Calculus II (MAC2312-02)
Test 1 (2015/05/28)
Name (PRINT):
Please show your work. An answer with no work receives no credit.
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Page
Points Score
2
17
3
24
4
30
5
14
6
15
Total:
100
Page 1 of 6
Calculus II (MAC2312-02)
Page 2 of 6
1. In each part, find the indefinite or definite integral given.
Z
(a) (7 points)
x sin(2x) dx
Solution: Put u := Rx and dv := Rsin(2x)dx. Then, du = dx and v = − 21 cos(2x).
Integrating by parts ( udv = uv − vdu) we get
Z
Z
1
1
x sin(2x) dx = − x cos(2x) − − cos(2x) dx
2
2
1
1
= − x cos(2x) + sin(2x) + C.
2
4
Z
(b) (10 points)
sin(2x) sin(x) dx
Solution: Using the identity sin(2x) ≡ 2 sin(x) cos(x) we have
Z
Z
sin(2x) sin(x) dx = 2 sin(x) cos(x) sin(x) dx
Z
= 2 sin2 (x) cos(x) dx,
which by temporarily putting u := sin(x), and noting that du = cos(x)dx, we can
integrate easily:
Z
= 2 u2 du
= 2u3 /3 + C
= 2 sin3 (x)/3 + C.
Calculus II (MAC2312-02)
π
12
Z
(c) (12 points)
Page 3 of 6
sin3 (3x) cos2 (3x) dx
0
Solution:
π
12
Z
3
Z
2
π
12
sin (3x) cos (3x) dx =
0
sin2 (3x) cos2 (3x) sin(3x) dx
0
Z
π
12
=
(1 − cos2 (3x)) cos2 (3x) sin(3x) dx
0
Z
π
12
=
(cos2 (3x) − cos4 (3x)) sin(3x) dx,
0
which we can integrate easily by putting u := cos(3x), which takes
√ x = 0 to u =
cos(3(0)) = 1 and x = π/12 to u = cos(3(π/12)) = cos(π/4) = 1/ 2, and noting that
du = −3 sin(3x)dx:
Z √
1 1/ 2 2
(u − u4 ) du
=−
3 1
√
1
1/ 2
= − (u3 /3 − u5 /5)|1
3
1
1
1
√ − √ − (1/3 − 1/5)
=−
3 6 2 20 2
1 7
= − ( √ − 2/15) ≈ 0.0169458.
3 60 2
π
6
Z
(d) (12 points)
tan3 (2x) sec(2x) dx
0
Solution:
Z
π
6
Z
3
π
6
tan (2x) sec(2x) dx =
0
tan2 (2x) sec(2x) tan(2x) dx
0
Z
=
π
6
(sec2 (2x) − 1) sec(2x) tan(2x) dx,
0
which we can integrate easily by putting u := sec(2x), which takes x = 0 to u =
sec(2(0)) = 1/ cos(2(0)) = 1 and x = π/6 to u = sec(2π/6) = 1/ cos(2π/6) =
1/ cos(π/3) = 1/(1/2) = 2, and noting that du = 2 sec(2x) tan(2x)dx
Z
1 2 2
=
(u − 1) du
2 1
1
= (u3 /3 − u)|21
2
1
= (8/3 − 2 − (1/3 − 1))
2
= 2/3.
Calculus II (MAC2312-02)
Page 4 of 6
Z
dt
√
t2 − 10t + 34
(Hint: Complete the square.)
(e) (15 points)
Solution:
Z
√
dt
=
t2 − 10t + 34
Z
dt
p
(t − 5)2 + 9
,
which by temprarily putting u := t − 5, and noting that du = dt, is converted to
Z
du
√
=
,
u2 + 9
which by temporarily putting u =: 3 tan(θ), −π/2 < θ < π/2, and noting that du =
3 sec2 (θ)dθ, we can integrate easily:
Z
Z
Z
3 sec2 (θ)dθ
3 sec2 (θ)dθ
sec2 (θ)dθ
p
p
=
=
=
,
| sec(θ)|
9 tan2 (θ) + 9
3 tan2 (θ) + 1
and since sec(θ) := 1/ cos(θ) > 0 when −π/2 < θ < π/2 we can drop the absolute value,
Z
Z
sec2 (θ)dθ
= sec(θ) dθ = ln | sec(θ) + tan(θ)| + C1
=
sec(θ)
√
u2 + 9 u
= ln |
+ | + C1
3
√ 3
2
= ln | u + 9 + u| + C1 − ln(3)
p
= ln | (t − 5)2 + 9 + t − 5| + C.
Z
dx
1 − cos(x)
(Hint: Use a “half-angle” trigonometric identity.)
(f) (15 points)
Solution: Using the identity 1 − cos(x) ≡ 2 sin2 (x/2) we get
Z
Z
dx
dx
=
1 − cos(x)
2 sin2 (x/2)
Z
1
=
csc2 (x/2) dx
2
= − cot(x/2) + C.
Calculus II (MAC2312-02)
Z
(g) (14 points)
Page 5 of 6
3x2 − 18x + 34
dx
(2x + 1)(x − 2)2
Solution: Since the degree of the numerator (2) is less than that of the denominator
(3), the integrand is a “proper” rational function and we can write it as the sum of
partial fractions:
B
C
A
3x2 − 18x + 34
+
+
≡
.
(2x + 1)(x − 2)2
2x + 1 x − 2 (x − 2)2
(*)
Multiplying (*) through by 2x + 1 and taking the limit at x = −1/2 we get
(2x + 1)(3x2 − 18x + 34)
A(2x + 1) B(2x + 1) C(2x + 1)
+
+
lim
= lim
x→−1/2
x→−1/2
(2x + 1)(x − 2)2
2x + 1
x−2
(x − 2)2
3x2 − 18x + 34
lim
= A.
x→−1/2
(x − 2)2
2
−18(−1/2)+34
So, A = 3(−1/2)(−1/2−2)
= 7. To find C, we multiply (*) through by (x − 2)2 and
2
take the limit at x = 2:
A(x − 2)2 B(x − 2)2 C(x − 2)2
(x − 2)2 (3x2 − 18x + 34)
= lim
+
+
lim
x→2
x→2
(2x + 1)(x − 2)2
2x + 1
x−2
(x − 2)2
3x2 − 18x + 34
lim
= C.
x→2
2x + 1
So, C =
3(2)2 −18(2)+34
2(2)+1
= 2. To find B we evaluate (*) at x = 0, e.g.:
A
B
C
34
= +
+
4
1
−2
4
2
B
+ ,
=7+
−2 4
which gives B = −2. Or, alternatively, we can multiply (*) by 2x and take the limit as
x → ∞:
2x(3x2 − 18x + 34)
2xA
2xB
2xC
lim
= lim
+
+
x→∞ (2x + 1)(x − 2)2
x→∞
2x + 1 x − 2 (x − 2)2
2x(3x2 )
lim
= A + 2B,
x→∞ 2x(x2 )
which gives 3 = 7 + 2B, i.e., B = −2. So,
Z
Z 3x2 − 18x + 34
7
−2
2
dx =
+
+
dx
(2x + 1)(x − 2)2
2x + 1 x − 2 (x − 2)2
7
2
= ln |2x + 1| − 2 ln |x − 2| −
+ C.
2
x−2
Calculus II (MAC2312-02)
Page 6 of 6
2. (15 points) Find the volume generated by rotating the region bounded by the curves y = ex ,
y = e−x , and x = 1 about the y-axis. Make sure to sketch the curves as well as your “area
element” whose position and size (width, height, etc.) are clearly labeled.
Solution:
Z
1
x(ex − e−x ) dx
V = 2π
0
Z
= 2π
1
x(ex − e−x ) dx,
0
which by putting u := x and dv := (ex − e−x )dx, and noting that du = dx and v = ex + e−x ,
we can integrate by parts:
Z 1
x
−x 1
x
−x
= 2π x(e + e )|0 −
(e + e ) dx
0
= 2π e1 + e−1 − (ex − e−x )|10
= 2π e1 + e−1 − (e1 − e−1 )
= 4π/e.