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Math 113 (Calculus II)
Final Exam Form A KEY
Multiple Choice. Fill in the answer to each problem on your scantron. Make sure your name, section and instructor is on your scantron.
- Find the area of the region enclosed by y = x and y = 5x − x^2.
a)
b)
c)
d)
e)
f)
Solution: e)
- Set up the integral representing the volume of the solid obtained by rotating the region bounded by y = x^2 + 1 and y = 3 − x^2 about the x-axis.
a)
− 1
π[(3 − x^2 )^2 − (x^2 + 1)^2 ] dx b)
− 1
2 πx [(3 − x^2 ) − (x^2 + 1)] dx
c)
− 1
π[(x^2 + 1)^2 − (3 − x^2 )^2 ] dx d)
−√ 2
2 πx [(x^2 + 1) − (3 − x^2 )] dx
e)
−√ 2
π[(x^2 + 1)^2 − (3 − x^2 )^2 ] dx f) none of the above
Solution: a)
- Evaluate
∫ π 2
0
sin^5 x cos^3 x dx.
a)
b)
c)
d) −
e) −
f) −
Solution: a)
- Determine whether
0
1 + x^2 dx is convergent or divergent. If convergent, evaluate the integral.
a) divergent b) 0, convergent c) π 4 , convergent
d) π 2 , convergent e) π, convergent f) 2 π, convergent
Solution: d)
- Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve y = e^2 x, 0 ≤ x ≤ 1 about the x-axis.
a)
0
2 πx
1 + e^4 x^ dx b)
0
2 πx
1 + 2e^2 x^ dx c)
0
2 πx
1 + 4e^4 x^ dx
d)
0
2 πe^2 x
1 + e^4 x^ dx e)
0
2 πe^2 x
1 + 2e^2 x^ dx f)
0
2 πe^2 x
1 + 4e^4 x^ dx
Solution: f)
- Find the sum of
∑^ ∞
n=
(−3)n−^1 4 n^
a)
b)
c)
d)
e)
Solution: b)
- What is the interval of convergence for
∑^ ∞
n=
xn 2 n^
a) (−
) b) [−
) c) [−
]
d) (− 2 , 2) e) [− 2 , 2) f) [− 2 , 2]
Solution: d)
- Find the first 4 terms of the power series for f (x) = e−x 2 centered at 0.
a) 1 − x +
x^2 −
x^3 b) 1 − x + x^2 − x^3 c) 1 − x^2 +
x^4 −
x^6
d) 1 − x^2 + x^4 − x^6 e) 1 + x^2 +
x^4 +
x^6 f) 1 + x^2 + x^4 + x^6
Solution: c)
- Which of the following is the graph of r = 3 cos θ?
Free response: Give your answer in the space provided. Answers not placed in this space will be ignored.
- (7 points) Find the volume of the solid obtained by rotating the region bounded by x = 1 + (y − 2)^2 and x = 2 about the x-axis. Solution: 2 = 1 + (y − 2)^2 , (y − 2)^2 = 1, y − 2 = ± 1 so y = 1, 3. Shell Method: ∫ (^3)
1
2 πy(2 − (1 + (y − 2)^2 )) dy = 2π
1
(−y^3 + 4y^2 − 3 y) dy = 2π(− y^4 4
y^3 −
y)|^31
16 π 3
- (7 points) Evaluate
∫ π 2
0
x^2 sin x dx.
Solution:
Form A Integration by parts: u = x^2 , du = 2x dx, dv = sin x dx, v = − cos x. ∫ π 2
0
x^2 sin x dx = −x^2 cos x|π/ 0 2 + 2
x cos x dx
Use integration by parts again: u = x, du = dx, dv = cos x dx, v = sin x. ∫ π 2
0
x^2 sin x dx = (^)
−x^2 cos x|π/ 0 2 + 2
x cos x dx
= 2x sin x|π/ 0 2 − 2
∫ (^) π/ 2
0
sin x dx = 2 π 2
Form B π^2 4
- (7 points) Evaluate
x^3 √ x^2 + 1
dx.
Solution: u = x^2 + 1, du = 2x dx, x^2 = u − 1. ∫ x^3 √ x^2 + 1
dx =
u − 1 √ u du =
u^1 /^2 − u−^1 /^2 du
u^3 /^2 − 2 u^1 /^2 + C =
(1 + x^2 )^3 /^2 −
1 + x^2 + C.
- (7 points) Evaluate
dx x^3 − 2 x^2 + x
Solution: 1 x^3 − 2 x^2 + x
x(x − 1)^2
A
x
B
(x − 1)
C
(x − 1)^2 1 = A(x − 1)^2 + Bx(x − 1) + Cx If x = 0 we see A = 1. If x = 1, we see C = 1. If x = −1, then
1 = 1(−2)^2 + B(−1)(−2) + (−1),
or 2 B = − 2 , B = − 1 Thus, (^) ∫ x^3 √ x^2 + 1
dx =
x dx −
x − 1 dx +
(x − 1)^2 dx
= ln |x| − ln |x − 1 | +
x − 1
+ C.
- (7 points) Show whether
− 1
x^4 dx is convergent or divergent. If convergent, evaluate the integral. Solution: (^) ∫ (^2)
− 1
x^4 dx =
− 1
x^4 dx +
0
x^4 dx ∫ (^2)
0
x^4 dx = lim b→ 0 +
b
x^4 dx = lim b→ 0 +
3 b^3
The integral is divergent.
- (7 points) Find the Taylor series for f (x) = x−^2 centered at a = 1.
Solution: f ′(x) = − 2 x−^3 , f ′′(x) = −2(−3)x−^4 , f ′′′(x) = −2(−3)(−4)x−^5. f (n)(1) = (−1)n(n + 1)! f (x) =
∑^ ∞
n=
(n + 1)(x − 1)n
- (7 points) Find the area enclosed by r = 3 + 2 sin θ.
Solution:
Form A (^) ∫ (^2) π
0
(3 + 2 sin θ)^2 dθ =
∫ (^2) π
0
9 + 12 sin θ + 4 sin^2 θ dθ
∫ (^2) π
0
9 + 12 sin θ + 2 − 2 cos(2θ) dθ =
(11θ − 12 sin θ − cos(2θ))|^20 π = 11π.
Form B (^) ∫ (^2) π
0
(4 + 2 sin θ)^2 dθ =
∫ (^2) π
0
16 + 16 sin θ + 4 sin^2 θ dθ
∫ (^2) π
0
16 + 16 sin θ + 2 − 2 cos(2θ) dθ =
(18θ − 16 sin θ − cos(2θ))|^20 π = 18π.
END OF EXAM
