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Sample examinations Calculus II (201-NYB-05) Autumn 2009
1. Given f (x) = 2x arctan 2x − 12 log(1 + 4x^2 ) + arcsin 23.
a. Find f ′(x) and simplify your answer. b. Evaluate f ′
` 1
2
2. Evaluate each of the following limits, using ∞ and −∞ when appropriate.
a. (^) xlim→∞
x
« 2 x b. lim x→ 0 +
`
e−^2 /x^ log x
c. lim x→ 0
e^6 x^ − 6 x − 1 x^2
3. Evaluate each of the following integrals.
a.
Z
2 x + 1 √ x − 3
dx b.
Z
9 x − 1 (x − 3)(x^2 + 4)
dx
c.
Z
x arcsec x dx d.
Z 1
4 π 0
sin^3 2 x cos^4 2 x dx
e.
Z 1
2 0
arcsin x √ 1 − x^2
dx f.
Z
e^3 x^ sin x dx g.
Z
dx √ 9 x^2 − 16
4. Evaluate each of the following improper integrals.
a.
Z 2
3
√ 3
1
dx x
x^2 − 1
b.
Z ∞
4
dx x log x
5. Solve the differential equation
2 y dy dx
= y^2 − 1; y(0) = 2.
6. Sketch the region enclosed by y = 2/x − 1 and y = 2 − x, and find its area. 7. Let R be the region enclosed by y = sin x^2 and the x-axis on [ 0,
π ]. a. Find the volume of the solid obtained by revolving R about the y-axis.
b. Set up, but do not evaluate, an integral that represents the volume of the solid obtained by revolving R about the x-axis.
8. Determine whether the sequence converges or diverges; if it converges, find its limit.
a.
1 + cos 12 (2n + 1)π
b.
(−1)n^
3 n^2 + n − 2 n^2
ff
9. Determine whether each statement is true or false. Justify each answer, with a proof or a counterexample, as appropriate). a. If lim|an| 6 = 0 then lim an 6 = 0.
b. If lim an = 0 then
X^ ∞
n=
sin an converges.
10. Find the sum of the series X^ ∞
n=
3 n+1^ + 2n 4 n^
11. Classify each of the following series as convergent or divergent, and justify your answers.
a.
X^ ∞
n=
n
n^2
b.
X^ ∞
n=
2 n − e n^2
« 2 n
c.
X^ ∞
n=
n^3 − 1 n^2 + 1
d.
X^ ∞
n=
(n!)^2 (2n)!
12. Classify each of the following series as absolutely convergent, conditionally convergent or divergent. Justify your answers.
a.
X^ ∞
n=
(−1)n^ arctan n n^3 + 1
b.
X^ ∞
n=
(−1)n^ cos
n
c.
X^ ∞
n=
(−1)n^ n n^2 + 1
13. Determine the radius and interval of convergence of the series X^ ∞
n=
3 n−^1 (x + 1)n n
n + 1
14. Let f (x) = log(1 + x). a. Write the first five non-zero terms of the Maclaurin series of f. b. Find a formula for the kth^ term of the Maclaurin series, and write the series using sigma notation.
Solution outlines
1. a. f ′(x) = 2 arctan 2x + 4 x 1 + 4x^2
4 x 1 + 4x^2
b. f ′
` 1
2
= 2 arctan 1 = 12 π.
2. a. One application of l’Hˆopital’s rule gives
lim x→∞
2 x log(1 + 4/x)
= 2 lim t→ 0 +
log(1 + 4t) t
= 8 lim t→ 0 +
1 + 4t
where t = 1/x, so the limit in question is equal to e^8.
b. One application of l’Hˆopital’s rule, after letting t = 1/x, gives
lim x→ 0 +
`
e−^2 /x^ log x
= − lim t→∞
log t e^2 t^
= − lim t→∞
2 te^2 t^
c. Two applications of l’Hˆopital’s rule gives
lim x→ 0
e^6 x^ − 6 x − 1 x^2
= 3 lim x→ 0
e^6 x^ − 1 x
= 18 lim x→ 0
e^6 x^ = 18.
3. a. Repeated partial integration (integrating the the fractional power and differ- entiating the polynomial) gives Z 2 x + 1 √ x − 3
dx = 2(2x + 1)
x − 3 − 83 (x − 3)^3 /^2 + C
= 23 (2x + 15)
x − 3 + C.
b. Resolving the integrand into partial fractions and then integrating term by term yields Z 9 x − 1 (x − 3)(x^2 + 4)
dx =
Z
x − 3
2 x − 3 x^2 + 4
ff dx
= log
(x − 3)^2 x^2 + 4
c. Partial integration gives Z x arcsec x dx = 12 x^2 arcsec x − (^12)
Z
x √ x^2 − 1
dx
= 12 x^2 arcsec x − (^12)
p x^2 − 1 + C. d. Changing the variable of integration to t = cos(2x) gives Z 1 4 π 0
sin^3 2 x cos^4 2 x dx = (^12)
Z 1
0
t^4 (1 − t^2 ) dt = 701 t^5 (7 − 5 t^2 )
1
0
e. Changing the variable of integration to t = arcsin x gives Z 1 2 0
arcsin x √ 1 − x^2
dx =
Z 1
6 π 0
t dt = 12 t^2
1 6 π 0
= 721 π^2.
f. Repeated partial integration (integrating the trigonometric function and differ- entiating the exponential function) gives Z e^3 x^ sin x dx = −e^3 x^ cos x + 3e^3 x^ sin x − 9
Z
e^3 x^ sin x dx,
and therefore Z e^3 x^ sin x dx = 101 e^3 x(3 sin x − cos x) + C.
g. Applying a standard integral formula gives Z dx √ 9 x^2 − 16
= 13 log| 3 x +
p 9 x^2 − 16 | + C.
Sample examinations (solution outlines) Calculus II (201-NYB-05) Autumn 2009
4. a. A standard integral formula gives Z 2 3
√ 3
1
dx x
x^2 − 1
= arcsec (^23)
3 − arcsec 1 = 16 π,
since arcsec is continuous on [ 1, (^23)
3 ].
b. One has (^) Z ∞
4
dx x log x
= lim t→∞ log log t − log log 4 = ∞
(so the integral diverges).
5. Separating variables and integrating gives Z 2 y y^2 − 1
dy =
Z
dx, or log|y^2 − 1 | = x + C, i.e. , y^2 = Aex^ + 1,
where A = ±eC^. Now y(0) = 2 gives A = 3 and y > 1 , and so y =
3 ex^ + 1.
6. Below is a sketch of the region in question.
y =
x y = 2 − x −^1
x
y
The curves meet where 2 − x = 2/x − 1 , or 0 = x^2 − 3 x + 2 = (x − 1)(x − 2), i.e. , where x = 1 or x = 2. On ( 1, 2 ) the line is above the hyperbola, so the area of the region in question is Z (^2)
1
(2 − x) − (2/x − 1)
dx =
Z 2
1
(3 − x − 2 /x) dx
= (3x − 12 x^2 − 2 log x)
2
1 = 32 − 2 log 2.
7. a. The solid obtained by revolving R about the y-axis can be decomposed into cylindrical shells of radius x and height sin x^2 , for 0 6 x 6
π, so its volume is equal to
2 π
Z √π
0
sin x^2 dx = −π cos x^2
√π
0
= 2π.
b. The solid obtained by revolving R about the x-axis can be decomposed into disks of radius sin x^2 , for 0 6 x 6
π, so its volume is represented by the integral
π
Z √π
0
sin^2 x^2 dx.
8. a. Since cos 12 (2n + 1)π = 0 for every natural number n, the given sequence converges to 1 (each of its terms is equal to 1 ). b. Let an denote the general term of the given sequence. Since lim a 2 n = 3 and lim a 2 n+1 = − 3 , it follows that {an} has no limit. 9. a. This statement is true. For if lim|an| 6 = 0, there is a positive real number ε 0 such that for any natural number N there is a natural number n > N for which ||an| − 0 | > ε 0 , i.e. , |an − 0 | > ε 0 , which means that lim an 6 = 0 by definition.
b. This statement is false. For example,
an = arcsin
n
→ 0 , and
X^ ∞
n=
sin an =
X^ ∞
n=
n
is the harmonic series, which diverges.
10. The given series is the sum of two geometric series, and in fact X^ ∞
n=
3 n+1^ + 2n 4 n^
X^ ∞
n=
` 3
4
´n
X^ ∞
n=
` 1
2
´n
11. a. The given series diverges because it is the difference of a divergent (p = 1) and a convergent (p = 2) p-series. (Alternatively, the given series diverges with the harmonic series because its terms are larger than 34 n−^1 if n > 2 .) b. Since
lim n
s˛ ˛ ˛ ˛
2 n − e n^2
2 n = lim (2 − e/n)^2 n^2
the series converges by the Root Test. c. If n > 1 then n^3 − 1 > 14 n^3 , n^2 + 1 < 2 n^2 , and therefore √ n^3 − 1 n^2 + 1
> 14 n−^1 /^2 ;
so the series in question diverges with
P
n−^1 /^2 (p = 12 ) by the Comparison Test. (Alternatively, the Limit Comparison Test could be used.) d. Since (n!)^2 (2n)!
2 n^
n(n − 1) · · · 2 · 1 (2n − 1)(2n − 3) · · · 3 · 1
2 n^
the given series converges by the Comparison Test. (Alternatively, the Ratio test could be used.)
12. a. Since 0 <
arctan n n^3 + 1
< 12 πn−^3 , for n > 0 , the series in question is absolutely convergent by the Comparison Test. b. Since lim cos (^) n^1 = 1, the series in question diverges by the vanishing criterion. c. Let an = n/(n^2 + 1). If n > 1 then an > 12 n−^1 , and so
P
(−1)n^ an is not absolutely convergent by the Comparison Test. However, an > 0 , {an} is decreasing since d dx
x x^2 + 1
ff
1 − x^2 (x^2 + 1)^2
< 0 if x > 1 ,
and lim an = lim
n
1 + 1/n^2
so
P
(−1)n^ an converges by the Alternating Series Test. Therefore,
P
(−1)n^ an is conditionally convergent.
13. Let un denote the general term of the series in question. Then
lim
un+ un
˛ =^
3 |x + 1| p (1 + 1/n)(1 + 2/n)
= 3|x + 1|,
so
P
un is absolutely convergent if |x + 1| < 13 , i.e. , − 43 < x < − 23 , by the Ratio Test. This means that the radius of convergence of
P
un is 13. If x = − (^43) or x = − 23 then |un| =
n
n + 1
< n−^3 /^2 ,
and so
P
un is (absolutely) convergent by the Comparison Test. Therefore, the interval of convergence of
P
un is
14. We have
f (x) = log(1 + x) =
Z (^) x
0
dt 1 + t
X^ ∞
k=
(−1)k
Z (^) x
0
tk^ dt
X^ ∞
k=
(−1)k−^1 k
xk^ (b.)
= x − 12 x^2 + 13 x^3 − 14 x^4 + 15 x^5 − · · · (a.)
