)2 = 1−101x100+100x101
(1−x)2
The right hand side is not defined for x = 1, but this case may be handled by usingl’Hopital’s rule to take the limit x → 1:
1 + 2x + 3x2 + 4x3 + · · ·+ 100x99∣
∣
∣
x=1= lim
x→11 + 2x + 3x2 + 4x3 + · · ·+ 100x99
= limx→1
1−101x100+100x101
(1−x)2
= limx→1
−10100x99+10100x100
2(x−1)
= limx→1
−999900x98+1010000x99
2= 5050
You can also use 1 + 2 + 3 + · · ·+ n = n(n+1)2 to handle this case.
b) Let x = 12 .
n∑
j=1
j2j =
n∑
j=1
jxj = xn
∑
j=1
jxj−1 = xddx
n∑
j=0
xj = xddx
1−xn+1
1−x
= x−(n+1)xn(1−x)+(1−xn+1)(1−x)2
Subbing x = 12
n∑
j=1
j2j = 1
2−(n+1)(1/2)n+1+(1−(1/2)n+1)
(1/2)2= 2
[
1− (n + 2)(
12
)n+1]= 2− n+2
2n
1
2) Find the area of a circle as a limit of the areas of inscribed polygons.
Solution. Inscribe a polygon with n equal sides in a circle of radiusr. The figure at the right illustrates such a polygon with n = 6.The polygon is built up out of n equal triangles. Each triangle hasone vertex, with angle 1
n2π, at the centre of the circle. Drop aperpendicular from that vertex to the centre of the opposite sideof the triangle. This perpendicular bisects the angle we were justtalking about. The length of the perpendicular is r cos π
n. We can
also view this as the height of the triangle. The base of the triangle
π/n
(the side opposite the centre of the circle) has length 2r sin πn . So one triangle has area
12(base)(height) = 1
2
(
r cos πn
)(
2r sin πn
)
= 12r2
(
2 sin πn cos π
n
)
= 12r2 sin 2π
n
The area of the polygon is the area of n triangles or n2 r2 sin 2π
n . The area of the circle is
limn→∞
n2 r2 sin 2π
n = limx→0
12
2πx r2 sin x = lim
x→0πr2 sin x
x = πr2 limx→0
sin xx
where we have substituted x = 2πn , or equivalently n = 2π
x . By L’Hopital’s rule,
limx→0
sin xx
= limx→0
cos x1
= 1, so the area of the circle is πr2 .
3) Evaluate the following integrals by interpreting it as an area.
a)∫ 2
−2
√4− x2 dx
b)∫ 3
0|3x− 5| dx
c)∫ 1
0
√4− x2 dx
Solution. Recall that the area between the x–axis and the curve y = f(x) with x running
from x = a to x = b is∫ b
af(x) dx.
a) The integral∫ 2
−2
√4− x2 dx represents the area between the x–axis and the curve
y =√
4− x2 with x running from −2 to 2. As the curve y =√
4− x2 is the top half ofthe curve y2 = 4−x2 or x2 +y2 = 4, and x = ±2 are the two values of x for which y = 0,
the integral∫ 2
−2
√4− x2 dx represents the area of the top half of the interior of the circle
x2 + y2 = 4. This circle has radius 2 and hence area π22 = 4π. The integral is half ofthis, or 2π .b) Observe that 3x− 5 takes the value zero for x = 5
3, is positive for x > 5
3and negative
for x < 53 . Hence
|3x− 5| ={
3x− 5 if x > 53
−(3x− 5) if x < 53
and
∫ 3
0
|3x−5| dx =
∫ 5/3
0
|3x−5| dx+
∫ 3
5/3
|3x−5| dx =
∫ 5/3
0
−(3x−5) dx+
∫ 3
5/3
(3x−5) dx
2
The first integral represents the area between the x–axis and the straight line y = 5− 3xwith x running from 0 (where y = 5) to 5
3(where y = 0). This is a triangle with height 5,
base 53 and area 1
25 53 = 25
6 . The second integral represents the area between the x–axisand the straight line y = 3x − 5 with x running from 5
3(where y = 0) to x = 3 (where
y = 4). This is a triangle with height 4, base 3− 53 = 4
3 and area 124 4
3 = 166 . All together
∫ 3
0
|3x− 5| dx =
∫ 5/3
0
−(3x− 5) dx +
∫ 3
5/3
(3x− 5) dx = 256 + 16
6 = 416
c) The integral∫ 1
0
√4− x2 dx represents the area be-
tween the x–axis and the top half of the circle x2+y2 = 4with 0 ≤ x ≤ 1. This regions consists of the cir-
cular sector OAB, which is the fraction π/62π = 1
12 ofa full disk of radius 2, and the triangle OBC, whichhas base 1 and height
√3. The total area is thus
112π22 + 1
2 × 1×√
3 = π3 +
√3
2 .
π3
B = (1,√
3)
C = (1, 0)
A = (0, 2)
O
2
4) Let h(x) =∫ sinx
−5t cos
(
t3)
dt. Evaluate h′(x).
Solution. Define
f(u) =
∫ u
−5
t cos(
t3)
dt and g(x) = sin x
Thendfdu = u cos
(
u3)
and dgdx = cos x
and h(x) = f(
g(x))
so that, by the chain rule,
h′(x) = f ′(
g(x))
g′(x) = u cos(
u3)
∣
∣
∣
u=g(x)cos x = sin x cos
(
sin3 x)
cos x
5) Find the area of the region bounded by the parabola y = x2, the tangent line to thisparabola at (1, 1) and the x–axis.
Solution. The slope of the parabola at x = 1 is m = ddxx2
∣
∣
∣
x=1= 2x
∣
∣
∣
x=1= 2. This is
also the slope of the tangent line to y = x2 at x = 1. So the equation of the tangent lineis y− 1 = m(x− 1) = 2(x− 1) or y = 2x− 1. We are to determine the area of the regionin the figure
x
y
(1, 1)
y = x2
y = 2x− 1
3
Use horizontal slices. On each slice, y is constant (with some fixed value between 0 and1) and x runs from
√y to (y + 1)/2. So, the area is
∫ 1
0
[
(y + 1)/2−√y]
dy =[
12
[
y2
2 + y]
− y3/2
3/2
]1
0= 1
2
[
12 + 1
]
− 13/23/2 = 34 − 2
3 = 112
6) Find the number b such that the line y = b divides the region bounded by the curvesy = x2 and y = 4 into two regions with equal area.
Solution. The area of the part of region with 0 ≤ y ≤ b is
A1 =
∫ b
0
√y dy = y3/2
3/2
∣
∣
∣
b
0= 2
3b3/2
The area of the part of region with b ≤ y ≤ 4 is
A2 =
∫ 4
b
√y dy = y3/2
3/2
∣
∣
∣
4
b= 2
343/2 − 23b3/2 x
y(2, 4)
y = x2
y = b
The two areas are equal when
23b3/2 = 2
343/2 − 2
3b3/2 ⇒ 2 2
3b3/2 = 2
343/2 ⇒ b3/2 = 1
243/2 ⇒ b = 4
22/3 = 22
22/3 = 24/3
7) Set up, but do not evaluate, an integral for the volume of the solid obtained by rotatingthe region bounded by the specified curves about the given line.a) y = ln x, y = 1, x = 1, about the x–axis.b) y = cos x, y = 0, x = 0, x = π
2 about the line y = 1.
Solution. a) When the vertical strip shown in the figure is rotated about the x–axisit forms a washer with inner radius ln x and outer radius 1. The area of this washer isπ12 − π(lnx)2. As x runs from 1 to e (which is where ln x = 1) the volume of the solid
is π∫ e
1[1− (ln x)2] dx .
x
yy = ln x
y = 1
x = 1
P1
P2
side view of slice
1
lnx
P1 P2
b) When the vertical strip shown in the figure below is rotated about the line y = 1 it formsa washer with inner radius 1− cos x and outer radius 1. The area of this washer is
π12 − π(1− cos x)2
4
As x runs from 0 to π2 , the volume of the solid is π
∫ π/2
0
[
12 − (1− cos x)2]
dx .
x
y
y = cos x
y = 1
8) Each of the following integrals represents the volume of a solid. Describe the solid.
a) π∫ 2
1y6 dy b) π
∫ 4
0
[
16− (x− 2)4]
dx
Solution. a) When the horizontal strip shown in the figure on the left below is rotatedabout the y–axis it forms a disk with radius f(y) and hence of area πf(y)2. The volumeof the region obtained when one rotates the curve x = f(y), a ≤ y ≤ b about the y–axis
is π∫ b
af(y)2 dy. This matches the given integral when a = 1, b = 2 and f(y) = y3.
So the integral represents the volume of the region obtained when one rotates the curvex = y3, 1 ≤ y ≤ 2 about the y–axis.
x
yx = f(y)
x
y
y = f(x)
y = g(x)
b) The integrand π[
16 − (x − 2)4]
= π42 − π[
(x − 2)2]2
is the area of a washer of outerradius 4 and inner radius (x− 2)2. Suppose in general that the region g(x) ≤ y ≤ f(x) isrotated about the x–axis. When the vertical strip shown in the figure on the right aboveis rotated about the x–axis it forms a washer with outer radius f(x) and inner radiusg(x) and hence of area π[f(x)2 − g(x)2]. The volume of the region obtained when one
rotates g(x) ≤ y ≤ f(x), a ≤ x ≤ b about the x–axis is π∫ b
a[f(x)2 − g(x)2] dx. This
matches the given integral when a = 0, b = 4, f(x) = 4 and g(x) = (x − 2)2. So theintegral represents the volume of the region obtained when one rotates (x− 2)2 ≤ y ≤ 4,0 ≤ x ≤ 4 about the x–axis.
5
9) Find the volume of a solid torus (donut) obtained by rotating the circle (x−R)2+y2 = r2
about the y–axis.
Solution.
(
R−√
r2 − y2, y) (
R +√
r2 − y2, y)
P1 P2
x
y
R
r
(x− R)2 + y2 = r2
top view of slice
R +√
r2 − y2
R−√
r2 − y2
P1 P2
The torus may be formed by rotating the circle in the figure on the left above about they–axis. Call this circle the “generating circle”. It is centred on (R, 0) and has radiusr and so has equation (x − R)2 + y2 = r2. Slice the donut horizontally. The figureon the right shows a top view of a typical slice. The points on one slice all have thesame value of y (with some −1 ≤ y ≤ 1). The slice looks like a washer. Our mainproblem is to determine the inner and outer radii of the washer. The portion of thewasher that lies in the generating circle is marked in both figures as a horizontal line.The end points of the line are denoted P1 and P2. Both P1 and P2 lie on the washerand hence have y–coordinate y. They also both lie on the generating circle. We maydetermine their x-coordinates by solving (x− R)2 + y2 = r2 for x. The x–coordinate of
P1 is x = R−√
r2 − y2 that of P2 is x = R +√
r2 − y2. So the washer has inner radius
R−√
r2 − y2 and outer radius R +√
r2 − y2. The area of the washer is
π(
R +√
r2 − y2)2 − π
(
R−√
r2 − y2)2
= π[
4R√
r2 − y2]
The volume of the donut is the integral of the cross–sectional area of the slice from y = −rto y = r. That is,
4πR
∫ r
−r
√
r2 − y2dy = 4πR(area of right half of a circle of radius r)
= 4πR(
12πr2
)
= 2π2Rr2
10) Evaluate∫ π
2
−π2
x2 sin x1+x6 dx.
Solution. The critical observations here are that the integrand is odd and the domainof integration is symmetric about x = 0. Recall that f(x) is odd if f(−x) = −f(x).I claim that, for any odd function f(x) and any positive number a,
∫ a
−af(x) dx = 0.
Intuitively, here is the reason. The part of the graph y = f(x) from x = 0 to x = aand part of the graph y = f(x) from x = 0 to x = −a are just upside down images of
6
y = f(x)
a−ax
y
each other. So the magnitude of the area between the x–axis and the part of the graphy = f(x) from x = −a to x = 0 is the same as the magnitude of the area between thex–axis and the part of the graph y = f(x) from x = 0 to x = a. The signed areas
are negatives of each other. That is∫ 0
−af(x) dx = −
∫ a
0f(x) dx, so that
∫ a
−af(x) dx =
∫ 0
−af(x) dx +
∫ a
0f(x) dx = 0. To see algebraically that
∫ 0
−af(x) dx = −
∫ a
0f(x) dx, just
make the substitution x = −t, dx = −dt:
∫ 0
−a
f(x) dx =
∫ 0
a
f(−t) (−dt) = −∫ 0
a
f(−t) dt =
∫ a
0
f(−t) dt = −∫ a
0
f(t) dt
The oddness of f was used in the last step. Choosing f(x) = x2 sinx1+x6 and a = π
2gives
∫ π2
−π2
x2 sinx1+x6 dx = 0 .
11) Evaluate
a)∫
dxex+e−x b)
∫ π/2
0
√1 + cos x dx c)
∫ π/2
0
√1− sin x dx
Solution. a) Make the substitution y = ex, dy = ex dx = y dx.
∫
dxex+e−x =
∫
1y+1/y
1y
dy =
∫
11+y2 dy = tan−1 y + C = tan−1 ex + C
b) Using the trig identity cos 2θ = 2 cos2 θ − 1 with 2θ = x
∫ π/2
0
√1 + cos x dx =
∫ π/2
0
√
2 cos2 x2 dx
Since cos x2 ≥ 0 for all 0 ≤ x ≤ π
2
∫ π/2
0
√
2 cos2 x2
dx =
∫ π/2
0
√2 cos x
2dx =
√2× 2 sin x
2
∣
∣
∣
π/2
0=√
2× 2 sin π4
= 2
b) (Second solution.)
∫ π/2
0
√1 + cos x dx =
∫ π/2
0
√
(1+cos x)(1−cos x)1−cos x dx =
∫ π/2
0
√
1−cos2 x1−cos x dx =
∫ π/2
0
sin x√1−cos x
dx
Now substitute y = 1− cos x, dy = sin x dx
∫ π/2
0
sinx√1−cos x
dx =
∫ 1
0
dy√y
= 2√
y∣
∣
∣
1
0= 2
7
c) Make the substitution y = π2 − x, dy = −dx.
∫ π/2
0
√1− sinx dx =
∫ 0
π/2
√
1− cos y (−dy) =
∫ π/2
0
√
1− cos y dy
By the trig identity cos 2θ = 1− 2 sin2 θ with θ = y2,
∫ π/2
0
√
1− cos y dy =
∫ π/2
0
√
2 sin2 y2
dy =√
2
∫ π/2
0
sin y2
dy = −2√
2 cos y2
∣
∣
π/2
0
= 2√
2(
1− cos π4
)
= 2(√
2− 1)
12) Find the volume of the solid obtained by rotating the region bounded by y = 4x − x2
and y = 8x− 2x2 about x = −2.
Solution. The region bounded by y = 4x − x2 andy = 8x − 2x2 is sketched on the right. Note that thetwo parabolas meet at (0, 0) and (4, 0). Consider thethin slice in the figure on the right. It runs verticallyfrom (x, 4x − x2) to (x, 8x − 2x2) and has width dx.When this slice is rotated about x = −2, it sweeps outa cylindrical shell. A radius for the shell is shown in the
x
y
−2 2 4
y = 8x− 2x2
y = 4x− x2
figure on the right. It is the horizontal line half way up the thin slice. The x–coordinateof the right hand end of the radius is x and the x–coordinate of the left hand end is−2. So the radius has length x− (−2) = x + 2. The height of the shell is the differencebetween the y–coordinates at the top and bottom of the thin slice. So the height of theshell is (8x− 2x2)− (4x− x2) = 4x− x2. The thickness of the shell is dx. So its volumeis 2π(x + 2)(4x− x2)dx. The total volume of the solid is
∫ 4
0
2π(x + 2)(4x− x2)dx = 2π
∫ 4
0
(8x + 2x2 − x3) dx = 2π[
4x2 + 23x3 − 1
4x4
]4
0
= 2π[
64 + 1283 − 64
]
= 2563 π
13) Find the volume of the solid obtained by rotating the region bounded by x = 4− y2 andx = 8− 2y2 about y = 5.
Solution. The region bounded by x = 4−y2 and x = 8−2y2 is sketched below. Note thatthe two parabolas meet when 4− y2 = 8− 2y2 or y2 = 4 or y = ±2. The correspondingx = 4− (±2)2 = 0. So, the two parabolas meet at (0,±2). Consider the thin slice in thefigure on the right. It runs horizontally from (4− y2, y) to (8− 2y2, y) and has width dy.When this slice is rotated about y = 5, it sweeps out a cylindrical shell, as illustrated inthe figure on the left. A radius for the shell is shown in the figure on the right. It is the
8
5− y
x
y
−2
2
5
x = 4− y2
x = 8− 2y2
vertical line half way along the thin slice. The y–coordinate of the top end of the radiusis 5 and the y–coordinate of the bottom end is y. So the radius has length 5 − y. Theheight of the shell is the difference between the x–coordinates at the right and left handends of the thin slice. So the height of the shell is (8 − 2y2) − (4 − y2) = 4 − y2. Thethickness of the shell is dy and its volume is 2π(5 − y)(4 − y2)dy. The total volume ofthe solid is
∫ 2
−2
2π(5− y)(4− y2)dy = 2π
∫ 2
−2
(20− 4y − 5y2 + y3) dy
For any odd power yn of y, and any a, the integral∫ a
−ayn dy = 0. This is because
the area with −a ≤ y ≤ 0 has the same magnitude but opposite sign as the area with
0 ≤ y ≤ a. See the figure on the left below. Thus the integrals∫ 2
−2y dy =
∫ 2
−2y3 dy = 0.
x
y y = x3
x
y y = x2
For any even power yn of y, and any a, the integral∫ a
−ayn dy = 2
∫ a
0yn dy. This is
because the area with −a ≤ y ≤ 0 has the same magnitude and same sign as the areawith 0 ≤ y ≤ a. See the figure on the right above.The volume of the solid is
2π
∫ 2
−2
(20− 4y − 5y2 + y3) dy = 2π
∫ 2
−2
(20− 5y2) dy = 4π
∫ 2
0
(20− 5y2) dy
= 20π
∫ 2
0
(4− y2) dy = 20π[
4y − 13y3
]2
0= 20π
[
8− 83
]
= 3203 π
14) The region below the curve y = 1√1−x2
, above the x–axis and between the y–axis and the
line x = 12 is rotated about the line x = 2.
a) Express the volume of the solid of revolution so obtained as a definite integral.
9
b) Evaluate the integral of part a.
Solution. a) When a point (x, y), between the y–axis and the line x = 12, is rotated
about the line x = 2, it sweeps out a circle of radius 2 − x and hence of circumference2π(2− x). So, by the method of cylindrical shells,
Vol =∫ 1/2
02π(2− x) 1√
1−x2dx
b)
Vol =
∫ 1/2
0
2π(2− x) 1√1−x2
dx = 4π
∫ 1/2
0
1√1−x2
dx− π
∫ 1/2
0
2x√1−x2
dx
For the second integral, sub in t = 1− x2, dt = −2xdx.
Vol = 4π arcsin x∣
∣
∣
1/2
0+ π
∫ 3/4
1
1√tdt = 4π π
6+ π
[ √t
1/2
]3/4
1= 2
3π2 +
√3π − 2π
15) Let R be the plane region bounded by x = 0, x = 1, y = 0 and y = c√
1 + x2, where cis a positive constant.(a) Find the volume V1 of the solid obtained by revolving R about the x–axis.(b) Find the volume V2 of the solid obtained by revolving R about the y–axis.(c) If V1 = V2, what is the value of c?
Solution. a) Let V1 be the solid obtained by revolving R about the x–axis. The portionof V1 with x–coordinate between x and x+ dx is a disk of radius c
√1 + x2 and thickness
dx. The volume of this disk is π(c√
1 + x2)2dx = πc2(1 + x2) dx. The total volume of V1
is
V1 =
∫ 1
0
πc2(1 + x2) dx = πc2[
x + x3
3
]1
0= 4
3πc2
b) Let V2 be the solid obtained by revolving R about the y–axis. Then V2 can be thoughtof as a union of cylindrical shells. The shell whose intersection with the x–axis is theinterval from x to x + dx has radius x, height c
√1 + x2 and thickness dx. The volume
of this shell is 2πx(c√
1 + x2) dx = 2πcx√
1 + x2 dx. The total volume of V2 is (makingthe substitution y = 1 + x2, dy = 2x dx)
V2 =
∫ 1
0
2πcx√
1 + x2 dx =
∫ 2
1
πc√
y dy = πc y3/2
3/2
∣
∣
∣
2
1= 2
3πc[23/2 − 1]
c) If V1 = V2,
43πc2 = 2
3πc[23/2 − 1] =⇒ c = 0 or c = 1
2[23/2 − 1]
16) A 45◦ notch is cut to the centre of a cylindrical log having radius 20 cm. One plane faceof the notch is perpendicular to the axis of the log. What volume of wood was removed?
10
Solution. Slice the notch into rectangles as in the figure on the right below.
Suppose that the base of the notch is in the xy–plane. Then the circular part of theboundary of the base of the notch has equation x2 + y2 = 202. If our coordinate systemis such that x is constant on each slice, then the slice has width 2y = 2
√202 − x2 and
height x (since the upper face of the notch is at 45◦ to the base). So the slice hascross–sectional area 2x
√202 − x2 and the volume is
V =
∫ 20
0
2x√
202 − x2 dx
Make the change of variables t = 202 − x2, dt = −2x dx.
V =
∫ 0
202
√t (−dt) = − t3/2
3/2
∣
∣
0
202 = 23203 = 16,000
3
Solution 2 Suppose that the base of the notch is in the xy–plane with the skinny edgealong the y–axis. Slice the notch into triangles parallel to the x–axis as in the figurebelow.
Then the circular part of the boundary of the base of the notch has equation x2+y2 = 202.Our coordinate system is such that y is constant on each slice, so that the slice has bothbase and height x =
√
202 − y2 (since the upper face of the notch is at 45◦ to the base).
So the slice has cross–sectional area 12
(√
202 − y2)2
and the volume is
V = 12
∫ 20
−20
(202 − y2) dy =
∫ 20
0
(202 − y2) dy =[
202y − y3
3
]20
0= 2
3203 = 16,0003
17) Evaluate the integrals
a)
∫
x tan−1 x dx b)
∫
sin(lnx) dx c)
∫ 4
1
e√
x dx
11
Solution. a) Integrate by parts with u = tan−1 x and dv = x dx. Then du = 11+x2 dx
and v = 12x2, so
∫
x tan−1 dx = 12x2 tan−1 x− 1
2
∫
x2
1+x2 dx
The integral on the right hand side is
∫
x2
1+x2 dx =
∫
1+x2−11+x2 dx =
∫
[
1− 11+x2
]
dx =
∫
dx−∫
11+x2 dx = x− tan−1 x + C
All together∫
x tan−1 dx = 12x2 tan−1 x + 1
2 tan−1 x− 12x + C ′
with C ′ = 12C.
b) First, substitute y = ln x, dy = 1x dx, in order to simplify the argument of sin. Because
dx = ey dy,∫
sin(ln x) dx =
∫
ey sin y dy
Integrate by parts using u = ey and dv = sin y dy so that du = ey dy and v = − cos y.
∫
ey sin y dy = −ey cos y −∫
(− cos y)ey dy = −ey cos y +
∫
ey cos y dy
Integrate by parts a second time using u = ey and dv = cos y dy so that du = ey dy andv = sin y.
∫
ey sin y dy = −ey cos y +
∫
ey cos y dy = −ey cos y + ey sin y −∫
(sin y)ey dy
Moving the −∫
(sin y)ey dy from the right hand side to the left hand side,
2
∫
ey sin y dy = −ey cos y + ey sin y =⇒∫
ey sin y dy = 12
[
− ey cos y + ey sin y]
All together
∫
sin(lnx) dx =
∫
ey sin y dy = 12
[
ey sin y − ey cos y]
+ C
= 12
[
x sin(lnx)− x cos(lnx)]
+ C
c) First, substitute y =√
x, dy = 12√
xdx, in order to simplify the argument of the
exponential. Because dx = 2√
x dy = 2y dy and y(1) = 1, y(4) = 2,
∫ 4
1
e√
x dx =
∫ 2
1
ey 2y dy = 2
∫ 2
1
yey dy
12
Integrate by parts using u = y and dv = ey dy so that du = dy and v = ey.
∫ 4
1
e√
x dx = 2
∫
yey dy = 2[
yey −∫ 2
1
ey dy]
= 2[
yey − ey]2
1
= 2[
(2e2 − e2)− (1e1 − e1)]
= 2e2
18) Evaluate∫
dx1−sinx .
Solution.
∫
dx1−sin x =
∫
1+sin x1−sin2 x
dx =
∫
1+sin xcos2 x dx =
∫
sec2 x dx +
∫
sinxcos2 x dx
= tanx−∫
d(cos x)cos2 x
= tan x + 1cos x
+ C
Solution 2 Using the trig identities sin y = cos(
π2− y
)
and cos(2y) = 1− 2 sin2 y,
∫
dx1−sin x
=
∫
11−cos( π
2−x)dx =
∫
11−[1−2 sin2( π
4− x2 )]
dx = 12
∫
1sin2( π
4− x2 )
dx
= cot(π4 − x
2 ) + C
The answers in the two solutions are really the same. Using the trig identities sin(a−b) =sin a cos b− cos a sin b and cos(a− b) = cos a cos b + sin a sin b,
cot(π4− x
2) =
cos( π4− x
2 )
sin( π4− x
2 )=
cos π4 cos x
2 +sin π4 sin x
2
sin π4 cos x
2−cos π4 sin x
2
Since sin π4
= cos π4
cot(π4− x
2) =
cos x2 +sin x
2
cos x2−sin x
2=
[cos x2 +sin x
2 ]2
cos2 x2−sin2 x
2=
1+2 sin x2 cos x
2
cos x= 1+sinx
cos x= sec x + tan x
19) Household electricity is supplied in the form of alternating current that varies from −155V to 155 V with a frequency of 60 cycles per second. The voltage is given by the equation
V (t) = 155 sin(120πt)
where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage,which is the square root of the average value of [V (t)]2 over one cycle. Calculate theRMS voltage of household current.
13
Solution. The voltage goes through 60 cycles in one second. So one cycle takes 1/60 secand
RMS voltage =
√
1
1/60
∫ 1/60
0
V (t)2 dt =
√
60
∫ 1/60
0
1552 sin2(120πt) dt
=
√
60(155)2∫ 1/60
0
1−cos(2×120πt)2
dt
=
√
1260(155)2
[
t− sin(2×120πt)240π
]1/60
0=
√
1260(155)2 1
60
= 155√2≈ 109.6 V
20) If a freely falling body starts from rest, then its displacement is given by s = 12gt2. Let
the velocity after a time T be vT . Show that if we compute the average of the velocitieswith respect to t we get vave = 1
2vT , but if we compute the average of the velocities with
respect to s we get vave = 23vT .
Solution. The speed at time t is dsdt = gt. In particular vT = gT . The average velocity
over the time interval 0 ≤ t ≤ T is
1T
∫ T
0
gt dt = 1T
[
12gt2
]T
0= 1
T12gT 2 = 1
2gT = 12vT
The body reaches displacement s at the time t that obeys s = 12gt2. That is, at t =
√
2s/g. At displacement s, the velocity of the body is g√
2s/g =√
2sg. The averagevelocity over the displacement interval 0 ≤ s ≤ 1
2gT 2 (the final displacement is the valueof s at time T ) is
1gT 2/2
∫ gT 2/2
0
√
2sg ds = 1gT 2/2
[√
2g s3/2
3/2
]gT 2/2
0= 1
gT 2/2
√
2g (gT 2/2)3/2
3/2= 1
gT 2/2g2T 3
3
= 23gT = 2
3vT
21) Show that if f(a) = f(b) = 0 and f is twice differentiable, then
∫ b
a
(x− a)(b− x)f ′′(x) dx = −2
∫ b
a
f(x) dx
Solution. Integrating by parts with u = (x − a)(b − x), du = [(b − x) − (x − a)] dx,dv = f ′′(x) dx and v = f ′(x)
∫ b
a
(x− a)(b− x)f ′′(x) dx = (x− a)(b− x)f ′(x)∣
∣
∣
b
a−
∫ b
a
(a + b− 2x)f ′(x) dx
= −∫ b
a
(a + b− 2x)f ′(x) dx
14
Integrating by parts with u = a + b− 2x, du = −2 dx, dv = f ′(x) dx and v = f(x)
∫ b
a
(x− a)(b− x)f ′′(x) dx = −∫ b
a
(a + b− 2x)f ′(x) dx
= −(a + b− 2x)f(x)∣
∣
∣
b
a− 2
∫ b
a
f(x) dx
= −2
∫ b
a
f(x) dx
22) Find the volume enclosed by the ellipsoid
x2
a2 + y2
b2+ z2
c2 = 1
Solution. Suppose that we compute the volume by slicing it into horizontal disks.
z
yx
h
c
ba
The height z of a disk varies from z = −c to z = c. The disk at height z has equationx2
a2 + y2
b2 = 1− z2
c2 . This is an ellipse with semi–axes a√
1− z2
c2 and b√
1− z2
c2 . Hence the
cross–sectional area of the disk is πa√
1− z2
c2 b√
1− z2
c2 = πab(
1 − z2
c2
)
. The volume of
the the ellipsoid is
V =
∫ c
−c
πab(
1− z2
c2
)
dz = 2πab
∫ c
0
(
1− z2
c2
)
dz = 2πab[
z − z3
3c2
]c
0= 4
3πabc
23) Evaluate the following integrals.
a)
∫ 2
0
x3√
4− x2 dx b)
∫
dx√x2+4x+8
c)
∫
dx(5−4x−x2)5/2
Solution. a) Make the substitution u = 4 − x2. Then du = −2x dx so thatx3√
4− x2 dx = x2√
4− x2 xdx = (4 − u)√
u du−2 . When x = 0, u = 4 and when x = 2,
u = 0. Hence∫ 2
0
x3√
4− x2 dx =
∫ 0
4
(4− u)√
u du−2 =
∫ 0
4
(
− 2√
u + 12u3/2
)
du =[
− 43u3/2 + 1
5u5/2]0
4
= 438− 1
532 =(
13 − 1
5
)
32 = 6415
15
b)∫
dx√x2+4x+8
=
∫
dx√x2+4x+4+4
=
∫
dx√(x+2)2+4
Sub in x + 2 = 2 tan θ. Then dx = 2 sec2 θ dθ and√
(x + 2)2 + 4 =√
4 tan2 θ + 4 =√4 sec2 θ = 2 sec θ. So
∫
dx√x2+4x+8
=
∫
2 sec2 θ dθ2 sec θ =
∫
sec θ dθ = ln∣
∣ sec θ + tan θ∣
∣ + C
Subbing in θ in terms of x
∫
dx√x2+4x+8
= ln∣
∣
√x2+4x+8
2 + x+22
∣
∣ + C
θ2
x + 2
√
(x + 2)2 + 4
c)∫
dx(5−4x−x2)5/2 =
∫
dx(9−4−4x−x2)5/2 =
∫
dx(9−(2+x)2)5/2
Sub in x + 2 = 3 sin θ. Then dx = 3 cos θ dθ and(
9 − (x + 2)2)5/2
=(
9− 9 sin2 θ)5/2
=(
9 cos2 θ)5/2
= 35 cos5 θ. So
∫
dx(5−4x−x2)5/2 =
∫
3 cos θ dθ35 cos5 θ = 1
81
∫
dθcos4 θ = 1
81
∫
sec2 θ sec2 θ dθ
= 181
∫
[1 + tan2 θ] sec2 θ dθ
Now sub in y = tan θ. Then dy = sec2 θ dθ so
∫
dx(5−4x−x2)5/2 = 1
81
∫
[1 + y2] dy = 181
[
y + y3
3
]
+ C
Subbing back in y = tan θ,
∫
dx(5−4x−x2)5/2 = 1
81
[
tan θ + tan3 θ3
]
+ C
Finally we have sub back in θ in terms of x, using x + 2 = 3 sin θ. From the triangle ofthe right, tan θ = x+2√
9−(x+2)2= x+2√
5−4x−x2so
∫
dx(5−4x−x2)5/2 = 1
81
[
x+2√5−4x−x2
+ (x+2)3
3(5−4x−x2)3/2
]
+ C
θx + 2
√
9− (x + 2)23
16
24) Find the area of the crescent shaped region bounded by arcs ofcircles with radii r and R, as in the figure to the right.
Solution. The large circle has equation x2 + y2 = R2. The smallone is centred at x = 0, y =
√R2 − r2 and has radius r, so its
equation is x2 + (y −√
R2 − r2)2 = r2. The area is
R
r
∫ r
−r
[
√
R2 − r2 +√
r2 − x2 −√
R2 − x2]
dx
=
∫ r
−r
√
R2 − r2 dx +
∫ r
−r
√
r2 − x2 dx−∫ r
−r
√
R2 − x2 dx
The first integral represents the area of a rectangle of width 2r and height√
R2 − r2. Sothe first integral equals 2r
√R2 − r2. The second integral represents the area of the top
half of a circle of radius r. So the second integral equals 12πr2. For the third integral
substitute x = R sin θ. Then dx = R cos θ dθ. When x = ±r, θ = ± sin−1 rR . So
∫ r
−r
√
R2 − x2 dx =
∫ sin−1 rR
− sin−1 rR
√
R2 − R2 sin2 θ R cos θ dθ
= 2
∫ sin−1 rR
0
R2 cos2 θ dθ = 2R2
∫ sin−1 rR
0
1+cos 2θ2 dθ
= R2[
θ + sin 2θ2
]sin−1 rR
0= R2
[
θ + sin θ cos θ]sin−1 r
R
0
When θ = sin−1 rR , sin θ = r
R and cos θ =√
1− r2
R2 so
∫ r
−r
√
R2 − x2 dx = R2[
sin−1 rR + r
R
√
1− r2
R2
]
and the area is
2r√
R2 − r2 + 12πr2 −R2 sin−1 r
R − R2 rR
√
1− r2
R2 = r√
R2 − r2 + 12πr2 −R2 sin−1 r
R
25) Let R be the region in the xy–plane bounded by y = x sinx, y = 0, x = 0 and x = π.a) Find the volume of the solid obtained by rotating R about the y–axis.b) Find the volume of the solid obtained by rotating R about the x–axis.
Solution. a) By cylindrical shells,
Vol =
∫ π
0
2πx(
x sin x)
dx = 2π
∫ π
0
x2 sin x dx
17
By integration by parts with u = x2, dv = sin x dx, du = 2x dx, v = − cos x
Vol = 2πx2(− cos x)∣
∣
∣
π
0+ 4π
∫ π
0
x cosx dx = 2π3 + 4π
∫ π
0
x cosx dx
By integration by parts, a second time, with u = x, dv = cos x dx, du = dx, v = sin x
Vol = 2π3 + 4πx sinx∣
∣
∣
π
0− 4π
∫ π
0
sin x dx = 2π3 − 4π(− cosx)∣
∣
∣
π
0= 2π3 − 8π
b)
Vol =
∫ π
0
π(
x sinx)2
dx = π
∫ π
0
x2 1−cos 2x2 dx
= π2
[
x3
3− 1
2x2 sin 2x− 1
2x cos 2x + 1
4sin 2x
]π
0= π
2
[
π3
3− 1
2π]
= π2
2
[
π2
3− 1
2
]
26) Let R be the region bounded by the line y = 2 and y = (x − 4)2 − 2. Calculate theposition of the centroid of R.Solution. The region is symmetric about x = 4, so x = 4 .The line and parabola intersect when (x − 4)2 − 2 = 2 or(x− 4)2 = 4 or x− 4 = ±2 or x = 2, 6. The area of R is
A =
∫ 6
2
[
2− [(x− 4)2 − 2]]
dx =
∫ 6
2
[
− x2 + 8x− 12]
dx
=[
− x3
3 + 4x2 − 12x]6
2= 0− [− 32
3 ] = 323
x
y
y = (x− 4)2 − 2
y = 2
We use vertical approximating rectangle to compute x. A typical rectangle has width dxand y running from (x − 4)2 − 2 to 2. Thus the average value of y on the rectangle is12
[
2 + [(x− 4)2− 2]]
= 12(x− 4)2 and the area of the rectangle is
[
2− [(x− 4)2− 2]]
dx =[
− x2 + 8x− 12]
dx.
y = 1A
∫ 6
2
12 (x− 4)2
[
− x2 + 8x− 12]
dx = 1A
∫ 6
2
12(x− 4)2(6− x)(x− 2) dx
Sub in x = t + 4, dx = dt.
y = 1A
∫ 2
−2
12 t2(2− t)(t + 2) dt = 3
64
∫ 2
−2
[4t2 − t4] dt = 332
∫ 2
0
[4t2 − t4] dt
= 332
[
4 t3
3 − t5
5
]2
0= 3
32
[
323 − 32
5
]
= 25
18
27) Find the x–coordinate of the centroid (centre of gravity) of the plane region R that lies inthe first quadrant x ≥ 0, y ≥ 0 and inside the ellipse 4x2 +9y2 = 36. (The area bounded
by the ellipse x2
a2 + y2
b2 = 1 is πab square units.)
Solution. In standard form 4x2 + 9y2 = 36 is x2
9+ y2
4= 1. The ellipse has a = 3 and
b = 2. So, on R, x runs from 0 to a = 3 and R has area A = 14π × 3× 2 = 3
2π. For each
fixed x, between 0 and 3, y runs from 0 to y(x) = 2√
1− x2
9 .
x = 1A
∫ 3
0
x y(x) dx = 1A
∫ 3
0
x 2
√
1− x2
9 dx = 43π
∫ 3
0
x
√
1− x2
9 dx
Sub in t = 1− x2
9 , dt = − 29x dx.
x = − 92
43π
∫ 0
1
√t dt = − 9
243π
[
t3/2
3/2
]0
1= − 9
243π
[
− 23
]
= 4π
28) a) Find the centroid of the quarter circular disk x ≥ 0, y ≥ 0, x2 + y2 ≤ a2.b) Find the centroid of the region R in the diagram.
(2, 2)2
1
21
2
R
quartercircle
Solution. a) By symmetry, x = y. The area of the quarter disk is A = 14πa2.
x = 1A
∫ a
0
x√
a2 − x2 dx
To evaluate the integral, sub in t = a2 − x2, dt = −2x dx.
∫ a
0
x√
a2 − x2 dx =
∫ 0
a2
√t dt−2 = − 1
2
[
t3/2
3/2
]0
a2= a3
3 (∗)
Sox = 4
πa2
[
a3
3
]
= 4a3π
19
b) Again, by symmetry, x = y. The area of the region is A = 2× 2− 14π = 16−π
4 .
x = 1A
[∫ 1
0
x[2−√
1− x2] dx +
∫ 2
1
x[2] dx
]
= 416−π
[
x2∣
∣
1
0+ x2
∣
∣
2
1−
∫ 1
0
x√
1− x2 dx
]
= 416−π
[
4− 13
]
by (∗) with a = 1
= 4448−3π
Solution 2 When the quarter disk is rotated about the y–axis a hemisphere is formed.The hemisphere has radius a and hence volume 1
243πa3 = 2
3πa3. The quarter disk has
area 14πa2 = 1
4πa2. By Pappus’s Theorem 23πa3 = V = 2πx A = 1
2π2a2 x, so thatx = 4a
3π. To get y = x either invoke symmetry or rotate about the x–axis.
When R is rotated about the y–axis a cylinder with a hemispherical hole is formed. Thecylinder has radius 2 and height 2 and hence volume π × 22 × 2 and the hole has radius1 and hence volume 1
243π × 12. So the volume of the solid formed by rotating R is 22
3π.
R has area 22 − 14π. By Pappus’s Theorem x = V
2πA= 22π/3
2π(16−π)/4= 44
48−3π. Again, to
get y = x either invoke symmetry or rotate about the x–axis.
29) Prove that the centroid of any triangle is located at the point of intersection of themedians.
Solution. Choose a coordinate system so that the vertices of the triangle are located at(a, 0), (0, b) and (c, 0). (In the figure, a is negative.)
(a, 0) (c, 0)
(0, b)
The line joining (a, 0) and (0, b) has equation bx + ay = ab. (Check that (a, 0) and (0, b)both really are on this line.) The line joining (c, 0) and (0, b) has equation bx + cy =bc. (Check that (c, 0) and (0, b) both really are on this line.) Hence for each fixed ybetween 0 and b, x runs from a − a
b y to c − cby. A thin strip at height y has length
`(y) =[
(c − cby) − (a − a
by)
]
= c−ab
(b − y). On this strip y has average value y and x
has average value 12
[
(a − ab y) + (c − c
by)]
= a+c2b (b − y). As the area of the triangle is
A = 12(c− a)b
y = 1A
∫ b
0
y`(y) dy = 2(c−a)b
∫ b
0
y c−ab (b− y) dy = 2
b2
∫ b
0
(by − y2) dy = 2b2
(
b b2
2 − b3
3
)
= 2b2
b3
6 = b3
20
x = 1A
∫ b
0
a+c2b
(b− y)`(y) dy = 2(c−a)b
∫ b
0
a+c2b
(b− y) c−ab
(b− y) dy = a+cb3
∫ b
0
(y − b)2 dy
= a+cb3
[
13(y − b)3
]b
0= a+c
b3b3
3= a+c
3
The midpoint of the line joining (0, b) and (c, 0) is 12(c, b). The point two thirds of the
way from (a, 0) to 12(c, b) is 1
3(a, 0) + 2
312(c, b) =
(
a+c3
, b3
)
= (x, y) as desired.
30) A water storage tanker has the shape of a cylinder with diameter 10 ft. It is mountedso that the circular cross–sections are vertical. If the depth of the water is 7 ft, whatpercentage of the total capacity is being used?
Solution. Here is an end view of the tank. The shaded part of the circle is filled withwater. The cross–sectional area of the the filled part of the tank is the area of a circle
θ 5
7
of radius 5, namely π52, minus the area of the unfilled cap. The area of the cap is
Area( )
= Area(
2θ 5
)
−Area(
θ 5
)
The first term is the fraction 2θ2π
of the area of a circle of radius 5. The triangle in thesecond has base 2× 5 cos θ and height 5 sin θ. So the area of the cap is
2θ2π π52 − 1
2(2× 5 cos θ)(5 sin θ) = 25θ − 25 sin θ cos θ
The tank is filled to a height of 7 ft, which is 2 ft above the centre of the tank. So
cos θ = 25 = 0.4 and sin θ =
√
1− 425 ≈ 0.917 and θ = arccos 2
5 ≈ 1.159. The area of
the cap is 25[
1.159− 0.4× 0.917]
≈ 19.82. The area of the full circle is A = 25π, so thefraction filled is
A−19.82A = 58.72
78.54 = 0.74 = 74.8%
31) Evaluate the following integrals.
a)
∫
18−2x−4x2
x3+4x2+x−6dx b)
∫
x3
(x+1)3dx c)
∫
2x3−xx4−x2+1
dx d)
∫ 1
0
√
x2 + 1 dx
21
Solution. a) First, we have to factor the denominator P (x) = x3 + 4x2 + x − 6. Thisis a polynomial with integer coefficients. So any integer roots of P (x) must factor theconstant term, 6, of P (x). The only possible candidates for integer roots of P (x) are±1, ±2, ±3, ±6. Try x = 1. Because P (1) = 0, x = 1 is a root of P (x) and x− 1 mustbe a factor of P (x). By long division,
P (x) = (x− 1)(x2 + 5x + 6) = (x− 1)(x + 2)(x + 3)
The degree of the numerator 18− 2x− 4x2 is 2, which is smaller than the degree of thedenominator, which is 3. So the integrand must be expressible in the form
18−2x−4x2
(x−1)(x+2)(x+3) = ax−1 + b
x+2 + cx+3
The easy way to find a is to multiply both sides of this equation by x− 1
a + (x− 1) bx+2 + (x− 1) c
x+3 = 18−2x−4x2
(x+2)(x+3)
and then set x = 1a = 18−2x−4x2
(x+2)(x+3)
∣
∣
∣
x=1= 18−2−4
3×4 = 1
Similarly,
b = 18−2x−4x2
(x−1)(x+3)
∣
∣
∣
x=−2= 18+4−16
(−3)×1 = −2
c = 18−2x−4x2
(x−1)(x+2)
∣
∣
∣
x=−3= 18+6−36
(−4)×(−1)= −3
To check this, put everything back over the common denominator (x− 1)(x + 2)(x + 3)
1x−1
+ −2x+2
+ −3x+3
= (x+2)(x+3)−2(x−1)(x+3)−3(x−1)(x+2)(x−1)(x+2)(x+3)
= (1−2−3)x2+(5−4−3)x+(6+6+6)(x−1)(x+2)(x+3)
This is the same as the original integrand. So
∫
18−2x−4x2
x3+4x2+x−6dx =
∫
[
1x−1
+ −2x+2
+ −3x+3
]
dx = ln |x− 1| − 2 ln |x + 2| − 3 ln |x + 3|+ C
b) Substitute t = x + 1, dt = dx.
∫
x3
(x+1)3dx =
∫
(t−1)3
t3dt =
∫
t3−3t2+3t−1t3
dt =
∫
(
1− 3t−1 + 3t−2 − t−3)
dt
= t− 3 ln |t| − 3 1t
+ 12t2
+ C = x− 3 ln |x + 1| − 3 1x+1
+ 12(x+1)2
+ C ′
with C ′ = C − 1.
c) Trick question! The numerator is, aside from a factor of 2, the derivative of thedenominator. So make the change of variables u = x4 − x2 + 1, du = 4x3 − 2x.
∫
2x3−xx4−x2+1 dx =
∫
du/2u = 1
2 ln |u|+ c = 12 ln |x4 − x2 + 1|+ c
22
d) Solution 1 Sub in x = tan θ. Then dx = sec2 θ dθ and√
x2 + 1 =√
tan2 θ + 1 =√sec2 θ = sec θ. When x = 0, tan θ = 0 so θ = 0. When x = 1, tan θ = 1, so θ = π
4 .
∫ 1
0
√
x2 + 1 dx =
∫ π/4
0
sec θ sec2 θ dθ =
∫ π/4
0
sec3 θ dθ
= 12
[
sec θ tan θ + ln∣
∣ sec θ + tan θ∣
∣
]π/4
0
= 12
[√2 + ln
∣
∣
√2 + 1
∣
∣
]
− 12
[
1× 0 + ln∣
∣1 + 0∣
∣
]
= 12
[√2 + ln
(√2 + 1
)
]
d) Solution 2 Sub in x = 12
(
ey − e−y)
. Then
dx = 12
(
ey + e−y)
dy√
x2 + 1 =[
14
(
ey − e−y)2
+ 1]1/2
=[
14
(
e2y − 2 + e−2y)
+ 1]1/2
=[
14
(
e2y + 2 + e−2y)
]1/2
=[
14
(
ey + e−y)2
]1/2
= 12
(
ey + e−y)
=⇒√
x2 + 1 dx = 14
(
ey + e−y)(
ey + e−y)
dy = 14
(
e2y + 2 + e−2y)
dy
=⇒∫
√
x2 + 1 dx = 14
∫
(
e2y + 2 + e−2y)
dy = 18e2y + 1
2y − 18e−2y + C
When x = 0, ey − e−y = 0, so ey = e−y so e2y = 1 so y = 0. When x = 1, ey − e−y = 2,so e2y − 1 = 2ey so e2y − 2ey − 1 = 0. View this as a quadratic equation in z = ey. Thesolutions to z2 − 2z − 1 = 0 are z = 1
2(2 ±
√4 + 4) = 1 ±
√2. Since ey > 0, only the
solution z =√
2 + 1 is allowed. So when x = 1, ey =√
2 + 1 and y = ln(√
2 + 1)
.
∫ 1
0
√
x2 + 1 dx =[
18e2y+ 1
2y− 1
8e−2y
]ln(√
2+1)
0= 1
8
(√
2+1)2− 1
8
(√
2+1)−2
+ 12
ln(√
2+1)
To simplify this use 1√2+1
=√
2−1(√
2+1)(√
2−1)=√
2− 1.
∫ 1
0
√
x2 + 1 dx = 18
(√
2 + 1)2 − 1
8
(√
2− 1)−2
+ 12 ln
(√
2 + 1)
= 12
√2 + 1
2 ln(√
2 + 1)
32) Evaluate
a)∫
ex+1ex−1
dx b)∫
dxex+1
c)∫ 1+
√x
1+ 3√
xdx d)
∫ √e2t − 9 dt
23
Solution. a) Make the substitution y = ex, dy = ex dx = y dx.
∫
ex+1ex−1 dx =
∫
y+1y−1
1y dy =
∫
[
2y−1 − 1
y
]
dy = 2 ln |y − 1| − ln |y|+ C
= 2 ln |ex − 1| − ln |ex|+ C
= 2 ln∣
∣ex/2 − e−x/2∣
∣ + C = ln∣
∣ex − 2 + e−x∣
∣ + C
b) Make the substitution y = ex, dy = ex dx = y dx.
∫
dxex+1
=
∫
1y+1
1y
dy =
∫
[
1y− 1
y+1
]
dy = ln |y| − ln |y + 1|+ C
= ln ex − ln(ex + 1) + C = − ln(
1 + e−x)
+ C
c) Let x = u6.∫
1+√
x1+ 3
√x
dx =
∫
1+u3
1+u2 6u5 du
By long divisionu8+u5
1+u2 = u6 − u4 + u3 + u2 − u− 1 + u+11+u2
Hence∫
1+√
x1+ 3
√x
dx = 6
∫
[
u6 − u4 + u3 + u2 − u− 1 + u+11+u2
]
du
= 6[
u7
7− u5
5+ u4
4+ u3
3− u2
2− u + 1
2ln(1 + u2) + tan−1 u
]
+ C
= 67x7/6 − 6
5x5/6 + 3
2x2/3 + 2x1/2 − 3x1/3 − 6x1/6 + 3 ln(1 + x1/3) + 6 tan−1 x1/6 + C
a) Make the substitution y = et, dy = et dt = y dt.
∫
√
e2t − 9 dt =
∫
√
y2 − 9dyy
Now substitute y = 3 sec u, dy = 3 sec u tanu du. Since√
9 sec2 u− 9 =√
9 tan2 u =3 tanu,
∫
√
e2t − 9 dt =
∫
√
9 sec2 u− 9 3 sec u tan u du3 sec u
= 3
∫
tan2 u du = 3
∫
(sec2 u− 1) du
= 3[
tanu− u]
+ C = 3[
√
sec2 u− 1− u]
+ C
= 3[
√
(
y3
)2 − 1− sec−1 y3
]
+ C =√
e2t − 9− 3 sec−1 et
3 + C
33) The circle x2 + (y − 1)2 = 2 and the parabola y = 12(x2 − 1) are tangent to each other.
24
a) Carefully plot these two curves and determine the vertex of the parabola and thepoints of tangency.
b) Find the area of the crescent shaped region bounded by these curves.
Solution. a) The parabola is symmetric about thex–axis. It hits the x–axis at y = −1/2. So the
vertex is (0,−1/2) . Any point of tangency must also
be a point of intersection. At a point of intersection,x2 = 2−(y−1)2 and x2 = 2y+1 so 2y+1 = 2−(y−1)2
or y2 = 0 or y = 0. The corresponding values of x arex = ±1. We are told that there is at least one point of
x
y
y = 12 (x2 − 1)
tangency. By symmetry about the x–axis ±(1, 0) are both points of tangency.
b) The bottom half of the circle has equation 1−√
2− x2. So the area is
∫ 1
−1
[
1−√
2− x2 − 12(x2 − 1)
]
dx =
∫ 1
0
[
3− x2 − 2√
2− x2]
dx
=[
3x− x3
3
]1
0− 2
∫ 1
0
√
2− x2 dx = 83− 2
∫ 1
0
√
2− x2 dx
For the remaining integral, sub in x =√
2 sin t, dx =√
2 cos t dt. When x = 0, t = 0 andwhen x = 1, sin t = 1√
2, so t = π
4. The area is
83− 2
∫ 1
0
√
2− x2 dx = 83− 2
∫ π/4
0
√
2− 2 sin2 t√
2 cos t dt = 83− 4
∫ π/4
0
cos2 t dt
= 83− 2
∫ π/4
0
[1 + cos 2t] dt = 83− 2
[
t + sin 2t2
]π/4
0
= 83 − 2
[
π4 + 1
2
]
= 53 − π
2
34) Let R be the region in the xy–plane bounded by y =√
x2+1x , y = 0, x = 1 and x =
√3.
a) Find the volume of the solid obtained by rotating R about the x–axis.b) Find the volume of the solid obtained by rotating R about the y–axis.
Solution. a)
Vol =
∫
√3
1
π(
√x2+1x
)2dx = π
∫
√3
1
(
1 + 1x2
)
dx = π[
x− 1x
]
√3
1= π
[√3− 1√
3
]
b) By cylindrical shells,
Vol =
∫
√3
1
2πx(
√x2+1x
)
dx = 2π
∫
√3
1
√
x2 + 1 dx
25
Sub in x = tan t, dx = sec2t dt, 1 = tan π4 ,√
3 = tan π3 .
Vol = 2π
∫ π/3
π/4
√
tan2 t + 1 sec2 t dt = 2π
∫ π/3
π/4
sec3 t dt = 2π
∫ π/3
π/4
cos tcos4 t dt
Sub in u = sin t, du = cos t dt, cos4 t = (1− sin2 t)2
= (1− u2)2, sin π
4 = 1√2, sin π
3 =√
32
Vol = 2π
∫
√3/2
1/√
2
1(1−u2)2
du = 2π
∫
√3/2
1/√
2
[
1/21−u
+ 1/21+u
]2
du
= π2
∫
√3/2
1/√
2
[
1(1−u)2
+ 2(1−u)(1+u)
+ 1(1+u)2
]
du
= π2
∫
√3/2
1/√
2
[
1(1−u)2 + 1
1−u + 11+u + 1
(1+u)2
]
du
= π2
[
11−u − ln |1− u|+ ln |1 + u| − 1
1+u
]
√3/2
1/√
2
= π2
[
2u1−u2 + ln |1+u|
|1−u|
]
√3/2
1/√
2= π
2
[ √3
1/4 + ln 2+√
32−
√3−
√2
1/2 − ln√
2+1√2−1
]
= 2π[√
3 + 14 ln 2+
√3
2−√
32+
√3
2+√
3−
√2
2 − 14 ln
√2+1√2−1
√2+1√2+1
]
= 2π[√
3 + 12 ln(2 +
√3)−
√2
2 − 12 ln(
√2 + 1)
]
35) A swimming pool has the shape shown in the figure below. The vertical cross–sectionsof the pool are squares. The distances in feet across the pool are given in the figure at 2foot intervals along the sixteen foot length of the pool.a) Denote by f(x) the width of the pool a distance x from the left hand end. Express
the volume of the pool as an integral in terms of f .b) Find an approximate value for the integral of part a.
10′ 12′10′
8′6′
8′10′
2′
26
Solution. a) The volume of the part of the pool with x–coordinate running from x tox + dx is f(x)2 dx. So the total volume is
V =∫ 16
0f(x)2 dx
b) Divide the domain of integration into 8 intervals each of length ∆x = 2. Denote byx∗i the right hand end point of interval number i. Thus x∗1 = 2, x∗2 = 4, · · · , x∗8 = 16.The corresponding values of f are f(x∗1) = f(2) = 10, f(x∗2) = f(4) = 12, f(x∗3) = 10,f(x∗4) = 8, f(x∗5) = 6, f(x∗6) = 8, f(x∗7) = 10 and f(x∗8) = 0. Then
V =
∫ 16
0
f(x)2 dx
≈ ∆x[
f(x∗1)2 + f(x∗2)
2 + f(x∗3)2 + f(x∗4)
2 + f(x∗5)2 + f(x∗6)
2 + f(x∗7)2 + f(x∗8)
2]
= 2[
102 + 122 + 102 + 82 + 62 + 82 + 102 + 02]
= 1216 ft3
36) Consider the function F (x) =∫ x2
0t2e−t2 dt −
∫ 0
−x2t4e−t4 dt. Find the values of x for
which F (x) takes its maximum and minimum values on 0 ≤ x ≤ 1 + a2. The number ahas the property that F (1 + a2) > 0.
Solution. We first compute F ′(x). To do so we write
F (x) = G(x2) + H(−x) with G(y) =
∫ y
0
t2e−t2 dt, H(y) =
∫ y
0
2t4e−t4 dt
By the Fundamental Theorem of Calculus,
G′(y) = y2e−y2
H ′(y) = 2y4e−y4
Hence, by the chain rule
F ′(x) = 2xG′(x2)−H ′(−x) = 2x5e−x4 − 2x4e−x4
= 2x4(x− 1)e−x4
Observe that F ′(x) < 0 for x < 1 and F ′(x) > 0 for x > 1. Hence F (x) is decreasing for
x < 1 and increasing for x > 1 and F (x) must take its minimum value when x = 1 . Itmay take its maximum value at x = 0 or x = 1 + a2. But F (0) = 0 and we are told that
F (1 + a2) > 0. Hence F (x) takes its maximum value when x = 1 + a2 .
37) Recall that, when n is odd, it is easy to integrate∫
sinm x cosn x dx using the substitutiony = sin x. The integral
∫
sec x dx is of the form∫
sinm x cosn x dx with m = 0 andn = −1, which is odd. Integrate
∫
sec x dx by substituting y = sinx.
27
Solution. Substitute y = sin x, dy = cos x dx. Then∫
sec x dx =
∫
cos x dxcos2 x =
∫
cos x dx1−sin2 x
=
∫
dy1−y2 = −
∫
dy(y−1)(y+1)
= −∫
[
1/2y−1 −
1/2y+1
]
dy = − 12 ln |y − 1|+ 1
2 ln |y + 1|+ C = 12 ln
∣
∣
∣
sinx+1sin x−1
∣
∣
∣+ C
To make the connection to the indefinite integral∫
sec x dx = ln | secx + tan x| + C,observe that
sin x+1sinx−1 = (sin x+1)2
sin2 x−1= − (sin x+1)2
cos2 x = −(
sin x+1cos x
)2
= −(
tan x + sec x)2
Hence∫
sec x dx = 12 ln
∣
∣
∣
sinx+1sin x−1
∣
∣
∣+ C = 1
2 ln∣
∣ tanx + sec x∣
∣
2+ C = ln
∣
∣ tan x + sec x∣
∣ + C
38) Evaluate, using the substitution x = tan θ2 .
a)∫
dθ2+sin θ b)
∫ π/2
0dθ
1+sin θ+cos θ c)∫
dθ3+2 cos θ
Solution. If x = tan θ2 , then, from the right–angled triangle with sides x and 1 and
hypotenuse√
1 + x2, sin θ2
= x√1+x2
and cos θ2
= 1√1+x2
so that
sin θ = 2 sin θ2 cos θ
2 = 2x1+x2
cos θ = 2 cos2 θ2 − 1 = 2 1
1+x2 − 1 = 1−x2
1+x2
dθ = 2 cos2 θ2 dx = 2
1+x2 dx
a)∫
dθ2+sin θ =
∫
1
2 + 2x1+x2
2
1 + x2dx =
∫
1
1 + x + x2dx =
∫
1
(x + 12)2 + 3
4
dx
Now sub x + 12 =
√3
2 y, dx =√
32 dy
∫
1
(x + 12)2 + 3
4
dx =
∫
134y2 + 3
4
√3
2 dy = 2√3
∫
1
y2 + 1dy = 2√
3tan−1
(
2√3
(
x + 12
)
)
+ C
= 2√3
tan−1(
2 tan θ2+1√3
)
+ C
b)
∫ π/2
0
dθ1+sin θ+cos θ
=
∫ 1
0
1
1 + 2x1+x2 + 1−x2
1+x2
2
1 + x2dx = 2
∫ 1
0
1
1 + x2 + 2x + 1− x2dx
=
∫ 1
0
11+x
dx = ln(1 + x)∣
∣
∣
1
0= ln 2
28
c)
∫
dθ3+2 cos θ
=
∫
1
3 + 2 1−x2
1+x2
2
1 + x2dx =
∫
2
3 + 3x2 + 2− 2x2dx = 2
∫
1
5 + x2dx
= 2√5
tan−1 x√5
+ C = 2√5
tan−1(
tan θ2√
5
)
+ C
39) Use cos θ = 12
(
eiθ + e−iθ)
and sin θ = 12i
(
eiθ − e−iθ)
to find identities for sin(3θ) andcos(3θ) in terms of sin θ and cos θ.
Solution. Note that
sin3 θ = − 18i
(
eiθ − e−iθ)3
= − 18i
(
ei3θ − 3eiθ + 3e−iθ − e−i3θ)
= 34
12i
(
eiθ − e−iθ)
− 14
12i
(
ei3θ − e−i3θ)
= 34 sin θ − 1
4 sin(3θ)
andcos3 θ = 1
8
(
eiθ + e−iθ)3
= 18
(
ei3θ + 3eiθ + 3e−iθ + e−i3θ)
= 34
12
(
eiθ + e−iθ)
+ 14
12
(
ei3θ + e−i3θ)
= 34 cos θ + 1
4 cos(3θ)
Hencesin(3θ) = 3 sin θ − 4 sin3 θ and cos(3θ) = 4 cos3 θ − 3 cos θ
40) For each natural number n and each nonzero complex number a, the equation zn = ahas precisely n distinct solutions, called the nth roots of a.a) Find the three cube roots of −1.b) Show that the sum of the n nth roots of 1 is zero.
Solution. a) Note that
(
eiπ/3)3
= eiπ = −1(
ei3π/3)3
= ei3π = eiπ = −1 since ei2π = 1(
ei5π/3)3
= ei5π = eiπ = −1 since ei4π = 1
Thus eiπ/3 = 12
+√
32
i, ei3π/3 = −1 and ei5π/3 = e−iπ/3 = 12−
√3
2i are three distinct cube
roots of −1.b) For all integers k, e2kπi = 1. Thus for every integer k
(
ei2kπ/n)n
= ei2kπ = 1
29
and ei2kπ/n is an nth root of 1. The n complex numbers
1 = ei0π/n, ei2π/n, ei2×2π/n, ei3×2π/n, · · · , , ei(n−1)×2π/n
are all different and so are the n nth roots of 1. Write zp = ei2π/n. The n nth roots of 1are 1, zp, z2
p, z3p , · · ·, zn−1
p . Using the formula for the partial sum of a geometric serieswith ratio zp
1 + zp + z2p + z3
p + · · ·+ zn−1p =
1−znp
1−zp
The numerator vanishes because zp is an nth root of 1.
41) Evaluate the given integrals, using complex numbers.
a)∫ 1
02x2−2
(x2+1)2dx b)
∫ π/2
0sin4 θ dθ
Solution. a) We use partial fractions.
2x2−2(x2+1)2
= 2x2−2(x+i)2(x−i)2 = a
x+i + b(x+i)2 + c
x−i + d(x−i)2
= a(x+i)(x−i)2+b(x−i)2+c(x−i)(x+i)2+d(x+i)2
(x+i)2(x−i)2
For this to be true for all x, we need a(x+i)(x−i)2+b(x−i)2+c(x−i)(x+i)2+d(x+i)2 =2x2 − 2 for all x. Setting x = i gives (2i)2d = −4 or d = 1. Setting x = −i gives(−2i)2b = −4 or b = 1. Subbing b = d = 1 back in and using b(x − i)2 + d(x + i)2 =(x− i)2 + (x + i)2 = 2x2 − 2 gives a(x + i)(x− i)2 + c(x− i)(x + i)2 = 0 or a = c = 0.Hence
∫ 1
0
2x2−2(x2+1)2
dx =
∫ 1
0
[
1(x+i)2
+ 1(x−i)2
]
dx =[
− 1x+i
− 1x−i
]1
0=
[
− x−i+x+i(x+i)(x−i)
]1
0
= − 2xx2+1
∣
∣
∣
1
0= −1
b) Since sin θ = 12i
(
eiθ − e−iθ)
,
∫ π/2
0
sin4 θ dθ = 116
∫ π/2
0
(
eiθ − e−iθ)4
dθ = 116
∫ π/2
0
(
e4iθ − 4e2iθ + 6− 4e−2iθ + e−4iθ)4
dθ
= 116
[
14i
e4iθ − 2ie2iθ + 6θ + 2
ie−2iθ − 1
4ie−4iθ
]π/2
0
= 116
[
12 sin(4θ)− 4 sin(2θ) + 6θ
]π/2
0= 1
166π2 = 3
16π
42) Solve
a) dxdt = 1 + t− x− tx b) dy
dt = ty+3tt2+1 , y(2) = 2 c) dy
dx = y2
30
Solution. a)
dxdt = 1 + t− x− tx = (1 + t)(1− x) ⇒ − dx
x−1 = (1 + t) dt ⇒ − ln |x− 1| = t + t2
2 + D
⇒ |x− 1| = eDe−t−t2/2 ⇒ x− 1 = ±eDe−t−t2/2 ⇒ x = 1 + Ce−t−t2/2
b)
dydt = ty+3t
t2+1 = (y + 3) tt2+1 ⇒ dy
y+3 = tt2+1 ⇒ ln |y + 3| = 1
2 ln(t2 + 1) + D
⇒ |y + 3| = eD√
t2 + 1 ⇒ y = −3 + C√
t2 + 1 where C = ±eD
To satisfy y(2) = 2, we need 2 = −3 + C√
5 or C =√
5, so y = −3 +√
5t2 + 5 .
c) When y 6= 0,
dydx = y2 =⇒ dy
y2 = dx =⇒ y−1
−1 = x + c =⇒ y = − 1x+c
When y = 0, this computation breaks down because dyy2 contains a division by 0. We can
check if the function y(x) = 0 satisfies the differential equation by just subbing it in:
y(x) = 0 =⇒ y′(x) = 0, y(x)2 = 0 =⇒ y′(x) = y(x)2
So y(x) = 0 is a solution and the full solution is
y(x) = 0 or y(x) = − 1x+c , for any constant c
43) Find an equation for the curve that passes through the point (1, 1) and whose slope at
(x, y) is y2
x3 .
Solution. We need to find a function y(x) obeying dydx = y2
x3 and y(1) = 1.
dydx = y2
x3 =⇒ dyy2 = dx
x3 =⇒ y−1
−1 = x−2
−2 + c
To satsify y(1) = 1, we need −1 = − 12 + c or c = − 1
2 . So − 1y = − 1
2x2 − 12 = − 1+x2
2x2 or
y = 2x2
x2+1 .
44) In 1986, the population of the world was 5 billion and was increasing at a rate of 2%per year. Using the logistic growth model with an assumed maximum population of 100billion, predict the population of the world in the years 2000, 2100 and 2500.
Solution. Let y(t) be the population of the world, in billions of people, at time 1986+ t.The logistic growth model assumes
y′ = ky(M − y)
31
We know that, if at time zero the population is below M , then as time increases thepopulation increases, approaching the limit M as t tends to infinity. So in this problemM is the maximum population. That is, M = 100. We are also told that, at time zero,
the percentage rate of change of population, 100 y′
y, is 2, so that, at time zero, y′
y= 0.02.
But, from the differential equation, y′
y= k(M−y). Hence at time zero, 0.02 = k(100−5),
so that k = 29500
. We now know k and M and can solve the differential equation
dydt = ky(M − y) ⇒ dy
y(M−y) = kdt ⇒∫
1M
[
1y − 1
y−M
]
dy = kt + C
⇒ 1M [ln |y| − ln |y −M |] = kt + C ⇒ ln |y|
|y−M| = kMt + CM ⇒∣
∣
∣
yy−M
∣
∣
∣= DekMt
with D = eCM . We know that y remains between 0 and M , so that∣
∣
∣
yy−M
∣
∣
∣= y
M−y =
DekMt. We are given that at t = 0, y = 5. Subbing this, and the values of M and k, in
5100−5 = De0 ⇒ D = 5
95 ⇒ y100−y = 5
95e2t/95 ⇒ y = (100− y) 595e2t/95
⇒ 95y = (500− 5y)e2t/95 ⇒ y =500e2t/95
95 + 5e2t/95=
100e2t/95
19 + e2t/95=
100
1 + 19e−2t/95
In the year 2000, t = 14 and y = 1001+19e−28/95 ≈ 6.6 billion . In the year 2100,
t = 114 and y = 1001+19e−228/95 ≈ 36.7 billion . In the year 2200, t = 514 and
y = 1001+19e−1028/95 ≈ 100 billion .
45) When a raindrop falls it increases in size so that its mass m(t), is a function of time t. Therate of growth of mass is km(t) for some positive constant k. According to Newton’s lawof motion, d
dt (mv) = gm, where v is the speed of the raindrop and g is the accelerationdue to gravity. Find the terminal velocity lim
t→∞v(t) of a raindrop.
Solution. First we determine m(t).
dmdt = km =⇒ dm
m = k dt =⇒ ln m = kt + c =⇒ m = dekt where d = ec
Define y(t) = m(t)v(t). Then
dydt
= gm(t) = gdekt =⇒ dy = gdekt dt =⇒ y = gd ekt
k+ C
Subbing back in y(t) = m(t)v(t) = dektv(t),
y = gd ekt
k + C ⇒ dektv(t) = gd ekt
k + C ⇒ v(t) = gk + C
dekt
Since k > 0, limt→∞
Cekt = 0 and lim
t→∞v(t) = g
k .
32
46) A population of rare South American monkeys has been found to have a populationgrowth rate which decreases with time. Specifically, the population size, y(t), satisfiesthe differential equation
y′(t) = y(t+1)2
a) Given that at time t = 0 the population is 100, solve for the population size y(t)after time t.
b) In the limit t →∞, what is the size of the population?
Solution. a)
y′(t) = y(t+1)2 ⇐⇒ dy = y
(t+1)2 dt ⇐⇒ dyy = dt
(t+1)2 ⇐⇒ ln y = (t+1)−1
−1 + C
⇐⇒ y = Ke−(t+1)−1
where K = eC . At t = 0, y = 100. Subbing this in 100 = Ke−1, so K = 100e and
y(t) = 100e−(t+1)−1+1 .
b) As t →∞, 1t+1
approaches 0. So
limt→∞
y(t) = limt→∞
100e−(t+1)−1+1 = 100e0+1 = 100e
47) Suppose that y(t) (measured in thousands of dollars – “kilobucks”) represents the balancein a savings account at time t years after the account was opened with an initial balance of2 kilobucks. Interest is being paid into the account continuously at a nominal rate of 5%per annum; that is to say, interest is flowing into the account at a rate of 5
100y(t) kilobucks
per year at time t. At the same time, the account is being continuously depleted by taxes
and other charges at a rate y(t)2
1000 kilobucks per year. No other deposits or withdrawalsare made on the account.a) What is balance in the account at time t = 20 years?b) How large can the balance in the account grow over time?
Solution. a) The differential equation determining the time dependence of y(t) is
dydt (t) = 5
100y(t)− y(t)2
1000 =⇒ −1000 dydt = −50y + y2 =⇒ 1000
y2−50y dy = − dt
The integral of the left hand side is
∫
1000y2−50y
dy =
∫
1000(y−50)y
dy =
∫
[
20y−50
− 20y
]
dy = 20 ln |y − 50| − 20 ln |y|+ C
= 20 ln∣
∣
y−50y
∣
∣ + C
Before continuing let’s discuss the sign of y−50y
. We start with y = 2. The rate of change
of y is determined by dydt (t) = 5
100y(t) − y(t)2
1000 = 11000y(50 − y). Note that the rate of
change is positive for 0 ≤ y ≤ 50 and negative otherwise. So, as we start with y = 2, y
33
increases at the beginning and will continue to do so as long as y ≤ 50. But y can neverget above 50, because dy
dtis negative, that is y is decreasing in time, for all y > 50. So y
must remain between 0 and 50 and∣
∣
y−50y
∣
∣ = 50−yy . Now back to the solution.
dydt
(t) = 5100
y(t)− y(t)2
1000=⇒ 1000
y2−50ydy = − dt =⇒ 20 ln 50−y
y+ C = −t
Dividing by 20 and then taking the exponential of both sides
50−yy = Ke−t/20
where K = e−C . At t = 0, y = 2, so K = 50−22 = 24.
50−yy = 24e−t/20 =⇒ 50− y = 24ye−t/20 =⇒
(
1 + 24e−t/20)
y = 50
=⇒ y =50
1 + 24e−t/20and y(20) ≈ 5.087
b) As t increases, 1 + 24e−t/20 decreases, approaching the limit 0 as t tends to infinity.Hence y increases over time, but remains below 50, approaching 50 as time tends toinfinity.
48 a) Let Tn and Mn be the n–step Trapezoidal and Midpoint Rule approximations, respec-
tively, for∫ 1
0cos(x2) dx. Find T4, T8, M4 and M8.
b) Estimate the errors involved in these approximations.
Solution. a)
T4 = 14
[
12
cos 0 + cos(
14
)2+ cos
(
24
)2+ cos
(
34
)2+ 1
2cos 1
]
= 0.895759
T8 = 18
[
12 cos 0 + cos
(
18
)2+ cos
(
28
)2+ cos
(
38
)2+ · · ·+ cos
(
78
)2+ 1
2 cos 1]
= 0.902333
M4 = 14
[
cos(
18
)2+ cos
(
38
)2+ cos
(
58
)2+ cos
(
78
)2]
= 0.908907
M8 = 18
[
cos(
116
)2+ cos
(
316
)2+ cos
(
516
)2+ · · ·+ cos
(
1516
)2]
= 0.905620
b)
f(x) = cos(x2) =⇒ f ′(x) = −2x sin(x2) =⇒ f ′′(x) = −2 sin(x2)− 4x2 cos(x2)
On the interval 0 ≤ x ≤ 1, | sin(x2)| is never bigger than 1 and | cos(x2)| is never biggerthan 1 and x2 is never bigger than 1, so |f ′′(x)| is never bigger than 2 × 1 + 4× 1 = 6.Hence
∣
∣ET4
∣
∣ ≤ 6(1−0)3
12×42 ≤ 0.0313∣
∣ET8
∣
∣ ≤ 6(1−0)3
12×82 ≤ 0.00782∣
∣EM4
∣
∣ ≤ 6(1−0)3
24×42 ≤ 0.0157∣
∣EM8
∣
∣ ≤ 6(1−0)3
24×82 ≤ 0.00391
34
49) How large should n be to ensure that the Simpson’s Rule approximation to∫ 1
0ex2
dx isaccurate to within 0.00001?
Solution.
f(x) = ex2
=⇒ f ′(x) = 2xex2
=⇒ f ′′(x) = 2ex2
+ 4x2ex2
= 2(1 + 2x2)ex2
=⇒ f (3)(x) = 2(4x)ex2
+ 4x(1 + 2x2)ex2
= 4(3x + 2x3)ex2
=⇒ f (4)(x) = 4(3 + 6x2)ex2
+ 8x(3x + 2x3)ex2
= 4(3 + 12x2 + 4x4)ex2
On the interval 0 ≤ x ≤ 1, ex2
is never bigger than e and (3+12x2 +4x4) is never biggerthan 3 + 12 + 4 = 19, so |f (4)(x)| is never bigger than 4× 19× e = 76e. Hence error in
Simpson’s rule is at most 76e(1−0)5
180n4 . This error is smaller than 10−5 if
76e180n4 ≤ 10−5 ⇐⇒ n4 ≥ 76e×105
180⇐⇒ n ≥ 4
√
76e×105
180= 18.4
For Simpson’s rule, n must be even so take n ≥ 20 .
50 a) State Simpson’s Rule to approximate∫ b
af(x) dx.
b) Use Simpson’s Rule to compute∫ 2
0x3 dx with n = 4.
Solution. a) Let n be a strictly positive even integer and h = b−an
. Define xj = a + jh.Simpson’s Rule is
∫ b
a
f(x) dx ≈ h3
[
f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+· · ·+2f(xn−2)+4f(xn−1)+f(xn)]
b) When a = 0, b = 2 and n = 4 we have h = 12.
∫ 2
0
x3 dx ≈ 16
[
f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + f(2)]
= 16
[
0 + 4(0.5)3 + 2(1)3 + 4(1.5)3 + 23]
= 4
51) Suppose that f(x) has a continuous second derivative that obeys |f ′′(x)| ≤ K for all
a ≤ x ≤ b. Let Mn be the result of applying the midpoint rule to∫ b
af(x) dx with n
subintervals. Prove that∣
∣
∣
∣
∫ b
a
f(x) dx−Mn
∣
∣
∣
∣
≤ K(b−a)3
24n2
Hint: use the error estimate for the tangent line approximation to show that
∣
∣f(x)− f(m)− f ′(m)(x−m)∣
∣ ≤ K2 (x−m)2
35
Solution. Set, for j = 0, 1, 2, · · · , n, xj = a + b−an j. This partitions a ≤ x ≤ b into n
equal subintervals, each of width b−an . Also, let mj = xj + b−a
2n be the midpoint of the jth
interval. The midpoint rule approximates∫ xj+1
xjf(x) dx by
∫ xj+1
xjf(mj) dx = f(mj)
b−an .
The error in this approximation is
ej =
∫ xj+1
xj
f(x) dx− f(mj)b−an =
∫ xj+1
xj
f(x) dx−∫ xj+1
xj
f(mj) dx =
∫ xj+1
xj
[
f(x)− f(mj)]
dx
Taylor expanding f(x) about mj to first order (with a second order error) gives
f(x) = f(mj) + f ′(mj)(x−mj) + 12f ′′(z)(x−mj)
2
for some z between x and mj . By hypothesis |f ′′(z)| ≤ K so that
∣
∣f(x)− f(mj)− f ′(mj)(x−mj)∣
∣ ≤ 12K(x−mj)
2
Furthermore, subbing y = x−mj ,
∫ xj+1
xj
f ′(mj)(x−mj) dx = f ′(mj)
∫b−a2n
− b−a2n
y dy = 0
So, the error arising in the jth interval
∣
∣ej
∣
∣ =
∣
∣
∣
∣
∫ xj+1
xj
[
f(x)− f(mj)]
dx
∣
∣
∣
∣
=
∣
∣
∣
∣
∫ xj+1
xj
[
f(x)− f(mj)− f ′(mj)(x−mj)]
dx
∣
∣
∣
∣
≤∫ xj+1
xj
∣
∣f(x)− f(mj)− f ′(mj)(x−mj)∣
∣ dx ≤∫ xj+1
xj
12K(x−mj)
2 dx
= 12K
(x−mj)3
3
∣
∣
∣
xj+1
xj
= 2× 12K
(xj+1−mj)3
3 = 13K
(
b−a2n
)3= 1
24K (b−a)3
n3
The total error, from all n intervals is
∣
∣
∣
∣
∫ b
a
f(x) dx−Mn
∣
∣
∣
∣
≤ |e1|+ |e2|+ · · ·+ |en| ≤ n 124
K (b−a)3
n3 ≤ K(b−a)3
24n2
36
52) Suppose that f(x) has a continuous second derivative that obeys |f ′′(x)| ≤ K for all
a ≤ x ≤ b. Let Tn be the result of applying the Trapezoidal Rule to∫ b
af(x) dx with n
subintervals. Use the generalization of the Mean Value Theorem given below to provethat
∣
∣
∣
∣
∫ b
a
f(x) dx− Tn
∣
∣
∣
∣
≤ K(b−a)3
12n2
Solution. Set, for j = 0, 1, 2, · · · , n, xj = a + b−an j. This partitions a ≤ x ≤ b into n
equal subintervals, each of width b−an . The Trapezoidal Rule approximates
∫ xj+1
xjf(x) dx
by 12[f(xj) + f(xj+1)]
b−an
, which is the area between xj and xj+1 under the straightline through
(
xj , f(xj))
and(
xj+1, f(xj+1))
. The straight line through(
xj , f(xj))
and(
xj+1, f(xj+1))
is y = p(x) = f(xj) +f(xj+1)−f(xj)
xj+1−xj(x− xj). Note that
∫ xj+1
xj
p(x) dx = f(xj)[xj+1−xj ]+f(xj+1)−f(xj)
xj+1−xj
12 (xj+1−xj)
2 = 12 [f(xj)+f(xj+1)][xj+1−xj ]
as desired. The error in this approximation is
ej =
∫ xj+1
xj
f(x) dx− 12 [f(xj)+f(xj+1)]
b−an =
∫ xj+1
xj
f(x) dx−∫ xj+1
xj
p(x) dx =
∫ xj+1
xj
[
f(x)−p(x)]
dx
By the generalization of the Mean Value Theorem given below, with n = 1, y0 = xj andy1 = xj+1
f(x)− p(x) = (x− xj)(x− xj+1)f ′′(cx)
2
for some cx between x and mj . By hypothesis |f ′′(cx)| ≤ K so that
∣
∣f(x)− p(x)∣
∣ ≤ 12K|x− xj ||x− xj+1| = 1
2K(x− xj)(xj+1 − x)
for all xj ≤ x ≤ xj+1. So, the error arising in the jth interval
∣
∣ej
∣
∣ =
∣
∣
∣
∣
∫ xj+1
xj
[
f(x)− p(x)]
dx
∣
∣
∣
∣
≤∫ xj+1
xj
12K(x− xj)(xj+1 − x) dx
Subbing t = x− xj , dt = dx, xj+1 − xj = b−an ,
∣
∣ej
∣
∣ ≤ 12K
∫b−a
n
0
t( b−an− t) dt = 1
2K
[
b−an
t2
2− t3
3
]
b−an
0= 1
2K (b−a)3
n3
[
12− 1
3
]
= 112
K (b−a)3
n3
The total error, from all n intervals is
∣
∣
∣
∣
∫ b
a
f(x) dx−Mn
∣
∣
∣
∣
≤ |e1|+ |e2|+ · · ·+ |en| ≤ n 112
K (b−a)3
n3 ≤ K(b−a)3
12n2
37
Theorem (Mean Value Theorem for Interpolation)Let a ≤ y0 < y1 < y2 < · · · < yn ≤ b. Let f(y) and its first n derivatives be continuous on
a ≤ y ≤ b and let f (n+1)(y) exist on a < y < b. Suppose that p(y) is a polynomial of degree
at most n that coincides with f(y) at y0, · · · , yn. That is, assume that
f(y0) = p(y0), f(y1) = p(y1), · · · , f(yn) = p(yn)
Then, for each a ≤ y ≤ b, there is a number a < cy < b, depending on y, such that
f(y)− p(y) = (y − y0)(y − y1) · · · (y − yn)f(n+1)(cy)
(n+1)!
Proof: If y is one of y0, · · · , yn then both of f(y) − p(y) and (y − y0)(y − y1) · · · (y −yn)
f(n+1)(cy)(n+1)! are zero, for any cy, and hence are equal. So it suffices to consider a y that is
different from all of y0, · · · , yn. Fix any such y and define
g(z) = f(z)− p(z)− (z − y0) · · · (z − yn)
(y − y0) · · · (y − yn)[f(y)− p(y)]
Note that g(z) is zero for at least n+2 different values of z, namely, z = y and z = y0, · · · , yn.The Mean Value Theorem implies that dg
dz has a zero between any two successive zeroes of
g(z). So dgdz (z) is zero for at least n+1 different values of z. By the Mean Value Theorem, d2g
dz2
has a zero between any two successive zeroes of dgdz . So d2g
dz2 (z) is zero for at least n different
values of z. Continuing in this way, we see that dn+1gdzn+1 has at least one zero. Call it cy.
As p(z) is a polynomial in z of degree at most n, dn+1pdzn+1 = 0. Recall that y is being held
fixed in this computation. So (z−y0)···(z−yn)(y−y0)···(y−yn) [f(y)− p(y)] is a polynomial in z of degree n + 1
whose degree n + 1 term is zn+1
(y−y0)···(y−yn) [f(y)− p(y)]. Hence
dn+1
dzn+1
(z−y0)···(z−yn)(y−y0)···(y−yn) [f(y)− p(y)] = (n+1)!
(y−y0)···(y−yn) [f(y)− p(y)]
and0 = dn+1g
dzn+1 (cy) = f (n+1)(cy)− (n+1)!(y−y0)···(y−yn) [f(y)− p(y)]
Moving f (n+1)(cy) to the other side of the equation and cross multiplying gives
f(y)− p(y) = (y−y0)···(y−yn)(n+1)!
f (n+1)(cy)
as desired
38
53) Let I =∫ π/2
π/6ln(sinx) dx.
a) Determine the maximum, K, of |f ′′(x)|, for π6 ≤ x ≤ π
2 , where f(x) = ln(sin x).b) How large should n be in order that the approximation I ≈ Mn be accurate to within
10−4?
Solution. a)
f(x) = ln(sin x) =⇒ f ′(x) = cos xsin x
= cotx =⇒ f ′′(x) = − csc2 x = − 1sin2 x
For π6 ≤ x ≤ π
2 , 12 = sin π
6 ≤ sin x ≤ sin π2 = 1. So the largest value of |f ′′(x)| on the
interval π6 ≤ x ≤ π
2 occurs at x = π6 and is 1
(1/2)2 = 4 = K .
b) We wish to choose n so thatK(b−a)3
24n2 ≤ 10−4
In this case K = 4, a = π6 and b = π
2 so
4(π/3)3
24n2 ≤ 10−4 ⇐⇒ n2 ≥ 4(π/3)3
24 104 = π3
2×34 104 ⇐⇒ n ≥ π3/2
32√
2102 = 43.75
So n ≥ 44 will do the job.
54) Find the values of p for which each of the following integrals converges and evaluate theintegrals for those values of p.
a)
∫ 1
0
dx
xpb)
∫ ∞
1
dx
xp
Solution. a)
∫ 1
0
dx
xp= lim
ε→0+
∫ 1
ε
dx
xp= lim
ε→0+
[
x1−p
1−p
]1
εif p 6= 1
lnx∣
∣
1
εif p = 1
= limε→0+
{[
11−p
− ε1−p
1−p
]
if p 6= 1
− ln ε if p = 1
=
11−p
if p < 1
∞ if p > 1∞ if p = 1
In conclusion,∫ 1
0dxxp = 1
1−p if p < 1 and diverges otherwise .
b)
∫ ∞
1
dx
xp= lim
R→∞
∫ R
1
dx
xp= lim
R→∞
[
x1−p
1−p
]R
1if p 6= 1
ln x∣
∣
R
1if p = 1
= limR→∞
{[
R1−p
1−p − 11−p
]
if p 6= 1
ln R if p = 1
=
− 11−p
if p > 1
∞ if p < 1∞ if p = 1
39
In conclusion,∫∞1
dxxp = 1
p−1 if p > 1 and diverges otherwise .
55) Evaluate∫ 1
−101
(2x+3)2 dx.
Solution. Trick question! The integrand has a singularity when 2x + 3 = 0 or x = − 32 .
The domain of integration has to be split up into two parts: −10 < x < − 32
and− 3
2 < x < 1.
∫ 1
− 32
1(2x+3)2 dx = lim
a→− 32+
∫ 1
a
1(2x+3)2 dx = lim
a→− 32+− 1/2
(2x+3)
∣
∣
∣
1
a
= lima→− 3
2+
[
1/22a+3 −
1/2(2+3)
]
= ∞
since 2a + 3 approaches zero as a tends to − 32 . So the integral diverges . Similarly,
∫ − 32
−10
1(2x+3)2
dx = lima→− 3
2−
∫ a
−10
1(2x+3)2
dx = lima→− 3
2−− 1/2
(2x+3)
∣
∣
∣
a
−10
= lima→− 3
2−
[
1/2(−20+3)
− 1/22a+3
]
= ∞
and the other part of the integral diverges too.
56) Determine if the following integrals converge or diverge. Justify your answers.
a)
∫ ∞
0
dx
x1/3(1 + x)b)
∫ ∞
0
dx
x1/3(1 + x1/2)
Solution. With both of these integrals, there are two potential problems which couldcause the integrals to diverge (i.e. become infinite). First, the integrands become infiniteat x = 0 and secondly x runs all the way to x = ∞. To separate these two potentialproblems, split up the domain of integration into 0 ≤ x ≤ 1 and 1 ≤ x ≤ ∞.
∫ ∞
0
dx
x1/3(1 + x)=
∫ 1
0
dx
x1/3(1 + x)+
∫ ∞
1
dx
x1/3(1 + x)∫ ∞
0
dx
x1/3(1 + x1/2)=
∫ 1
0
dx
x1/3(1 + x1/2)+
∫ ∞
1
dx
x1/3(1 + x1/2)
First consider the domain of integration 0 ≤ x ≤ 1. On this domain 0 ≤ 1x1/3(1+x)
≤ 1x1/3
and 0 ≤ 1x1/3(1+x1/2)
≤ 1x1/3 . So both integrals
∫ 1
0
dx
x1/3(1 + x),
∫ 1
0
dx
x1/3(1 + x1/2)≤
∫ 1
0
dx
x1/3=
x2/3
2/3
∣
∣
∣
1
0=
3
2
40
So both∫ 1
0dx
x1/3(1+x)and
∫ 1
0dx
x1/3(1+x1/2)converge. Now for the second domain. For x ≥ 1,
0 ≤ 1x1/3(1+x)
≤ 1x1/3x
= 1x4/3 . So
∫ ∞
1
dx
x1/3(1 + x)≤
∫ ∞
1
dx
x4/3=
x−1/3
−1/3
∣
∣
∣
∞
1= 3
and∫∞0
dxx1/3(1+x)
converges . On the other hand, for x ≥ 1, we have 1x1/3(1+x1/2)
≥1
x1/3(x1/2+x1/2)= 1
2x5/6 and hence
∫ ∞
1
dx
x1/3(1 + x1/2)≥
∫ ∞
1
dx
2x5/6=
1
2
x1/6
1/6
∣
∣
∣
∞
1= ∞
and∫∞0
dxx1/3(1+x1/2)
diverges .
57) Find the length, L, of the curve y = ex, 0 ≤ x ≤ 1.
Solution.
L =
∫ 1
0
√
1 +(
dydx
)2dx =
∫ 1
0
√
1 + e2x dx
Make the change of variables u =√
1 + e2x, du = e2x√
1+e2xdx = u2−1
u dx.
L =
∫
√1+e2
√2
u2
u2−1 du =
∫
√1+e2
√2
[
1 + 1/2u−1 −
1/2u+1
]
du
=[
u + 12 ln |u− 1| − 1
2 ln |u + 1|]
√1+e2
√2
=√
1 + e2 −√
2 + 12
ln√
1+e2−1√1+e2+1
− 12
ln√
2−1√2+1
58) Find the length of the curve 12xy = 4y4 + 3 from(
712 , 1
)
to(
6724 , 2
)
.
Solution. On this curve x = 13y3 + 1
4y. So
dxdy
= y2 − 14y2
1 +(
dxdy
)2= 1 + y4 − 1
2 + 116y4 = y4 + 1
2 + 116y4 =
(
y2 + 14y2
)2
Hence
length =
∫ 2
1
√
1 +(
dxdy
)2dy =
∫ 2
1
(
y2 + 14y2
)
dy =[
y3
3 − 14y
]2
1= 59
24
41
59) Find the length of the curve x2/3 + y2/3 = 1.
Solution. This equation does not change if you substitute x → −x or y → −y. So thecurve is invariant under reflection in the x– or y–axes. It suffices to find the length ofthe part of the curve in the first quadrant and multiply by 4.
x
y
On this curve y =(
1− x2/3)3/2
. So
dydx = 3
2
(
1− x2/3)1/2(− 2
3x−1/3)
= −(
1− x2/3)1/2
x−1/3
1 +(
dydx
)2= 1 + (1− x2/3)x−2/3 = x−2/3
Hence
length = 4
∫ 1
0
√
1 +(
dydx
)2dx = 4
∫ 1
0
x−1/3 dx = 4 32x2/3
∣
∣
∣
1
0= 6
60) Set up two integrals, one of which is improper, representing the part of the curve y3 = x2
between (0, 0) and (1, 1). Evaluate both integrals.
Solution. Since 3y2 dy = 2x dx
dydx = 2x
3y2 = 23x1/3
dxdy = 3y2
2x = 32y1/2
L =
∫ 1
0
√
1 +(
dydx
)2dx L =
∫ 1
0
√
1 +(
dxdy
)2dy
=
∫ 1
0
√
1 + 49x2/3 dx =
∫ 1
0
√
1 + 94y dy
To evaluate the integral over y, substitute z = 1 + 94y:
L =
∫ 13/4
1
√z 4
9dz = 49
[
23z3/2
]13/4
1= 8
27
[
133/2
8 − 1]
To evaluate the integral over x, which is improper because of the singularity at x = 0,substitute x = y3/2, dx = 3
2y1/2 dy. This yields the y–integral that we just evaluated.
61) Find the area of the surface obtained by rotating the given curve about the x–axis.a) y = 1
4x2 − 12 ln x, 1 ≤ x ≤ 4 b) 8y2 = x2(1− x2), 0 ≤ x ≤ 1
42
Solution. a)
dydx = 1
2x− 12x
(
dydx
)2= 1
4
(
x2 − 2 + 1x2
)
1 +(
dydx
)2= 1 + 1
4
(
x2 − 2 + 1x2
)
= 14
(
x2 + 2 + 1x2
)
= 14
(
x + 1x
)2
So
Area =
∫ 4
1
2πy(x)
√
1 +(
dydx
)2dx = 2π
∫ 4
1
(
14x2 − 1
2 ln x)
12
(
x + 1x
)
dx
= π2
∫ 4
1
(
12x3 + 1
2x− x lnx− 1
xlnx
)
dx
= π2
[
x4
8 + x2
4 − x2
2 lnx + x2
4 − 12 (lnx)2
]4
1
= π[
31516 − 8 ln 2− (ln 2)2
]
b) Differentiating both sides of 8y(x)2 = x2(1− x2) with respect to x,
16y dydx = 2x− 4x3 = 2x(1− 2x2) =⇒ dy
dx = x8y (1− 2x2)
(
dydx
)2= x2
64y2 (1− 2x2)2 = (1−2x2)2
8(1−x2)
1 +(
dydx
)2= 1 + (1−2x2)2
8(1−x2) = 8−8x2+1−4x2+4x4
8(1−x2) = 9−12x2+4x4
8(1−x2) = (3−2x2)2
8(1−x2)
So
Area =
∫ 1
0
2πy(x)
√
1 +(
dydx
)2dx = 2π
∫ 1
0
x√
1−x2√8
3−2x2√8(1−x2)
dx
= π4
∫ 1
0
(3x− 2x3) dx = π4
[
32x2 − 1
2x4]1
0
= π4
62) Let L be the length of the part of the curve y = e−x lying between x = 0 and x = 4. Usethe Trapezoidal Rule with four subintervals to estimate L.
Solution. Let f(x) = e−x. Then f ′(x) = −e−x so L =∫ 4
0
√
1 + f ′(x)2 dx =∫ 4
0
√1 + e−2x dx . Denoting g(x) =
√1 + e−2x, the Trapezoidal Rule with ∆x = 1
gives
L ≈ ∆x[
12g(0)+g(1)+g(2)+g(3)+ 1
2g(4)]
≈ .707+1.066+1.009+1.001+0.500 = 4.283
63) At what point does the curve x = 1− 2 cos2 t, y = (tan t)(1− 2 cos2 t) cross itself. Findthe equations of both tangents at that point.
43
Solution. Suppose that the curve is at the same point at time t1 as it is at time t2. Then
x(t1) = x(t2) ⇐⇒ 1−2 cos2 t1 = 1−2 cos2 t2 ⇐⇒ cos2 t1 = cos2 t2 ⇐⇒ cos t1 = ± cos t2
andy(t1) = y(t2) =⇒ (tan t1)(1− 2 cos2 t1) = (tan t2)(1− 2 cos2 t2)
This is the case if 1−2 cos2 t1 = 0 or cos t1 = 1√2, since then 1−2 cos2 t2 = 0 automatically
and both y(t1) and y(t2) are zero. For example t1 = π4 and t2 = 3π
4 does the job. At
both these times the curve is at (0, 0) . Since
x′(t) = 4 sin t cos t y′(t) = (sec2 t)(1− 2 cos2 t) + (tan t)(4 sin t cos t)
=⇒ dydx = (sec2 t)(1−2 cos2 t)
4 sin t cos t + tan t
When t = π4
or 3π4
, the term (sec2 t)(1−2 cos2 t)4 sin t cos t
= 0. Hence
dydx
∣
∣
∣
t=π/4= tan π
4 = 1 dydx
∣
∣
∣
t=3π/4= tan 3π
4 = −1
The line through (0, 0) with slope +1 is y = x and the line through (0, 0) with slope −1
is y = −x .
64) A string is wound around a circle of radius r and then unwound while being held taut.Find the equation of the curve traced by the point P at the end of the string if the initialposition of P is (r, 0).
Solution.
x
y
r
O
P
ST
θ
α
Use the angle θ in the figure above as the parameter. The point T has coordinates(r cos θ, r sin θ). The length of the line TP is the same as the length of the arc of thecircle from (r, 0) to T . This arc is the fraction θ
2π of the full circle, which has circumference
2πr. So the line TP has length θ2π2πr = θr. The line OT is perpendicular to the line
TP and the line TS is parallel to the x–axis. So
θ + π2
+ α = π =⇒ α = π2− θ
44
Consequently, the x–coordinate of P is the x–coordinate of T , which is r cos θ plus thelength of TS which is |TP | cosα = rθ sin θ. The y–coordinate of P is the y–coordinateof T , which is r sin θ minus the length of SP which is |TP | sinα = rθ cos θ. So
x = r(
cos θ + θ sin θ)
y = r(
sin θ − θ cos θ)
65) A cow is tied to a silo with radius r by a rope that is just long enough to reach theopposite side of the silo. Find the area available for grazing by the cow.
silocow
Solution. Imagine the cow walking with the rope pulled taut, starting from the oppositeside of the silo. At first, the cow follows the parametrized path that was found in Problem
silo
#64 above. This happens until the parameter value θ reaches π. At his point, the rope is avertical straight line segment that is tangent to the silo (as illustrated in the figure above).From this point the cow just swings in a semicircle until the rope is pointing straightdown. The length of the rope is πr, half the circumference of the silo. So the top half ofthe grazing area is (a) the area between
{
r(
cos θ+θ sin θ, sin θ−θ cos θ)
∣
∣ 0 ≤ θ ≤ π}
andthe line x = −r, plus (b) the area of one quarter of a circle of radius πr minus (c) the areaof one half of a circle of radius r (since the cow cannot graze inside of the silo itself). Thethin horizontal strip shown in the figure above has length x+r = r
(
cos θ+θ sin θ+1)
and
width dy = dydθ dθ. So the the area between
{
r(
cos θ+θ sin θ, sin θ−θ cos θ)
∣
∣ 0 ≤ θ ≤ π}
and the line x = −r is∫ π
0
[
x(θ) + r]
dydθ dθ =
∫ π
0
r2[
cos θ + θ sin θ + 1](
cos θ − cos θ + θ sin θ)
dθ
= r2
∫ π
0
(
θ sin θ + θ sin θ cos θ + θ2 sin2 θ)
dθ
= 12r2
∫ π
0
(
2θ sin θ + θ sin(2θ)− θ2 cos(2θ) + θ2)
dθ
= 12r2
[
− 12θ2 sin(2θ)− θ cos(2θ) + 1
2 sin(2θ)− 2θ cos θ + 2 sin θ + θ3
3
]π
0
= 12r2
[
− π + 2π + π3
3
]
= 12r2
[
π + π3
3
]
45
So the area of the top half is
12r2
[
π + π3
3
]
+ 14π(πr)2 − 1
2πr2 = 512π3r2
The full area is twice this, or 56π3r2 .
66) At 2:00 a.m. the animal, studentia enubriatus, having contracted the disease, munchia,leaves its dwelling to forage for food. At time t, its x–coordinate (in m) is 60et sin 2πtand its y–coordinate (in m) is 60et cos 2πt. (The time t is in hours and t = 0 correspondsto 2:00 a.m.) If it still has not found any food at 4:00 a.m.a) sketch the path that it has followed andb) determine how far has it traveled.
Solution. a)
x
y
−300 300
−300
300
b)
dxdt = 60 et sin 2πt + 120π et cos 2πtdydt = 60 et cos 2πt− 120π et sin 2πt
⇒(
dxdt
)2+
(
dydt
)2=
(
60 et)2
[
sin2 2πt + 4π sin 2πt cos 2πt + 4π2 cos2 2πt
+ cos2 2πt− 4π sin 2πt cos 2πt + 4π2 sin2 2πt]
=(
60 et)2
[
1 + 4π2]
Hence
L =
∫ 2
0
√
(
dxdt
)2+
(
dydt
)2dt = 60
√
1 + 4π2
∫ 2
0
et dt = 60√
1 + 4π2[
e2 − 1]
67) Find the length of the parametric curve x = 34t4, y(t) = 1
6t6, t ≥ 0 from the point (0, 0)
to the point(
12, 323
)
.
46
Solution. The points (0, 0) and(
12, 323
)
correspond to t = 0 and t = 2 respectively. So,the length is
∫ 2
0
√
(
dxdt
)2+
(
dydt
)2dt =
∫ 2
0
√
9t6 + t10 dt =
∫ 2
0
t3√
9 + t4 dt
Make the change of variables u = t4 +9, du = 4t3 dt. When t = 0, u = 9 and when t = 2,u = 25, so the length is
∫ 25
9
√u du
4 = 14
u3/2
3/2
∣
∣
∣
25
9= 1
6 [53 − 33] = 493
68) A curve C has parametric representation x = 6t + 1, y = t3 − 2t.a) Find all points on C at which the tangent line is perpendicular to 3x + 5y − 8 = 0.b) Does the curve have a vertical tangent at any point?c) Find all local minima and maxima for y.
Solution. a) 3x + 5y − 8 = 0, or equivalently, y = 85− 3
5x, is a straight line with slope
− 35 . To be perpendicular to it, another line must have slope 5
3 . The slope of the curve isdydx = y′(t)
x′(t) = 3t2−26 . This slope is 5
3 when
3t2−26 = 5
3 ⇐⇒ 3t2 − 2 = 10 ⇐⇒ t2 = 4 ⇐⇒ t = ±2
The points on C have t = ±2 are (−11,−4) and (13, 4) .
b) The curve C has vertical tangent when dydx = y′(t)
x′(t) = 3t2−26 is infinite. This never
happens.
c) Observe that dydx
= y′(t)x′(t)
= 3t2−26
is zero for t = ±√
2/3, positive for t < −√
2/3
and t >√
2/3 and negative for −√
2/3 < t < −√
2/3. That is, y(t) increases for
−∞ < t < −√
2/3 then decreases for −√
2/3 < t <√
2/3 and then increases for
t >√
2/3. So y has a local maximum when t = −√
2/3 (so that y = 4√
69
) and a local
minimum when t =√
2/3 (so that y = − 4√
69 ).
69 a) Sketch r = 3 + 2 sin θ.b) Describe r = 2(sin θ + cos θ).Solution. a) As• θ runs from 0 to π
2 , sin θ runs from 0 to 1, so r runs from 3 to 5.• θ runs from π
2 to π, sin θ runs from 1 to 0, so r runs from 5 to 3.• θ runs from pi to 3π
2 , sin θ runs from 0 to −1, so r runs from 3 to 1.• θ runs from 3π
2 to 2π, sin θ runs from −1 to 0, so r runs from 1 to 3.This gives the figure
47
x
y
b)
r = 2(sin θ + cos θ) =⇒ r2 = 2(r sin θ + r cos θ) =⇒ x2 + y2 = 2(y + x)
=⇒ (x− 1)2 + (y − 1)2 = 2
This is a circle of radius√
2 centered on (1, 1).
70) Find the area of the region formed by the curve parametrized by the equations x =4a cos t, y = a
√2 sin t cos t for −π
2 ≤ t ≤ π2 .
Solution.
x
yNote that as t runs from −π
2to 0, x runs from 0 to 4a and y =
a√2
sin(2t) runs from 0 to − a√2
(at t = −π4 ) and then back to 0. As t
runs from 0 to π2, x runs from 4a back to 0 and y runs from 0 to a√
2
(at t = π4 ) and then back to 0. This curve is symmetric about the x
axis, because x is even under t → −t while y is odd under t → −t.So the area is twice the area above the x axis. The latter area is
obtained by have t run from π2 to 0. The total area is
2
∫ 4a
0
y dx = 2
∫ 0
π/2
y(t)x′(t) dt = 8a2√
2
∫ 0
π/2
sin t cos t(− sin t) dt
Making the substitution u = sin t, du = cos t dt
Area = −8a2√
2
∫ 0
1
u2 du = −8a2√
2u3
3
∣
∣
∣
0
1= 8
√2
3 a2
71) Find the area of the region lying inside both r = 1 and r2 = 2 sin 2θ.
Solution.
x
y
r2 = 2 sin 2θ
The two curves intersect when r2 = 1 = 2 sin 2θ. At these pointssin 2θ = 1
2 , so 2θ = π6 , 5π
6 , · · · or θ = π12 , 5π
12 , · · ·. By symmetry, we
may find the area inside both r = 1 and r =√
2 sin 2θ with 0 ≤ θ ≤ π4
and then multiply by 4. For 0 ≤ θ ≤ π12
, 2 sin 2θ is smaller than 1 so
the outer boundary of the region is r =√
2 sin 2θ. For π12 ≤ θ ≤ π
4 ,
48
2 sin 2θ is larger than 1 so the outer boundary of the region is r = 1.The area of the region is
A = 4[
12
∫ π/12
0
[√
2 sin 2θ ]2 dθ + 12
∫ π/4
π/12
12 dθ]
= 4[
∫ π/12
0
sin(2θ) dθ + 12
π6
]
= 4[
− cos 2θ2
∣
∣
π/12
0+ π
12
]
= 4[
12−
√3/22
+ π12
]
= 2−√
3 + π3
72) Consider the cardiod r = 1− cos θ and the unit circle r = 1.a) Graph both curves on the same graph.b) Find the area which lies inside the cardioid, but outside the circle.
Solution.
x
y The two curves intersect when r = 1 = 1− cos θ or r = 1, cos θ = 0or r = 1, θ = ±π
2 or x = 0, y = ±1. We are to find the area of theregion with π
2≤ θ ≤ 3π
2, 1 ≤ r ≤ 1− cos θ. This region has area
12
∫ 3π/2
π/2
[
(1− cos θ)2 − 12]
dθ = 12
∫ 3π/2
π/2
[
cos2 θ − 2 cos θ]
dθ
By the trig identity cos2 θ = 1+cos(2θ)2
, the area is
12
∫ 3π/2
π/2
[
12 + 1
2 cos(2θ)− 2 cos θ]
dθ =[
14 + 1
8 sin(2θ)− sin θ]3π/2
π/2= π
4 + 2
73) Sketch and find the area of the plane region that lies inside both the circle with polarequation r = sin θ and the circle with polar equation r =
√3 cos θ.
Solution.
x
y
r =√
3 cos θ
r = sin θ
The two circles intersect when r = 0 and when r = sin θ =√
3 cos θ.At the latter point tan θ =
√3, so θ = π
3 . For 0 ≤ θ ≤ π3 , the
outer boundary of the region is r = sin θ. For π3≤ θ ≤ π
2, the outer
boundary of the region is r =√
3 cos θ. The area of the region is
A = 12
∫ π/3
0
sin2 θ dθ + 12
∫ π/2
π/3
3 cos2 θ dθ
Subbing in sin2 θ = 1−cos 2θ2 and cos2 θ = 1+cos 2θ
2 ,
A = 14
∫ π/3
0
(1− cos 2θ) dθ + 34
∫ π/2
π/3
(1 + cos 2θ) dθ
= 14
[
θ − sin 2θ2
]π/3
0+ 3
4
[
θ + sin 2θ2
]π/2
π/3
= 14
π3 − 1
8 sin 2π3 + 3
4π6 − 3
8 sin 2π3
= 5π24 −
√3
4
49
74) Find the area of the region lying inside both r = 1 and r = 2 sin 2θ.
Solution.
x
y
r = 2 sin 2θ
By symmetry, we may find the area inside both r = 1 and r = 2 sin 2θwith 0 ≤ θ ≤ π
4 and then multiply by 8. The two curves intersectwhen r = 1 = 2 sin 2θ. At these points sin 2θ = 1
2, so 2θ = π
6, 5π
6, · · ·
or θ = π12 , 5π
12 , · · ·. For 0 ≤ θ ≤ π12 , 2 sin 2θ is smaller than 1 so the
outer boundary of the region is r = 2 sin 2θ. For π12≤ θ ≤ π
4, 2 sin 2θ
is larger than 1 so the outer boundary of the region is r = 1. The areaof the region is
A = 8[
12
∫ π/12
0
4 sin2 2θ dθ + 12
∫ π/4
π/12
12 dθ]
= 8[
∫ π/12
0
(1− cos 4θ) dθ + 12
π6
]
= 8[
π12 − sin 4θ
4
∣
∣
π/12
0+ π
12
]
= 4π3 −
√3
75) Evaluate∫
tan−1(x2) dx as a series.
Solution.
11+x = 1− x + x2 − x3 + · · · =
∞∑
n=0
(−1)nxn
11+x2 = 1− x2 + x4 − x6 + · · · =
∞∑
n=0
(−1)nx2n (replaced x by x2)
tan−1 x =
∫
dx1+x2 = C + x− x3
3+ x5
5− x7
7+ · · · = C +
∞∑
n=0
(−1)n x2n+1
2n+1
Because tan−1(0) = 0 the constant C must be zero and
tan−1 x = x− x3
3 + x5
5 − x7
7 + · · · =
∞∑
n=0
(−1)n x2n+1
2n+1
tan−1(x2) = x2 − x6
3+ x10
5− x14
7+ · · · =
∞∑
n=0
(−1)n x4n+2
2n+1(replace x by x2)
∫
tan−1(x2) dx = C + x3
3 − x7
3×7 + x11
5×11 − x15
7×15 + · · · = C +∑∞
n=0(−1)n x4n+3
(2n+1)(4n+3)
76) Find the sum of∞∑
n=0
(−1)n
62nπ2n
(2n)! .
50
Solution.
cos t = 1− t2
2! + t4
4! − t6
6! + · · · =
∞∑
n=0
(−1)n t2n
(2n)!
√3
2= cos π
6= 1− π2
2!62 + π4
4!64 − π6
6!66 + · · · =∞∑
n=0
(−1)n π2n
(2n)! 62n
77) Show that ex > 1 + x for all x > 0.
Solution.ex = 1 + x + 1
2x2 + 13!x
3 + · · ·+ 1n!x
n + · · ·
When x > 0, 1n!
xn > 0 for every n. Hence
A = 12x2 + 1
3!x3 + · · ·+ 1
n!xn + · · · > 0 =⇒ ex = 1 + x + A > 1 + x
78) Use series to evaluate the limitlimx→0
1−cos x1+x−ex
Solution.cos x = 1− x2
2!+ x4
4!− x6
6!+ · · ·
1− cos x = x2
2! − x4
4! + x6
6! − · · ·
= x2(
12! − x2
4! + x4
6! − · · ·)
ex = 1 + x + 12x2 + 1
3!x3 + · · ·
1 + x− ex = − 12x2 − 1
3!x3 − · · ·
= −x2(
12 − 1
3!x− · · ·)
So
limx→0
1−cos x1+x−ex = lim
x→0
x2(
12!−x2
4!+
x4
6!−···
)
−x2(
12−
13!x−···
) = − limx→0
12!−
x2
4! +x4
6! −···12− 1
3!x−···
= −12!12
= −1
79) Let I(x) =∫ x
0t3e−t2 dt.
a) Find a Maclaurin series for I(x).b) Use this series to approximate I(1), with an error less than 0.01.
51
Solution. a)
ex = 1 + x + x2
2! + x3
3! + x4
4! + · · · =
∞∑
n=0
xn
n!
e−t2 = 1− t2 + t4
2!− t6
3!+ t8
4!+ · · · =
∞∑
n=0
(−1)n t2n
n!(subbed in x = −t2)
t3e−t2 = t3 − t5 + t7
2! − t9
3! + t11
4! + · · · =
∞∑
n=0
(−1)n t2n+3
n!
I(x) = x4
4 − x6
6 + x8
2!8 − x10
3!10 + x12
4!12 + · · · =∞∑
n=0(−1)n x2n+4
n!(2n+4)
b)
I(1) = 14− 1
6+ 1
2!8− 1
3!10+ 1
4!12+· · · = 0.25−0.16+0.0625−0.016+0.000145−· · · = 0.129
The series for I(1) is an alternating series with decreasing successive terms. So approxi-mating I(1) by 1
4 − 16 + 1
2!8 − 13!10 introduces an error between 0 and 1
4!12 = 0.000145.
80) Let I(x) =∫ x
0sin t
tdt.
a) Find a Maclaurin series for I(x).b) Use this series to calculate an approximate value for I(0.5), with an error less than
10−6.
Solution. a)
sin t = t− t3
3!+ t5
5!− t7
7!+ · · · =
∞∑
n=0
(−1)n t2n+1
(2n+1)!
sin tt = 1− t2
3! + t4
5! − t6
7! + · · · =∞∑
n=0
(−1)n t2n
(2n+1)!
I(x) =
∫ x
0
sin tt dt = x− x3
3!3 + x5
5!5 − x7
7!7 + · · · =∞∑
n=0(−1)n x2n+1
(2n+1)!(2n+1)
b)I(0.5) = 1
2 − 1233!3 + 1
255!5 − 1277!7 + · · ·
= 0.5− 0.0069444 + 0.0000521− 0.0000002− · · ·= 0.4931076
The series for I(0.5) is an alternating series with decreasing successive terms. So approx-imating I(0.5) by 1
2− 1233!3 + 1
255!5 introduces an error between 0 and − 1277!7 = −2×10−7.
52
81) Use a suitable series expansion to calculate
∫ 1/4
0
1√t(1+t2)
dt
to within 10−4.
Solution. Define I(x) =∫ x
01√
t(1+t2)dt. Then
11+x = 1− x + x2 − x3 + · · · =
∞∑
n=0
(−1)nxn
11+t2
= 1− t2 + t4 − t6 + · · · =∞∑
n=0
(−1)nt2n (subbed in x = −t2)
1√t(1+t2)
= 1√t− t
32 + t
72 − t
112 + · · · =
∞∑
n=0
(−1)nt2n−0.5
I(x) = x12
1/2 − x52
5/2 + x92
9/2 − x132
13/2 + · · · =∞∑
n=0(−1)n x2n+0.5
2n+0.5
In particular, since(
14
)12 = 2−1,
I( 14) = 2−1
1/2− 2−5
5/2+ 2−9
9/2− 2−13
13/2+ · · ·
= 1− 2−4
5+ 2−8
9− 2−12
13+ · · ·
= 1− 0.0125 + 0.000434− 0.000019 · · · = 0.9879
The series for I( 14 ) is an alternating series with decreasing successive terms. So approxi-
mating I( 14 ) by 1− 2−4
5 + 2−8
9 introduces an error between 0 and − 2−12
13 = −0.000019.
82) Find the Maclaurin series for ln(1− x) and use it to calculate ln(0.9), with an error lessthan 10−5. Justify that your error is less than 10−5.
Solution. Let f(x) = ln(1− x). Then
ddx ln(1− x) = − 1
1−xd2
dx2 ln(1− x) = − 1(1−x)2
d3
dx3 ln(1− x) = − 2(1−x)3
d4
dx4 ln(1− x) = − 2×3(1−x)4
d5
dx5 ln(1− x) = − 2×3×4(1−x)5
dn
dxn ln(1− x) = − (n−1)!(1−x)n
In particular
ln(1− x)∣
∣
x=0= 0 d
dx ln(1− x)∣
∣
x=0= −1 d2
dx2 ln(1− x)∣
∣
x=0= −1
d3
dx3 ln(1− x)∣
∣
x=0= −2 d4
dx4 ln(1− x)∣
∣
x=0= −3! dn
dxn ln(1− x)∣
∣
x=0= −(n− 1)!
53
Henceln(1− x) = −x− 1
2x2 − 1
3x3 − · · · − 1
nxn + Rn(x)
whereRn(x) = f(n+1)(z)
(n+1)! xn+1 = − 1n+1
1(1−z)n+1 xn+1
for some z between 0 and x. In particular, with x = 0.1,
ln(0.9) = − 110 − 1
21
102 − 13
1103 − · · · − 1
n1
10n + Rn(x)
whereRn(0.1) = − 1
n+11
(1−z)n+11
10n+1
for some z between 0 and 0.1. So
∣
∣Rn(0.1)∣
∣ ≤ 1n+1
1(1−0.1)n+1
110n+1 = 1
n+11
9n+1
When n = 4∣
∣R4(0.1)| ≤ 15
195 = 3.4× 10−6
Hence
ln(0.9) = − 110− 1
21
102 − 13
1103 − 1
41
104 ± 3.4× 10−6 = −0.105358± 5× 10−6
83) Use a suitable series expansion to calculate
∫ 1/4
0
et2 dt
to within 10−5.
Solution. Since every derivative of ex is also ex
ex = 1 + x + 12x2 + 1
3!x3 + · · ·+ 1
n!xn + Rn(x)
whereRn(x) = 1
(n+1)!ezxn+1
for some z between 0 and x. In particular, with 0 ≤ x = t2 ≤ 142 ,
et2 = 1 + t2 + 12t4 + 1
3!t6 + · · ·+ 1
n!t2n + Rn(t2)
whereRn(t2) = 1
(n+1)!ezt2(n+1) ≤ 1
(n+1)!e1/16 1
42(n+1)
since z is between 0 and 116
and t is between 0 and 14. When n = 3
∣
∣Rn(t2)| ≤ 14!e
1/16 148 = 7× 10−7
54
Hence
∫ 1/4
0
et2 dt =
∫ 1/4
0
[
1 + t2 + 12t4 + 1
3!t6 ± 7× 10−7
]
dt
=[
t + 13t3 + 1
10t5 + 1
42t7 ± 7× 10−7t
]1/4
0
= 14 + 1
3×43 + 110×45 + 1
42×47 ± 2× 10−7 = 0.255307± 2× 10−6
84 a) Show that the series∞∑
n=1
1n4+1 converges.
b) How many terms would be needed to compute the sum with an error less than 0.001?
c) Does the series∞∑
n=1
√n2n
n2n−n2+1converge? Justify your answer.
Solution. a) Observe that 1n4+1 ≤ 1
x4+1 for all x ≤ n. In particular, 1n4+1 ≤
∫ n
n−11
x4+1dx. This forces
∞∑
n=N
1n4+1 ≤
∞∑
n=N
∫ n
n−1
1x4+1 dx =
∫ ∞
N−1
1x4+1 dx <
∫ ∞
N−1
1x4 dx = x−3
−3
∣
∣
∣
∞
N−1= 1
3(N−1)3
In particular S = 1 +∑∞
n=21
n4+1 ≤ 1 + 13(2−1)3 = 4
3 , so that the series converges.
b) The tail∞∑
n=N+1
1n4+1 < 0.001 if
13N3 < 0.001 ⇐⇒ N3 > 1
3×0.001 ⇐⇒ N > 3
√
10.003 ⇐⇒ N > 6.93
If we keep at least 7 terms, the truncation error will be at most 0.001.c)
∞∑
n=1
√n2n
n2n−n2+1 ≥∞∑
n=1
√n2n
n2n =
∞∑
n=1
1√n
>
∫ ∞
1
1√x
dx = limT→∞
2√
x∣
∣
∣
T
1= ∞
so the series diverges .
85 For each of the following functions, find a power series representation of the function.
a) f(x) = 14+x2 b) f(x) = 1+x2
1−x2 c) f(x) = 3x−22x2−3x+1
d) f(x) = ln(
1+x1−x
)
e) f(x) = x2
(1−2x)2
Solution. a)
f(x) = 14+x2 = 1
41
1−y
∣
∣
y=−x2/4= 1
4
∞∑
n=0
yn∣
∣
y=−x2/4= 1
4
∞∑
n=0(−1)n x2n
4n
55
b)
f(x) = 1+x2
1−x2 = (1 + x2) 11−y
∣
∣
y=x2 = (1 + x2)
∞∑
n=0
yn∣
∣
y=x2 = (1 + x2)
∞∑
n=0
x2n
Combining
∞∑
n=0
x2n = 1 + x2 + x4 + x6 + · · ·
x2∞∑
n=0
x2n = x2 + x4 + x6 + · · ·
=⇒ f(x) = 1 + 2x2 + 2x4 + 2x6 + · · · = 1 +∞∑
n=12x2n
c)
f(x) = 3x−22x2−3x+1 = 3x−2
(1−2x)(1−x) = − 11−2x − 1
1−x = −∞∑
n=0
(2x)n −∞∑
n=0
xn
= −∞∑
n=0(1 + 2n)xn
d) We start with
ln(1 + x) =
∫ x
0
11+t dt =
∫ x
0
[
1− t + t2 − t3 − · · ·]
dt = x− x2
2 + x3
3 − x4
4 − · · ·
So
f(x) = ln(1 + x)− ln(1− x) = x− x2
2 + x3
3 − x4
4 − · · · −[
− x− x2
2 − x3
3 − x4
4 − · · ·]
= 2x + 2x3
3+ 2x5
5+ · · · =
∞∑
n=02x2n+1
2n+1
e)
f(x) = x2
(1−2x)2 = x2
2ddx
11−2x = x2
2ddx
[
1 + 2x + 22x2 + 23x3 + · · ·+ 2nxn + · · ·]
= x2
2
[
2 + 2(22)x + 3(23)x2 + · · ·+ n(2n)xn−1 + · · ·]
= x2 + 22x3 + 3(22)x4 + · · ·+ n(2n−1)xn+1 + · · ·
=∞∑
n=1n(2n−1)xn+1
56
