Sample Calculus II Exam #2
No calculators or notes.
Mastery Exam
Compute the following antiderivatives. You need not simplify your answers. You must
answer at least 7 of 10 correctly. No partial credit given.
Z
7
1. (8 points)
3x3 + − 5x−2/3 dx.
x
Solution:
Z
2. (8 points)
Solution:
Z
3 4
x + 7 ln(x) − 15x1/3 + C.
4
sin2 (2x) dx.
2
1 − cos(4x)
dx
2
Z
Z
1
1
1
1
dx −
cos(4x) dx = x − sin(4x) + C.
=
2
2
2
8
Z
sin (2x) dx =
Z
3. (8 points)
arcsin(5x) dx.
Solution: Integration by Parts using u(x) = arcsin(5x) and v 0 (x) = 1 gives
!
Z
Z
5
arcsin(5x) dx = x arcsin(5x) − x p
dx.
1 − (5x)2
Letting u = 1 − (5x)2 one has
Z
p
√
5x du
= x arcsin(5x)− √
= x arcsin(5x)−2 u+C = x arcsin(5x)−2 1 − (5x)2 +C.
u 5x
Calculus 2
Sample Exam #2, Page 2 of 5
Z
4. (8 points)
x2 ln(x) dx.
0
2
Solution: Integration
Z by Parts using u(x) = ln(x) and v (x) = x gives
1 3
1 1 3
1
1
x ln(x) −
x dx = x3 ln(x) − x3 + C.
3
x 3
3
9
Z
5. (8 points)
x2
Z
Solution:
3
dx.
− 2x − 3
3
dx =
2
x − 2x − 3
Z
3
dx =
(x + 1)(x − 3)
Z
A
B
+
dx. Solvx+1 x−3
ing
3 = A(x − 3) + B(x + 1) = (A + B)x + (B − 3A)
3
3
and B = − . Continuing, one gets
4
4
Z
Z
3
1
3
1
3
3
dx −
dx = ln(|x + 1|) − ln(|x − 3|) + C.
4
x+1
4
x−3
4
4
one obtains A =
Z
6. (8 points)
x3 e2x dx.
Solution: Using Integration by Parts with u(x) = x3 and v 0 (x) = e2x one gets
Z
Z
Z
1
1
3
3 2x
3 1 2x
e
− (3x2 ) e2x dx = x3 e2x −
x2 e2x dx.
x e dx = x
2
2
2
2
Using Integration by Parts again with u(x) = x2 and v 0 (x) = e2x one gets
Z
Z
1 3 2x 3 2 1 2x
1 2x
1 3 2x 3
2 2x
x e −
x
e
− (2x) e dx
x e dx = x e −
2
2
2
2
2
2
Z
1
3
3
= x3 e2x − x2 e2x +
xe2x dx.
2
4
2
Using Integration by Parts yet again with u(x) = x and v 0 (x) = e2x one gets
Z Z
1 3 2x 3 2 2x 3
1 3 2x 3 2 2x 3
1 2x
1 2x
2x
x e − x e +
xe dx = x e − x e +
x
e
− 1
e
dx
2
4
2
2
4
2
2
2
1
3
3
3
= x3 e2x − x2 e2x + xe2x − e2x + C.
2
4
4
8
Calculus 2
Sample Exam #2, Page 3 of 5
Z
7. (8 points)
5x
dx.
2 + 8x2
Solution: Letting u = 2 + 8x2 one has
Z
Z
5x
1
5
5
5
dx
=
du =
ln(|u|) + C =
ln(2 + 8x2 ) + C.
2
2 + 8x
16
u
16
16
Z
8. (8 points)
2−x
dx.
x2 + 4
Solution: Letting u = x/2 and w = x2 + 4 one has
Z
Z
Z
1
x
2−x
dx = 2
dx −
dx
2
2
2
x +4
x +4
x +4
Z
Z
1
x
1
dx −
dx
=
2
2
2
(x/2) + 1
x +4
Z
Z
1
1
1
=
du −
dw
2
u +1
2
w
1
= arctan(u) − ln(|w|) + C
2
1
= arctan(x/2) − ln(x2 + 4) + C.
2
Z
9. (8 points)
cos3 (x) dx.
Solution: Letting u = sin(x) one has
Z
Z
cos3 (x) dx = (1 − sin2 (x)) cos(x) dx
Z
= 1 − u2 du
1
= u − u3 + C
3
1
= sin(x) − sin3 (x) + C.
3
Calculus 2
Sample Exam #2, Page 4 of 5
Z
10. (8 points)
sec2 (x)e2 tan(x) dx.
Solution: Letting u = 2 tan(x) one has
Z
Z
1
1
1
2
2 tan(x)
sec (x)e
dx =
eu du = eu + C = e2 tan(x) + C.
2
2
2
Rest of Exam
11. (20 points) Find the area bounded by 2 + sin(θ).
Solution:
1
2
Z
2π
(2 + sin(θ))2 dθ =
0
=
=
=
=
=
Z
1 2π
4 + 4 sin(θ) + sin2 (θ) dθ
2 0
Z
1 2π
1 − cos(2θ)
4 + 4 sin(θ) +
dθ
2 0
2
Z
1 2π 9
1
+ 4 sin(θ) − cos(2θ) dθ
2 0 2
2
2π
1 9
1
θ − 4 cos(θ) − sin(2θ)
2 2
4
0
1
[(9π − 4 − 0) − (0 − 4 − 0)]
2
9π
.
2
EXTRA CREDIT
12. (20 points) Find the solution to y ln(x) − xy 0 = 0 so that y(1) = 2.
Calculus 2
Sample Exam #2, Page 5 of 5
Solution:
y ln(x) − xy 0 = 0
xy 0 = y ln(x)
1 0 ln(x)
y =
y
x
Z
Z
1
ln(x)
dy =
dx
y
x
ln(y) = ln2 (x)/2 + C
y = eln
2
(x)/2+C
= eC eln
2
(x)/2
= Keln
2
(x)/2
Now plugging in (1, 2) one has 2 = K so the answer is y = 2eln
2
(x)/2 .