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Math 113 (Calculus II)
Final Exam KEY
Short Answer. Fill in the blank with the appropriate answer.
- (10 points) a. Let y = f (x) for x ∈ [a, b]. Give the formula for the length of the curve formed by the graph of this function.
∫ (^) b a
√ 1 + (f ′(x))^2 dx
b. Find the vertex of the graph of y = x^2 + 4x. (− 2 , −4)
c. The equation 2x^2 + 3x + y^2 − y = 7 is called a ellipse.
d. If r = f (θ) , θ ∈ [a, b] is an equation of a curve in polar coordinates, give the formula for the area enclosed by this curve.
∫ (^) b a
f 2 (θ) dθ
e. What does the ratio test predict with regard to the convergence of the series ∑^ ∞ n=
n^2 n^4 + 1 ? The ratio test fails.
f. What is the radius of convergence of the series ∑^ ∞ n=
4 n n
xn?
g. The integral
∫ (^) ∞ 0 dx 1+x^2 equals^
π 2
h. Here is an antiderivative: ∫ √ 9 −x 2 x^2 dx.^ Tell what substitution to use in order to find this
antiderivative.x = 3 sin θ
i. The integral
∫ (^1) 0
x^2 /^3 dx equals 3
j. The antiderivative
∫ x sin (x) dx equals (sin x − x cos x + C)
Multiple Choice. In the grid below fill in the correct answer to each question.
2 A B C D E F G H I J
3 A B C D E F G H I J
4 A B C D E F G H I J
5 A B C D E F G H I J
6 A B C D E F G H I J
7 A B C D E F G H I J
8 A B C D E F G H I J
9 A B C D E F G H I J
- Find the area enclosed by the polar curve r = 2 + sin θ for θ ∈ [0, 2 π].
a) 52 π b) 113 π c) 2 π
d) 4 π e) 92 π f) None of the above.
- Identify the equation which goes with the polar graph, 0.
−0. −0.
−1.
0.0 0.4 0.8 1.
a) r = 1 + 2 cos θ b) r = 1 + cos θ c) r = 2 + sin θ
d) r = 2 sin (2θ) e) r = 2 cos (2θ) f) None of the above.
- Which of the following series has as its interval of convergence (−∞, 1]?
a) ∑ x^2 /n! b) ∑ e−nx^2 n^ c) ∑ (^) (x−1)n 2 n
d) ∑ xn−^1 /n^2 e) ∑ (x + 1)n+1/nn^ f) None of these
Free Response. For problems 10 - 17, write your answers in the space provided. Use the back of the page if needed, indicating that fact. Neatly show all work.
- (8 points) A ball of radius 5 has a hole of radius 3 drilled completely through it such that the axis of the hole is a diameter of the ball. What is the volume of what is left? Solution: 2 π ∫ (^5) 3 2 x
25 − x^2 dx = 2563 π You could also do it by cross sections, (washers). 2
∫ (^4) 0 π
25 − y^2
) 2 − 9
) dy = 2563 π
- (8 points) A spherical tank having radius 10 feet is filled with a fluid which weighs 100 pounds per cubic foot. This tank is half full. Find the work in foot pounds needed to pump the fluid out of a hole in the top of the tank. Solution:
∫ (^0) − 10 (10^ −^ y) 100π^ (100^ −^ y^2 )^ dy^ =^ 2750 000 3 π
- (6 points) Determine whether the following series converges and explain your answer.
∑^ ∞ n=
3 n^2 + 2
Solution: Notice that
nlim→∞
1 3 n^2 + 1 n^2
= lim n→∞ n^2 3 n^2 + 2
Since (^) ∞ ∑ n=
n^2 converges (p series, p=2¿1), the original series converges by the limit comparison test. The integral test or comparison test could also be used.
- (6 points) The economic trickle-down effect is based on the idea that injection of money into the economy reaches far beyond the initial receiver of funds. Say that one person receives a dollar and spends 4/5 of it. The person who receives that portion spends 4/5 of it in turn and so on and on “forever.” The total cash flow from the first dollar is the sum of all these passed-on funds including the first dollar. What is that sum?
Solution: Add up ∑∞ n=0(4/5)n^ a geometric series. This equals 5.
- (6 points) Compute the first 4 nonzero terms of the MacLaurin series for the function f (x) = ln(x^2 + 1). Solution: One can use the formula ∑ (^) f n(0) n! xn
to get x^2 − x^4 2
x^6 3
x^8 4 Or, Since students know the series for
ln(1 + x) =
∑^ ∞ n=
(−1)n+ n xn,
they only need to replace x by x^2 to get
ln(1 + x^2 ) =
∑^ ∞ n=
(−1)n+ n x^2 n.
- (8 points) A cube with 2 foot long sides is sitting on the bottom of an aquarium in which the water is 5 feet deep. Find the hydrostatic force on one of the sides of the cube. Use 62. pounds per cubic foot as the weight density of water. Solution: (^) ∫ (^5)
3
5 h · 2 dh = 62. 5 h^2 | 35 = 16 · 62 .5 = 990.
Find the following
(a) (6 points)
∫ (^) π/ 2 0 sin^3 (x) cos^3 (x)^ dx Solution: ∫ (^) π/ 2 0 sin^3 (x) cos^3 (x)^ dx^ =^121 (b) (6 points)
∫ (^) x+ x^2 − 1 dx. Solution: Using partial fractions, we see that x + 5 x^2 − 1
A
x − 1
B
x + 1
or x + 5 = A(x + 1) + B(x − 1). By plugging in ±1 for x, we see that A = 3 and B = −2. Thus, ∫ (^) x + 5 x^2 − 1 dx =
∫ (
x − 1
x + 1 dx = 3 ln |x − 1 | − 2 ln |x + 1|.
(c) (6 points)
∫ (^1) x^2 √x^2 +4 dx Solution: Let x = 2 tan θ. Then, dx = 2 sec^2 θ. The integral becomes ∫ (^) 2 sec (^2) θ tan^2 θ · 2 sec θ
dθ =
∫ csc θ cot θ dθ = − csc θ + C.
