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