Solutions to Math 152 Review Problems for Exam 1
(1) If A(x) is the area of the rectangle
formed when the solid is sliced at x perpendicular
√
to the x-axis, then A(x) = |x|(2 1 −√x2 ), because the height of the rectangle is |x| and
the base of the rectangle has length 2 1 − x2 . Therefore, the volume of the solid is
Z
1
|x|(2
p
1−
x2 ) dx
0
Z
|x|(2
=
−1
−1
0
Z
=
p
1−
x2 ) dx
Z
1
|x|(2
+
p
1 − x2 ) dx
0
Z
p
2
(−x)(2 1 − x ) dx +
−1
1
p
x(2 1 − x2 ) dx.
0
If we use the substitution u = 1 − x2 then we see that the integral over [−1, 0] is equal
to 2/3, and that the integral over [0, 1] is also equal to 2/3. Therefore, the volume of the
solid is 2/3 + 2/3 = 4/3.
√
R1 √
(2) The average of f (x) = 1 − x2 over [−1, 1] is 12 −1 1 − x2 dx = 21 π2 = π4 . We used
R1 √
the following fact: −1 1 − x2 dx is half of the area of a circle of radius 1. Now we have
p
√
to solve 1 − c2 = f (c) = π4 . The solutions are c = ± 1 − π 2 /16. The existence of at
least one c is guaranteed by the Mean Value Theorem for Integrals.
(3)(a) This type of cone is obtained when the region bounded by y = 0, x = H, y = (R/H)x
RH
is rotated about the x-axis. The volume is π 0 ((R/H)x)2 dx = πR2 H/3.
(3)(b) This type of cone is obtained when the region bounded by x = 0, y = H, y = (H/R)x
RR
is rotated about the y-axis. The volume is 0 2πx(H − (H/R)x) dx = πR2 H/3.
(4)(a) The equation y = 2(x − 3) is equivalent to x = 3 + y/2. The method of washers
gives a volume of
Z
2
2
2
Z
(3 + y/2 − 1) − (3 − 1) dy = π
π
0
2
2y + y 2 /4 dy = π(4 + 2/3) = 14π/3.
0
(4)(b) The method of shells gives a volume of
Z
4
4
Z
−x2 + 5x − 4 dx = 4π(7/6) = 14π/3.
2π(x − 1)(2 − 2(x − 3)) dx = 4π
3
3
Z
du
cos x dx
=
= ln |u| + C = ln | sin x| + C.
(5)(a) u = sin x, cot x dx =
sin x
u
Z
Z
Z
sin x dx
du
(5)(b) u = cos x,
tan x dx =
=−
= − ln |u| + C = − ln | cos x| + C,
cos x
u
where the last expression equals ln | sec x| + C.
Z
Z
1
Z
du
tan x sec x dx
=
= ln | sec x| + C.
(5)(c) u = sec x, tan x dx =
sec x
u
Z
Z
Z
sinh x dx
du
(5)(d) u = cosh x,
tanh x dx =
=
= ln |u| + C = ln | cosh x| + C,
cosh x
u
which equals ln(cosh x) + C because cosh x is always positive for real values of x.
Z
Z
Z
du
cosh x dx
=
= ln |u| + C = ln | sinh x| + C.
(5)(e) u = sinh x,
coth x dx =
sinh x
u
Z
Z
(6)(A) When u = csc x + cot x we get
Z
Z
csc x + cot x
csc2 x + csc x cot x
csc x dx = csc x ·
dx =
dx
csc x + cot x
csc x + cot x
Z
du
=−
= − ln |u| + C = − ln | csc x + cot x| + C.
u
Z
(6)(B) When u = csc x − cot x we get
Z
Z
csc x − cot x
csc2 x − csc x cot x
csc x dx = csc x ·
dx =
dx
csc x − cot x
csc x − cot x
Z
du
=
= ln |u| + C = ln | csc x − cot x| + C.
u
Z
(6)(C) The computation below uses csc2 x − cot2 x = 1 at the end:
1
csc
x
−
cot
x
= ln − ln | csc x + cot x| = ln csc x + cot x
(csc x + cot x)(csc x − cot x) csc x − cot x = ln | csc x − cot x|.
= ln 2
csc x − cot2 x (7) For the first integral, we do an integration by parts and get
Z
Z
2
sin x dx = sin x sin x dx
Z
= (− cos x)(sin x) − (− cos x)(cos x) dx
Z
= − cos x sin x + cos2 x dx
Z
= − cos x sin x + (1 − sin2 x) dx
Z
= − cos x sin x + x − sin2 x dx.
2
After adding
R
sin2 x dx to both sides, we discover
Z
2 sin2 x dx = − cos x sin x + x + C,
and this is equivalent to
Z
sin2 x dx =
1
(x − cos x sin x) + C.
2
The second integral is very similar. Integration by parts produces
Z
Z
2
cos x dx = cos x cos x dx
Z
= (sin x)(cos x) − (sin x)(− sin x) dx
Z
= sin x cos x + sin2 x dx
Z
= sin x cos x + (1 − cos2 x) dx
Z
= sin x cos x + x − cos2 x dx.
We add
R
cos2 x dx to both sides and get
Z
2 cos2 x dx = sin x cos x + x + C,
and this is equivalent to
Z
cos2 x dx =
1
(x + sin x cos x) + C.
2
R
The Rmethod in this problem can be used to derive the reduction formulas for cosn x dx
and sinn x dx. In order to get the cos reduction formula, we proceed as follows:
Z
Z
n
cos x dx = cos x cosn−1 x dx
Z
n−1
= (sin x)(cos
x) − (sin x) (n − 1) cosn−2 x(− sin x) dx
Z
n−1
= sin x cos
x + (n − 1) sin2 x cosn−2 x dx
Z
n−1
= sin x cos
x + (n − 1) (1 − cos2 x) cosn−2 x dx
Z
Z
n−1
n−2
= sin x cos
x + (n − 1) cos
x dx − (n − 1) cosn x dx.
3
R
After adding (n − 1) cosn x dx to both sides, we get
Z
Z
n
n−1
n cos x dx = sin x cos
x + (n − 1) cosn−2 x dx,
and this is equivalent to
Z
Z
n−1
1
n−1
n
x+
cosn−2 x dx.
cos x dx = sin x cos
n
n
(8) If we use the substitution u = tan x then we find
Z
Z
Z
4
2
2
tan x sec x dx = tan x sec x sec x dx = tan x(1 + tan2 x) sec2 x dx
Z
Z
u2
u4
tan2 x tan4 x
2
+
+C =
+
+ C.
= u(1 + u ) du = u + u3 du =
2
4
2
4
On the other hand, we can use the substitution u = sec x and obtain
Z
Z
Z
u4
sec4 x
4
3
3
tan x sec x dx = sec x(tan x sec x) dx = u du =
+C =
+ C.
4
4
To see that these two answers are really the same, note
(sec2 x)2
(1 + tan2 x)2
1 + 2 tan2 x + tan4 x
sec4 x
+C =
+C =
+C =
+C
4
4
4
4
tan2 x tan4 x
1
tan2 x tan4 x
=
+
+
+C =
+
+C
2
4
4
2
4
because
1
4
+ C is an arbitrary constant.
2
x
(9) We use integration by parts and long division (applied to 1+x
2 ) to conclude
Z
Z
x2
x2
1
−1
−1
x tan x dx =
tan x −
·
dx
2
2 1 + x2
Z
1
1
x2
1
x2
−1
tan x −
1−
dx
=
tan−1 x − (x − tan−1 x) + C.
=
2
2
2
1+x
2
2
(10) There are three integrations by parts:
Z
Z
3(ln x)2
3
3
(ln x) dx = x(ln x) − x ·
dx
x
Z
3
= x(ln x) − 3 (ln x)2 dx
Z
2 ln x
3
2
= x(ln x) − 3 x(ln x) − x ·
dx
x
Z
3
2
= x(ln x) − 3x(ln x) + 6 ln x dx
Z
x
3
2
= x(ln x) − 3x(ln x) + 6 x ln x −
dx
x
= x(ln x)3 − 3x(ln x)2 + 6x ln x − 6x + C.
4
Of course, we could have derived a reduction formula for
times.
R
(ln x)n dx and used it three
(11) The first integration by parts gets us as far as
Z
Z
1
1
cos(ax) cos(bx) dx =
sin(ax) cos(bx) −
sin(ax) − b sin(bx) dx
a
a
Z
1
b
= sin(ax) cos(bx) +
sin(ax) sin(bx) dx.
a
a
R
Now we focus on sin(ax) sin(bx) dx and do another integration by parts:
Z
Z
1
1
sin(ax) sin(bx) dx = − cos(ax) sin(bx) −
− cos(ax) b cos(bx) dx
a
a
Z
1
b
= − cos(ax) sin(bx) +
cos(ax) cos(bx) dx.
a
a
When this equation is substituted into the previous equation, we get
Z
cos(ax) cos(bx) dx
Z
1
b
1
b
= sin(ax) cos(bx) +
− cos(ax) sin(bx) +
cos(ax) cos(bx) dx ,
a
a
a
a
which can be rewritten as
Z
b2
b
1
1− 2
cos(ax) cos(bx) dx = sin(ax) cos(bx) − 2 cos(ax) sin(bx) + C.
a
a
a
This is equivalent to
Z
a sin(ax) cos(bx) − b cos(ax) sin(bx)
+ C.
cos(ax) cos(bx) dx =
a2 − b2
One can show that this answer is equivalent to equation 25 on page 410 of the textbook.
(12) Integration by parts and the trigonometric identity 1 + tan2 x = sec2 x allow us to
write
Z
Z
Z
3
2
sec x dx = sec x sec x dx = tan x sec x − tan x(tan x sec x) dx
Z
= tan x sec x − tan2 x sec x dx
Z
= tan x sec x − (sec2 x − 1) sec x dx
Z
Z
3
= tan x sec x − sec x dx + sec x dx
Z
= tan x sec x − sec3 x dx + ln | tan x + sec x|.
5
After adding
R
sec3 x dx to both sides, we get
Z
2
sec3 x dx = tan x sec x + ln | tan x + sec x| + C.
After dividing by 2, we get
Z
1
1
sec3 x dx = tan x sec x + ln | tan x + sec x| + C.
2
2
R
The Rmethod in this problem can be used to derive the reduction formulas for secm x dx
and cscm x dx. In order to get the sec reduction formula, we proceed as follows:
Z
m
Z
sec2 x secm−2 x dx
Z
m−2
= (tan x)(sec
x) − (tan x) (m − 2) secm−3 x tan x sec x dx
Z
m−2
= tan x sec
x − (m − 2) tan2 x secm−2 x dx
Z
m−2
= tan x sec
x − (m − 2) (sec2 x − 1) secm−2 x dx
Z
Z
m−2
m
= tan x sec
x − (m − 2) sec x dx + (m − 2) secm−2 x dx.
sec x dx =
R
After adding (m − 2) secm x dx to both sides, we get
Z
(m − 1)
m
sec x dx = tan x sec
and this is equivalent to
Z
secm x dx =
m−2
Z
x + (m − 2)
1
m−2
tan x secm−2 x +
m−1
m−1
Z
secm−2 x dx,
secm−2 x dx.
(13) We use the substitution x = 3 tan θ and the result of problem (12):
Z p
Z
Z
2
2
9 + x dx = (3 sec θ)(3 sec θ) dθ = 9 sec3 θ dθ
9
9
tan θ sec θ + ln | tan θ + sec θ| + C
2
2
√
√
2
9 x
9+x
9 x
9 + x2 = · ·
+ ln +
+C
2 3
3
2 3
3
p
xp
9
=
9 + x2 + ln(x + 9 + x2 ) + C,
2
2
=
6
where we used
√
p
9
9
9 x
9 + x2 +
C
=
9 + x2 | − ln 3 + C,
ln +
ln
|x
+
2 3
3
2
2
√
the fact that − 29 ln 3 + C is an arbitrary constant C, and the positivity of x + 9 + x2 for
any x.
(14) Here the substitution x = 3 sin θ is appropriate and we use problem (7):
Z
Z
Z p
2
9 − x dx = 3 cos θ · 3 cos θ dθ = 9 cos2 θ dθ
!
x x √9 − x 2
9
9
= (θ + sin θ cos θ) + C =
sin−1
+ C.
+ ·
2
2
3
3
3
(15) Here the substitution x = tan θ is appropriate and we also use problem (7). The trick
tan θ
sin θ cos θ = sec
2 θ is helpful at the end when we rewrite everything in terms of x.
Z
sec2 θ dθ
=
(1 + tan2 θ)2
Z
Z
dθ
sec2 θ dθ
=
= cos2 θ dθ
(sec2 θ)2
sec2 θ
1
1
tan θ
= (θ + sin θ cos θ) + C =
θ+
+C
2
2
sec2 θ
1
x
−1
=
tan x +
+ C.
2
1 + x2
dx
=
(1 + x2 )2
Z
Z
(16) The correct partial fractions setup is
Z
3x2 − 3x − 2
dx =
(x2 − 1)(x − 1)
Z
3x2 − 3x − 2
dx =
(x + 1)(x − 1)2
Z
A
B
C
+
+
dx.
x + 1 x − 1 (x − 1)2
Now we have to solve for A, B, C in the equation
3x2 − 3x − 2
A
B
C
=
+
+
.
(x + 1)(x − 1)2
x + 1 x − 1 (x − 1)2
Since this can be rewritten as
A(x − 1)2 + B(x − 1)(x + 1) + C(x + 1)
3x2 − 3x − 2
=
,
(x + 1)(x − 1)2
(x + 1)(x − 1)2
we have to solve for A, B, C in the equation
3x2 − 3x − 2 = A(x − 1)2 + B(x − 1)(x + 1) + C(x + 1).
7
If we plug in x = 1 and x = −1 into this last equation, then we get C = −1 and A = 1,
respectively. If we compare the coefficients of x2 in this same last equation, then we get
3 = A + B. Now we conclude B = 2. Finally,
Z
Z
2
−1
1
3x2 − 3x − 2
dx
dx =
+
+
2
(x − 1)(x − 1)
x + 1 x − 1 (x − 1)2
1
= ln |x + 1| + 2 ln |x − 1| +
+ C.
x−1
(17) The correct partial fractions setup is
Z
Z
x2 + 3x
Ax + B
C
dx
=
+
dx.
(x2 + 1)(x + 1)
x2 + 1
x+1
Now we have to solve for A, B, C in the equation
x2 + 3x
Ax + B
C
(Ax + B)(x + 1) + C(x2 + 1)
=
+
=
.
(x2 + 1)(x + 1)
x2 + 1
x+1
(x2 + 1)(x + 1)
This is the same as solving for A, B, C in the equation
(∗)
x2 + 3x = (Ax + B)(x + 1) + C(x2 + 1) = (A + C)x2 + (B + A)x + (B + C).
Equating coefficients, we get
(∗∗)
A+C =1 ,
B+A=3 ,
B + C = 0.
There is a short cut for solving these three equations in three unknowns: If we go back to
the equation x2 + 3x = (Ax + B)(x + 1) + C(x2 + 1) in (∗) and plug in x = −1, then we
obtain −2 = 2C, hence C = −1. Plugging C = −1 into part A + C = 1 of (∗∗) we get
A = 2. Plugging A = 2 into B + A = 3 of (∗∗) we get B = 1. Our results are consistent
with B + C = 0 of (∗∗). This is a way of checking our arithmetic. Plugging these A, B, C
into our partial fractions setup, we get
Z
Z
2x + 1
−1
x2 + 3x
dx =
+
dx = ln(x2 + 1) + tan−1 x − ln |x + 1| + C.
2
2
(x + 1)(x + 1)
x +1 x+1
1
1
(18) We know 0 < <
for 0 < x ≤ 1 and we know that
dx diverges. All this
x
0
Z 1 x
e
implies that
dx diverges.
0 x
Z ∞
Z
1 ∞ −u2
−x4
Turning to the other improper integral, we obtain
xe
dx =
e
du when we
2 0
Z0 ∞
2
use the substitution u = x2 . If we can show that
e−u du converges then we will be
1
x
Z
ex
x
0
8
Z
∞
Z
−x4
xe
dx converges. It is clear that
able to conclude that
0
Z ∞
2
can show that
e−u du is finite then it will be clear that
1
2
e−u du is finite. If we
0
1
Z
∞
−u2
e
0
1
Z
du =
e
−u2
∞
Z
0
2
e−u du
du +
1
−u2
−u
is finite, and hence convergent. If u ≥ 1 then 0 < e
≤ e . In addition,
Z ∞
2
e−1 . The last two sentences imply that
e−u du is finite.
Z
∞
e−u du =
1
1
1
A
B
=
+
. This is equivalent to A(3x + 1) +
(2x + 1)(3x + 1)
2x + 1
3x + 1
B(2x + 1) = 1. The substitution x = −1/3 gives B = 3. The substitution x = −1/2 gives
A = −2. Now we get
Z
Z
3x + 1 3
2
dx
+C.
=
−
dx = ln |3x+1|−ln |2x+1|+C = ln (2x + 1)(3x + 1)
3x + 1 2x + 1
2x + 1 (19) We solve
The computation
Z
leads to
4
∞
3R + 1 3
lim ln = ln = ln(3/2)
R→∞
2R + 1
2
3R + 1 3 · 4 + 1
dx
− ln = lim ln 2 · 4 + 1 = ln(3/2) − ln(13/9).
(2x + 1)(3x + 1) R→∞ 2R + 1 cos2 x
1
(20) We know 0 ≤
≤ 3 for x ≥ 1. Since
3
x
x
Z ∞
cos2 x
dx converges.
x3
1
9
Z
1
∞
1
dx converges, we conclude that
x3