Final Examination
201-NYB-05
December 2016
1. Evaluate the following integrals.
Z
cos3 x
√
(a)
dx
sin x
Z
x arcsin(x2 )
√
dx
(b)
1 − x4
Z
x+6
(c)
dx
2
x(x + 2x + 3)
Z
(d)
sin(ln x) dx
Z
1
√
dx
(e)
x3 x2 − 4
Z r
3+x
dx
(f)
3−x
2. Evaluate the following limits.
ln(sin x)
(a) lim
+
x→0 ln(sin(2x))
(b) lim (tan x)2x−π
x→π/2−
(c) lim
x→0
1
3
− 3x
x e −1
3. Evaluate each improper integral or show that it diverges.
Z ∞
(a)
(−xe−x ) dx
0
Z 2
1
(b)
dx
2/3
0 (x − 1)
4. Give the solution of the differential equation cos x
p
dy
= sin x y 2 + 4 which satisfies y = 0 if x = 0.
dx
5. Find the area of the region bounded by y1 = x3 + x2 + 3x + 1 and y2 = x3 + x + 4.
6. Let R be the region bounded by x = 0, f (x) = 1 + x and g(x) = x3 + x. Set up, but do not evaluate, an
integral which represents the volume obtained by revolving R about:
(a) the x-axis;
(b) the line x = 3.
7. Find the arc length function for the curve x = 41 y 2 − 12 ln y, taking
1
4, 1
as the starting point.
8. Determine with justification, whether the sequence {an } converges or diverges. If a sequence converges, find
its limit.
3n + 1 n
(a) an =
3n − 1
(b) an =
n3 (2n)!
(2n + 2)!
9. For the telescoping series
∞ X
1
n=1
1
−
,
n n+2
(a) give a formula for sn , the sum of the first n terms of the series, and (b) find the sum of the series.
Final Examination
201-NYB-05
December 2016
10. Determine whether each series is convergent or divergent. Justify your answers.
∞ √ 2
X
n +3
(a)
3n2 + 7
(b)
n=0
∞
X
ln n
n3/2
n=1
11. Determine whether each series is absolutely convergent, conditionally convergent or divergent. Justify your
answers by displaying proper solutions.
(a)
(b)
(c)
∞
X
(−1)n
n=1
∞
X
n=1
∞
X
n!
1 · 3 · 5 · 7 · · · · · (2n + 1)
(−1)n nn
3n+1
cos(nπ)
√
5n + 3
n=1
12. Find the radius and interval of convergence of the power series
∞
X
(−1)n (x + 1)n
√
.
5n n
n=1
1
, find the Taylor series around x = 1. Write the first four terms of the series
2+x
explicitly, and express the series using appropriate sigma notation.
13. For the function f (x) =
Solutions
√
1. a. If y = sin x then 2y dy = cos x dx and cos2 x = 1 − y4 , so
Z
Z
√
cos3 x
√
dx = 2 (1 − y4 ) dy = 2 y − 15 y5 + a = 25 (5 − sin2 x) sin x + a.
sin x
b. If y = arcsin(x2 ) then dy = 2x(1 − x4 )−1/2 dx, so
Z
Z
2
x arcsin(x2 )
√
dx = 12 y dy = 41 arcsin(x2 ) + b.
4
1−x
c. The resolution into partial fractions of the integrand is
1√
2 2(x + 1)
+ c.
d. Repeated partial integration and absorbtion gives
Z
Z
sin(log x) dx = x sin(log x) − x cos(log x) − sin(log x) dx
= 21 x (sin(log x) − cos(log x)) + d.
√
e. If y = x2 − 4, then y2 = x2 − 4, so y dy = x dx, or dx/(xy) = dy/(x2 ), and thus
y
dy
2y dx
dy
2(x2 − 4)
8 dx
dy
d 2 = 2 −
= 2 −
dx = 3 − 2
.
x
x
x3
x
x3 y
x y
y +4
Therefore,
√
Z
Z
p
y
dy
dx
x2 − 4
1
1
√
=
+
=
+ 16
arctan 21 x2 − 4 + E.
8
8x2
y2 + 4
8x2
x3 x2 − 4
√
f. Multiplying and dividing by 3 + x omits −3 from the domain of the integrand, and gives
Z
Z
p
3+x
√
dx = 3 arcsin 13 x − dy = 3 arcsin 31 x − 9 − x2 + f,
2
9−x
√
where in the second term y = 9 − x2 , so that dy = −(x/y) dx.
2. a. Revising the expression in the limit and using the fact that lim sinϑ ϑ = 1 gives
ϑ→0
1 + log sinx x /(log x)
log(sin x)
lim
= lim
= 1.
x→0+ log(sin 2x)
x→0+ 1 + log 2 sin 2x /(log x)
2x
b. If y = 12 π − x and z = 1/y then
lim (tan x)2x−π = lim (cot y)−2y = lim
x→ 21 π−
y→0+
y→0+
sin y
y cos y
2y
· lim e−2(log z)/z = 1
z→∞
by elementary properties of the logarithm (the definition of log z implies that 0 < log z < z if
z > 1, which immediately gives 0 < (log z)a /zb < (2a/b)a z−b/2 for z > 1 and a, b > 0).
c. Combining terms and using the Maclaurin expansion of the exponential function gives
lim
x→0
9
2
+ 29 x + 27
e3x − 1 − 3x
8 x + ···
= lim 2
= 23 .
x→0 3 + 9 x + 9 x2 + · · ·
x(e3x − 1)
2
2
(Alternatively, two applications of l’Hôpital’s rule could be used.)
∞
t+1
−
1
= −1,
= lim
t→∞
et
0
via basic properties of the exponential function (the inequality proved in Part c of Question 2
1
c
immediately gives 0 < ya /eby < (2a/(bc))a/c e− 2 by for a, b, c, y > 0).
b. Integrating by inspection gives
s
2
Z2
√
√
dx
3
3
+
lim
3
=
lim
3
x
−
1
x
−
1
= 6.
(x − 1)2/3
t→1+
s→1−
0
0
where the coefficient over x is found by inspection (covering and evaluating) and the coefficients
over x2 + 2x + 3 are obtained by comparing the quadratic and constant terms of the numerator.
The integral of the second partial fraction is
Z
√
√
1
2x + 2
+
dx = log(x2 + 2x + 3) + 21 2 arctan 12 2(x + 1) + c,
x2 + 2x + 3
(x + 1)2 + 2
√
x+6
x2
dx = log 2
− 21 2 arctan
x(x2 + 2x + 3)
x + 2x + 3
t→∞
0
c
x+6
2
2x + 3
= − 2
,
x
x(x2 + 2x + 3)
x + 2x + 3
and therefore
Z
3. a. Partial integration gives
∞
Z
(−xe−x ) dx = lim (x + 1)e−x
t
4. For − 21 π < x < 21 π, the equation in question is equivalent to
p
1
dy
p
= tan x,
or
log y + y2 + 4 = log(sec x) + C,
2
dx
y +4
p
2 + 4 = A sec x (in which A = eC ). The initial condition y = 0
which is equivalent to y + yp
if x = 0 gives A = 2, so y + y2 + 4 = 2 sec x. To express y as a function of x, observe that
subtracting y and squaring gives 4 sec2 x − 4y sec x = 4, or y = sec x − cos x = sin x tan x.
5. If y = x3 + x + 4 and y = x3 + x2 + 3x + 1 then y − y = −x2 − 2x + 3 = (3 + x)(1 − x),
which is positive if −3 < x < 1 and vanishes if x is −3 or 1. So the area of the region enclosed by
the curves is
Z1
Z1
1
32
(y − y) dx = (−x2 − 2x + 3) dx = − 31 x3 − x2 + 3x = − 28
3 + 8 + 12 = 3 .
−3
−3
−3
x3
1 − x3 ,
6. If y = x + 1 and y =
+ x then y − y =
which is positive if 0 < x < 1 and vanishes
if x = 1. The solid obtained by revolving R about the x axis consists of annuli of inner radius
x3 + x and outer radius x + 1, for 0 6 x 6 1, so its volume is equal to
π
Z1
0
(x + 1)2 − (x3 + x)2 dx.
The solid obtained by revolving R about the line defined by x = 3 consists of concentric cylindrical
shells of radius 3 − x and radius 1 − x3 , for 0 6 x 6 1, so its volume is equal to
Z1
2π (3 − x)(1 − x3 ) dx.
0
7. If x = 41 y2 − 12 log y, then
dx 2
2
2
1+
= 1 + 12 y − 12 y−1 = 14 y2 + 21 + 14 y−2 = 12 y + 12 y−1 ,
dy
and hence
s
Zy
Zy dx 2
1+
dη = 12
η + η−1 dη = 14 (y2 − 1) + 21 log y,
dη
1
1
which is the length of the curve between 41 , 1 and (x, y) if y > 1, and is −1 times the length of
1
the curve between 4 , 1 and (x, y) if 0 < y < 1.
8. a. Since
lim (1 + t)1/t = e,
t→0
and
an =
3n + 1
3n − 1
it follows that lim an = e2/3 .
b. Since
n
=
1+
lim
n→∞
2n
=
3n − 1
2
3n − 1
3n−1 ·
2
12. If x 6= −1 and
2
3
2n
3n−1
αn =
,
n→∞
n3 (2n)!
n3
n
=
=
,
(2n + 2)!
(2n + 2)(2n + 1)
2(1 + 1/n)(2 + 1/n)
it follows that the sequence {an } diverges to ∞.
an =
(−1)n (x + 1)n
,
n√
5
n
P
so the ratio test implies that
αn is absolutely convergent if |x + 1| < 5, i.e., −6 < x < 4,
P
P −1/2
and divergent if x < −6 or x > 4. If x = −6 then
is a divergent p-series
αn =
n
P
P
1
n
−1/2
(p = 2 6 1), and if x = 4 then αn = (−1) n
, which is convergent by the alternating
−1/2 > (n+1)−1/2 if n > 1, and lim n−1/2 = 0). Therefore, the radius of convergence
series
P test (n
P
of αn is 5, and the interval of convergence of αn is ( −6, 4 ].
13. From the expansion 1/(1 + t) =
−
=
1
1
−
.
n+1
n+2
Hence, the sum of the series is lim (a1 + a2 + · · · + an ) = 23 .
n→∞
10. a. If n > 1 then
√
√ 2
n
n2 + 3
1
> 0,
an =
>
=
2+7
10n
3nP
3n2 + 7n2
so the comparison test implies that an diverges with the harmonic series.
d
(x−1/4 log x) = 14 x−5/4 (4 − log x) is positive if 0 < x < e4 and negative if x > e4 ,
b. Since dx
it follows that
log n
log n
1
1
4
0 6 an = 3/2 = 1/4 · 5/4 < · 5/4 ,
e n
n
n
n
P
P
for n > 1. Therefore, the comparison test implies that an converges with the p-series n−5/4 .
P
11. a. Since 2−n is a convergent geometric series, and
1 n
n!
1 2 3
n
0 < an =
= 1 · · · ···
<
,
1 · 3 · 5 · 7 · · · (2n + 1)
3 5 7
2n + 1
2
P
n
for n > 1, the comparison test implies that (−1) an is absolutely convergent.
b. If n > 3 then
nn
1 n n n
n
n
·
·
···
>
> 0,
so
lim an = ∞.
an = n+1 = ·
n→∞
3 3 3 3
3
9
3
P
n
Hence, the vanishing condition implies that (−1) an is divergent.
c. If n > 1 then
√
1
1
1
2
an = √
> √
=
· √ > 0,
4
n
5n + 3
5n + 3n
P
P
so the comparison test implies that an diverges with the p-series n−1/2 . On the other hand,
1
1
an = √
> √
= an+1
5n + 3
5n + 8
P
if n > 1, and lim anP= 0, so the alternating
series test implies that (−1)n an is convergent.
P
n
Therefore, the series cos(nπ)an = (−1) an is conditionally convergent.
(−1)k tk (a geometric series), it follows that
∞
1
1
1
1 X (−1)k
= ·
=
(x − 1)k
1
2+x
3 1 + 3 (x − 1)
3
3k
k=0
1
1
1
1
−
and
An =
+
,
an =
n
n+2
n
n+1
then an = An − An+1 for n > 1, and the sum of the first n terms of the series is
3
2
∞
P
k=0
9. If
a1 + a2 + · · · + an = A1 − An+1 =
α
1
|x + 1| = 51 |x + 1|,
lim n+1 = lim p
αn
5 1 + 1/n
then
=
which is valid if
1
3 |x − 1|
∞
X
(−1)k
(x − 1)k
3k+1
k=0
2
1
1
1
3 − 9 (x − 1) + 27 (x − 1)
−
< 1, or equivalently, −2 < x < 4.
3
1
81 (x − 1)
+ ···,