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Math132 Exam1 Solution
1. (2× 7% = 14%)
(a) Given the function f(x) =
∫ x2
1
tan√t dt. Use part I of the Fundamental Theo-
rem of Calculus and the chain rule to find f ′(x). You do not need to algebraically
simplify your answer.
Solution:
Set u = x2. Since dfdx
= dfdu
dudx, we have
f ′(x) =d
du[
u∫1
tan√t dt] · du
dx
= tan√u · (2x)
= tan√x2 · (2x)
(b) Evaluate
∫(2ex + 4 cos x+ 6 sec2 x+
8
x2 + 1) dx.
Solution:
Split the integral into four terms and then evaluate each:∫(2ex + 4 cos(x) + 6 sec2(x) +
8
x2 + 1) dx
=
∫2ex dx+
∫4 cos(x) dx+
∫6 sec2(x) dx+
∫8
x2 + 1dx
= 2ex + 4 sin(x) + 6 tan(x) + 8 arctan(x) + C
1
2. (2× 7% = 14%) The velocity function (in meters per second) is given v(t) = t2 − t for
a particle moving along a line.
(a) Find the displacement of the particle during the given time interval: 0 ≤ t ≤ 2.
Solution:
Let d denote the displacement of the particle. Then
d =
2∫0
v(t) dt
=
2∫0
t2 − t dt
=t3
3− t2
2
∣∣∣∣20
=2
3
(b) Find the distance traveled by the particle during the given time interval: 0 ≤ t ≤
2.
Solution:
Let D denote the distance traveled by the particle. Then
D =
2∫0
|v(t)| dt
=
2∫0
|t2 − t| dt
=
1∫0
t− t2 dt+
2∫1
t2 − t dt
=t2
2− t3
3
∣∣∣∣10
+t3
3− t2
2
∣∣∣∣21
=[12− 1
3
]+[(8
3− 2)− (
1
3− 1
2)]
= 1
2
3. (2× 7% = 14%)
(a) Evaluate
∫(ln x)2
xdx.
Solution:
Set u = ln x. Then du = 1xdx and∫(lnx)2
xdx =
∫u2 du
=u3
3+ C
=(lnx)3
3+ C
(b) Evaluate
∫esin x cos x dx.
Solution:
Set u = sin x. Then du = cos x dx and∫esinx cosx dx =
∫eu du
= eu + C
= esinx + C
3
4. (2× 7% = 14%)
(a) Evaluate
∫x2 ln x dx.
Solution:
Set u = ln x and dv = x2dx. Then we have du = 1xdx and v = x3
3. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
lnx d(x3
3) =
x3
3· lnx−
∫x3
3d(lnx)
=x3
3· lnx−
∫x3
3· 1xdx
=x3
3· lnx−
∫x2
3dx
=x3
3· lnx− 1
9x3 + C
(b) Evaluate
∫arcsin x dx.
Solution:
Set u = arcsin x and dv = dx. Then we have du = 1√1−x2dx and v = x. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
arcsinx dx = x · arcsinx−∫
x√1− x2
dx
Now use u-substitution to solve the second integral. Let u = 1 − x2. Then
du = −2xdx (in particular, xdx = −12du) and we have∫
x√1− x2
dx =
∫(−1
2du)
√u
= −1
2
∫u− 1
2 du
= −1
2· 2u
12 + C
= −√1− x2 + C
Therefore,∫arcsin x dx = x · arcsinx+
√1− x2 + C
4
5. (2× 7% = 14%)
(a) Evaluate
∫1√
1 + x2dx.
Solution:
i. Set x = tan θ, −π2< θ < π
2
ii.√1 + x2 =
√1 + (tan θ)2 = sec θ and dx = sec2 θdθ
iii. ∫1√
1 + x2dx =
∫1
sec θsec2 θ dθ
=
∫sec θ dθ
= ln | tan θ + sec θ|+ C
= ln |x+√1 + x2|+ C
(b) Evaluate
∫1√
x2 − 1dx.
Solution:
i. Set x = sec θ, 0 < θ < π2
ii.√x2 − 1 =
√(sec θ)2 − 1 = tan θ and dx = sec θ tan θdθ
iii. ∫1√
x2 − 1dx =
∫1
tan θsec θ tan θ dθ
=
∫sec θ dθ
= ln | tan θ + sec θ|+ C
= ln |√x2 − 1 + x|+ C
5
6. (2× 7% = 14%)
(a) Evaluate
∫sin4 x cos3 x dx.
Solution:
Set u = sin x. Then du = cos x dx, and we have∫sin4 x cos3 x dx =
∫sin4(x)(1− sin2 x) d(sinx)
=
∫u4(1− u2) du
=
∫u4 − u6 du
=u5
5− u7
7+ C
=sin5 x
5− sin7 x
7+ C
(b) Evaluate
∫sec4 x tan xdx.
Solution:
Set u = tan x. Then du = sec2 x dx, and we have∫sec4 x tanx dx =
∫sec2 x tanx sec2 x dx
=
∫(1 + tan2 x) tan x d(tan x)
=
∫(1 + u2)u du
=
∫u+ u3 du
=u2
2+
u4
4+ C
=tan2 x
2+
tan4 x
4+ C
6
7. (2× 7% = 14%) Let R be the region enclosed by the curves y =√x and y = x4.
(a) Find the area of the region R.
Solution:
i. Find the common points:
√x = x4 ⇒ x = x8
⇒ x− x8 = 0
⇒ x(1− x7) = 0
⇒ x = 0, 1
ii. Calculate the area:
A(R) =
1∫0
√x− x4 dx
=2x
32
3− x5
5
∣∣∣∣10
=2
3− 1
5
=7
15
7
(b) Suppose a solid S is created by revolving R around the x-axis. Calculate the
volume of S.
Solution:
i. ro =√x
ri = x4
ii. Calculate the volume:
V (S) =
1∫0
π(r2o − r2i ) dx
=
1∫0
π(x− x8) dx
= π[x2
2− x9
9]
∣∣∣∣10
= π[1
2− 1
9]
Math132 Make-Up Exam1 Solution
1. (2× 7% = 14%)
(a) Given the function f(x) =
∫ sin x
1
√1 + t2 dt. Use part I of the Fundamental Theo-
rem of Calculus and the chain rule to find f ′(x). You do not need to algebraically
simplify your answer.
Solution:
0
Set u = sin x. Since dfdx
= dfdu
dudx, we have
f ′(x) =d
du[
u∫1
√1 + t2 dt] · du
dx
=√1 + u2 · (cosx)
=√1 + sin2 x · (cosx)
(b) Evaluate
∫(3ex + 5 sin x+ 7 csc2 x+
9
x2 + 1) dx.
Solution:
Split the integral into four terms and then evaluate each:∫(2ex + 5 sin(x) + 7 csc2(x) +
9
x2 + 1) dx
=
∫2ex dx+
∫5 sin(x) dx+
∫7 csc2(x) dx+
∫9
x2 + 1dx
= 2ex − 5 cos(x)− 7 cot(x) + 9 arctan(x) + C
1
2. (2 × 7% = 14%) The velocity function (in meters per second) is given v(t) = t2 − 2t
for a particle moving along a line.
(a) Find the displacement of the particle during the given time interval: 1 ≤ t ≤ 3.
Solution:
Let d denote the displacement of the particle. Then
d =
3∫1
v(t) dt
=
3∫1
t2 − 2t dt
=t3
3− 2t2
2
∣∣∣∣31
=2
3
(b) Find the distance traveled by the particle during the given time interval: 1 ≤ t ≤
3.
Solution:
Let D denote the distance traveled by the particle. Then
D =
3∫1
|v(t)| dt
=
3∫1
|t2 − 2t| dt
=
2∫1
2t− t2 dt+
3∫2
t2 − 2t dt
= t2 − t3
3
∣∣∣∣21
+t3
3− t2
∣∣∣∣32
= 2
2
3. (2× 7% = 14%)
(a) Evaluate
∫(ln x)3
xdx.
Solution:
Set u = ln x. Then du = 1xdx and∫(lnx)3
xdx =
∫u3 du
=u4
4+ C
=(lnx)4
4+ C
(b) Evaluate
∫ecos x sin x dx.
Solution:
Set u = cos x. Then du = − sinx dx and∫ecosx sinx dx = −
∫eu du
= −eu + C
= −ecosx + C
3
4. (2× 7% = 14%)
(a) Evaluate
∫x3 ln x dx.
Solution:
Set u = ln x and dv = x3dx. Then we have du = 1xdx and v = x4
4. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
lnx d(x4
4) =
x4
4· lnx−
∫x4
4d(lnx)
=x4
4· lnx−
∫x4
4· 1xdx
=x4
4· lnx−
∫x3
4dx
=x4
4· lnx− 1
16x4 + C
(b) Evaluate
∫arctan x dx.
Solution:
Set u = arctan x and dv = dx. Then we have du = 11+x2dx and v = x. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
arctanx dx = x · arctanx−∫
x
1 + x2dx
Now use u-substitution to solve the second integral. Let u = 1 + x2. Then
du = 2xdx and we have∫x
1 + x2dx =
∫(12du)
u
=1
2
∫u−1 du
=1
2ln |u|+ C
=1
2ln |1 + x2|+ C
Therefore,∫arctanx dx = x · arctanx− 1
2ln |1 + x2|+ C
4
5. (2× 7% = 14%)
(a) Evaluate
∫1
x√1 + x2
dx.
Solution:
i. Set x = tan θ, −π2< θ < π
2
ii.√1 + x2 =
√1 + (tan θ)2 = sec θ and dx = sec2 θdθ
iii. ∫1
x√1 + x2
dx =
∫1
tan θ sec θsec2 θ dθ
=
∫sec θ
tan θdθ
=
∫csc θ dθ
= ln | − cot θ + csc θ|+ C
= ln | − 1
x+
√1 + x2
x|+ C
(b) Evaluate
∫1
x√1− x2
dx.
Solution:
i. Set x = sin θ
ii.√1− x2 =
√1− (sin θ)2 = cos θ and dx = cos θdθ
iii. ∫1
x√1− x2
dx =
∫1
sin θ cos θcos θ dθ
=
∫csc θ dθ
= ln | − cot θ + csc θ|+ C
= ln | −√1− x2
x+
1
x|+ C
5
6. (2× 7% = 14%)
(a) Evaluate
∫sin3 x cos4 x dx.
Solution:
Set u = cos x. Then du = − sinx dx, and we have∫sin3 x cos4 x dx = −
∫cos4(x)(1− cos2 x) d(cosx)
= −∫
u4(1− u2) du
= −∫
u4 − u6 du
= −u5
5+
u7
7+ C
= −cos5 x
5+
cos7 x
7+ C
(b) Evaluate
∫sec x tan3 xdx.
Solution:
Set u = sec x. Then du = sec x tanx dx, and we have∫sec x tan3 x dx =
∫sec x tanx(sec2 x− 1) dx
=
∫(u2 − 1) du
=u3
3− u+ C
=sec3 x
3+ sec x+ C
6
7. (2× 7% = 14%) Let R be the region enclosed by the curves y =√x and y = x2.
(a) Find the area of the region R.
Solution:
i. Find the common points:
√x = x2 ⇒ x = x4
⇒ x− x4 = 0
⇒ x(1− x3) = 0
⇒ x = 0, 1
ii. Calculate the area:
A(R) =
1∫0
√x− x2 dx
=2x
32
3− x3
3
∣∣∣∣10
=2
3− 1
3
=1
3
7
(b) Suppose a solid S is created by revolving R around the y-axis. Calculate the
volume of S.
Solution: Integrate with respect to y:
i. ro =√y
ri = y2
ii. Calculate the volume:
V (S) =
1∫0
π(r2o − r2i ) dx
=
1∫0
π(y − y4) dx
= π[y2
2− y5
5]
∣∣∣∣10
= π[1
2− 1
5]
Math132 Special Make-Up Exam1 Solution
1. (2× 7% = 14%)
(a) Given the function f(x) =
∫ cos x
1
√1 + t3 dt. Use part I of the Fundamental The-
orem of Calculus and the chain rule to find f ′(x). You do not need to algebraically
simplify your answer.
Solution:
8
Set u = cos x. Since dfdx
= dfdu
dudx, we have
f ′(x) =d
du[
u∫1
√1 + t3 dt] · du
dx
=√1 + u3 · (sinx)
=√1 + cos3 x · (sinx)
(b) Evaluate
∫(ex +
3
x+ 5 sec x tan x+
7√1− x2
) dx.
Solution:
Split the integral into four terms and then evaluate each:∫(ex +
3
x+ 5 sec(x) tan(x) +
7√1− x2
) dx
=
∫ex dx+
∫3
xdx+
∫5 sec(x) tan(x) dx+
∫7√
1− x2dx
= ex + 3 log |x|+ 5 sec(x) + 7 arcsin(x) + C
9
2. (2 × 7% = 14%) The velocity function (in meters per second) is given v(t) = t2 − 3t
for a particle moving along a line.
(a) Find the displacement of the particle during the given time interval: 2 ≤ t ≤ 4.
Solution:
Let d denote the displacement of the particle. Then
d =
4∫2
v(t) dt
=
4∫2
t2 − 3t dt
=t3
3− 3t2
2
∣∣∣∣42
=2
3
(b) Find the distance traveled by the particle during the given time interval: 2 ≤ t ≤
4.
Solution:
Let D denote the distance traveled by the particle. Then
D =
4∫2
|v(t)| dt
=
4∫2
|t2 − 3t| dt
=
3∫2
3t− t2 dt+
4∫3
t2 − 3t dt
=3t2
2− t3
3
∣∣∣∣32
+t3
3− 3t2
2
∣∣∣∣43
= 3
10
3. (2× 7% = 14%)
(a) Evaluate
∫(ln x)5
xdx.
Solution:
Set u = ln x. Then du = 1xdx and∫(lnx)5
xdx =
∫u5 du
=u6
6+ C
=(lnx)6
6+ C
(b) Evaluate
∫xex
2
dx.
Solution:
Set u = ex2. Then du = 2xex
2dx and∫xex
2
dx =
∫1
2du
=u
2+ C
=1
2ex
2
+ C
11
4. (2× 7% = 14%)
(a) Evaluate
∫x5 ln x dx.
Solution:
Set u = ln x and dv = x5dx. Then we have du = 1xdx and v = x6
6. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
lnx d(x6
6) =
x6
6· lnx−
∫x6
6d(lnx)
=x6
6· lnx−
∫x6
6· 1xdx
=x6
6· lnx−
∫x5
6dx
=x6
6· lnx− 1
36x6 + C
(b) Evaluate
∫x2ex dx.
Solution:
Set u = x2 and dv = exdx. Then we have du = 2xdx and v = ex. Then by
integration by parts,∫u dv = uv −
∫v du, we have∫
x2ex dx = x2 · ex −∫
2xex dx
Now use integration by parts again.. Let u = x and dv = exdx. Then du = dx
and v = ex. Then by integration by parts, we have∫xex dx = xex −
∫ex dx
= xex − ex
Therefore,∫x2ex dx = x2ex − 2xex + 2ex + C
12
5. (2× 7% = 14%)
(a) Evaluate
∫ √1− x2 dx.
Solution:
i. Set x = sin θ
ii.√1− x2 =
√1− (sin θ)2 = cos θ and dx = cos θdθ. Recall, sin(2θ) =
2 sin θ cos θ and cos2 θ = 1+cos(2θ)2
iii. ∫ √1− x2 dx =
∫cos2 θ dθ
=
∫1 + cos(2θ)
2dθ
=1
2θ +
sin(2θ)
4+ C
=1
2θ +
sin θ cos θ
2+ C
=1
2arcsinx+
x ·√1− x2
2+ C
13
(b) Evaluate
∫ √1 + x2 dx.
Solution:
i. Set x = tan θ, −π2< θ < π
2
ii.√1 + x2 =
√1 + (tan θ)2 = sec θ and dx = sec2 θdθ
iii. ∫ √1 + x2 dx =
∫sec θ sec2 θ dθ
=
∫sec3 θ dθ
Now do integration by parts. Set u = sec θ and dv = sec2 θdθ. Then∫sec3 θ dθ = sec θ · tan θ −
∫sec θ tan2 θ dθ
= sec θ · tan θ −∫
sec θ(sec2 θ − 1) dθ
= sec θ · tan θ −∫
sec3 θ dθ +
∫sec θ dθ
= sec θ · tan θ −∫
sec3 θ dθ + ln | sec θ + tan θ|
Thus ∫sec3 θ dθ =
1
2
[sec θ · tan θ + ln | sec θ + tan θ|
]=
1
2
[√1 + x2 · x+ ln |
√1 + x2 · x|+ C
]
14
6. (2× 7% = 14%)
(a) Evaluate
∫sin6 x cos3 x dx.
Solution:
Set u = sin x. Then du = cos x dx, and we have∫sin6 x cos3 x dx =
∫sin6(x)(1− sin2 x) d(sinx)
=
∫u6(1− u2) du
=
∫u6 − u8 du
=u7
7− u9
9+ C
=sin7 x
7− sin9 x
9+ C
(b) Evaluate
∫sec3 x tan xdx.
Solution:
Set u = sec x. Then du = sec x tanx dx, and we have∫sec3 x tan dx =
∫sec2 x secx tanx dx
=
∫u2 du
=u3
3+ C
=sec3 x
3+ C
15
7. (2× 7% = 14%) Let R be the region enclosed by the curves y = x and y =√x.
(a) Find the area of the region R.
Solution:
i. Find the common points:
√x = x ⇒ x = x2
⇒ x− x2 = 0
⇒ x(1− x) = 0
⇒ x = 0, 1
ii. Calculate the area:
A(R) =
1∫0
√x− x dx
=2x
32
3− x2
2
∣∣∣∣10
=2
3− 1
2
=1
6
16
(b) Suppose a solid S is created by revolving R around the y-axis. Calculate the
volume of S.
Solution:
Integrate with respect to y:
i. ro = y
ri = y2
ii. Calculate the volume:
V (S) =
1∫0
π(r2o − r2i ) dy
=
1∫0
π(y2 − y4) dy
= π[y3
3− y5
5]
∣∣∣∣10
= π[1
3− 1
5]
17