1. No category

Calculus I–Sample Final exam
Solutions
[1] Compute the following integrals:
Z 4
dx
(a)
2
2 x ln (x)
Solution. Substituting u = ln x,
ln 4
Z 4
Z ln 4
dx
1
1
−2
−1 −
.
=
u
du
=
−u
=
2
ln 2 ln 4
2 x ln (x)
ln 2
ln 2
Taking common denominator, using properties of the logarithm, one can get the
answer in the more compact form
Z 4
1
dx
=
.
2
2 ln 2
2 x ln (x)
Z
2
(b)
sin(2θ)esin (θ) dθ
Solution.
Z
(c)
Substituting u = sin2 θ, du = 2 sin θ cos θ dθ = sin(2θ) dθ,
Z
Z
2
sin2 (θ)
sin(2θ)e
dθ = eu du = eu + C = esin θ + C.
dx
√
x2 + 1
Solution. Substituting x = tan θ, x2 + 1 = sec2 θ, dx = sec2 θ dθ,
Z
Z
Z
dx
sec2 θ
cos θ
1
x
√
=
dθ =
dθ = −
+C = √
+ C.
2
2
tan θ sec θ
sin θ
sin θ
x2 x2 + 1
x2 + 1
x2
Z
(d)
π
x2 sin x dx
0
Solution. Integrating by parts,
π
Z π
Z π
Z π
2
2
2
x sin x dx = − x cos x + 2
x cos x dx = π + 2
x cos x dx
0
0
0
0
π Z π
Z π
2
2
= π + 2 x sin x −
sin x dx = π − 2
sin x dx
0
0
0
π
2
= π + 2 cos x = π 2 − 4.
0
2
Z
(e)
t2 + 8
dt
t2 − 5t + 6
Solution. Since the numerator has degree equal to that of the denominator, we
have to divide to get
t2 + 8
5t + 2
=1+ 2
.
2
t − 5t + 6
t − 5t + 6
5t + 2
Next we decompose 2
into partial fractions:
t − 5t + 6
t2
5t + 2
5t + 2
A
B
=
=
+
.
− 5t + 6
(t − 2)(t − 3)
t−2 t−3
Using Heaviside’s method we see at once that A = −12, B = 17 so that, all in all,
t2 + 8
17
12
=1+
−
.
2
t − 5t + 6
t−3 t−2
Thus
Z
t2 + 8
dt =
t2 − 5t + 6
Z
(f )
Z 17
12
1+
−
dt = t + 17 ln |t − 3| − 12 ln |t − 2| + C.
t−3 t−2
1/2
2x + 2
dx
x2 − 2x + 1
0
Solution. By partial Fractions
x2
2x + 2
A
B
A(x − 1) + B
2x + 2
=
=
+
=
.
2
2
− 2x + 1
(x − 1)
x − 1 (x − 1)
(x − 1)2
Equating the numerators of the expression at the extreme right with that at the
extreme left we get Ax + (B − A) = 2x + 2, thus A = 2, B − A = 2, thus B = 4.
Now
Z 1/2
Z 1/2 2x + 2
2
4
dx =
+
dx
x2 − 2x + 1
x − 1 (x − 1)2
0
0
1/2
4
= 4 + 2 ln 1 .
=
2 ln |x − 1| −
x−1 2
0
[2] Find the volume of the solid generated by revolving the region bounded by the
following curves about the x-axis. Draw a picture of the region to be revolved.
√
{y = 2 x}, {y = 2}, {x = 0}
Solution.
The region in question is
3
By slicing–washers.
Z
V =π
1
√
(22 − (2 x)2 ) dx = π ∈10 (4 − 4x) dx = 2π. .
0
By the shell method. Here we describe the region in terms of y; it is the region
between the y − axis and the curve x = y 2 /4 for 0 ≤ y ≤ 2.
Z
Z 2 2
y
π 2 3
y dy = 2π.
V = 2π
y
dy =
4
2 0
0
[3] Find the center of mass of a thin plate of constant density δ covering the region
delimited by
{y = x2 }, {y = 2}.
Solution.
The region in red is the region covered by the plate:
4
By symmetry, x¯ = 0. Now (assume, as one may, that δ = 1)
√
1
Mx =
2
Z
M =
Z
2
1
(22 − (x2 )2 ) dx =
√
2
− 2
√
√
2
8 2
(2 − x2 ) dx =
.
√
3
− 2
√
2
16
,
(4 − x4 ) dx =
√
5
− 2
Z
√
2
6
Thus y¯ = Mx /M = 6/5. The center of mass is at 0,
.
5
[4] It took 1800 J of work to stretch a spring from its natural length of 2 m to a length
of 5m. Find the spring force constant
Solution.
Z 3
9k
1800 = W = k
x dx =
.
2
0
Solving for k, k = 400.
[5]
A tank in the shape of a circular cone with its vertex at the bottom, base at the
top, is filled with a liquid that happens to weigh 1 Newton/cubic meter. The tank is 12
meters high, the radius of the base of the cone is 4 meters. Determine the works it takes
to empty the tank if the liquid needs to be pumped to a height 2 meters above the top
of the tank.
Suggestion: Use the axis of the cone as the y-axis, with the origin at the base of the
cone. The top of the tank is then at y = 12 and the water needs to be raised to y = 14.
Solution. Consider a “slab” of liquid at level y of thickness dy. By similar triangles this is an infinitesimal cylinder of radius y/3, height dy, thus of volume dV =
π(y/3)2 dy = πy 2 dy/9. Since the liquid weighs 1 Newton per cubic foot, dV equals the
weight of the slab in Newtons. The slab needs to be raised 14 − y meters. Thus
Z
π 14
(14 − y)y 2 dy = 320π.
W =
9 0
[6] Determine the nature (convergence or divergence) of the following integrals and give
reasons in each case.
Z 2
s+1
√
(a)
ds.
4 − s2
0
Solution.
b !
Z 2
Z b
s+1
s+1
s
√
√
ds = lim
ds = lim
− (4 − s2 )1/2 + arcsin b→2− 0
b→2−
2 0
4 − s2
4 − s2
0
b
π
= lim −(4 − b2 )1/2 + 2 + arcsin
=2+
b→2−
2
2
Converges
5
Z
∞
2
dx.
−x
2
Solution.
(b)
x2
This integral can be computed. But, since
2
x2 −x
lim 1
x→∞
x2
and
Z
R∞
2
∞
(c)
= 2 6= 0, ∞
(1/x2 ) dx converges being a p integral wit p = 2 > 1, the integral converges
3
x2 e−x dx. Converges. One can compare to e−x , or one can calculate it.
0
[7] Determine the following limits, or declare that the corresponding sequence diverges.
en + 1
.
n→∞ en − 1
Solution.
(a) lim
en + 1
= 1.
n→∞ en − 1
lim
n
(b) lim
n→∞
n+1
Solution.
n
.
lim
n→∞
100n
(c) lim
.
n→∞ n!
Solution.
n
n+1
n
= e−1 .
100n
= 0.
n→∞ n!
lim
6
[8] Determine the nature (convergence or divergence) of the following series. Give reasons
in each case.
(a)
∞
X
ln(n)
n=1
n
.
Solution.
Since
ln n
n
n→∞ 1
n
lim
= lim ln n = ∞
n→∞
the series diverges by limit comparison with the divergent harmonic series
(b)
∞
X
n=2
P∞
n=1
1/n.
(1/n)
q
.
2
ln(n) ln (n) − 1
P
p
Solution. One approach is to state that we saw in class that the series ∞
n=2 1/[n(ln n) ]
converges if and only if p > 1 (not exactly a pPseries!), thus the series in question
2
converges by limit comparison with the series ∞
n=2 1/[n(ln n) ]. Or one can apply
the integral test directly; the terms of the series decrease:
Z ∞
Z b
dx
dx
p
p
= lim
b→∞ 2
2
x ln x ln2 x − 1
x ln x ln2 x − 1
Z ln b
du
√
(u = ln x)
= lim
b→∞ ln 2 u u2 − 1
ln b
π
= lim arcsec u = − arcsec ln 2.
b→∞
2
ln 2
The series converges .
(c)
∞
X
lnn (n)
n=1
nn
.
Solution.
(d)
n
∞ X
n−2
n=2
n
Solution.
The series converges by the root test.
.
The series diverges by the n-th term test.
∞
X
√
(e)
( n6 + n + 1 − n3 ).
n=1
Solution.
Converges. Done in class.
[9] Compute
∞
X
n=1
6
(2n − 1)(2n + 1)
7
Solution.
This is a telescoping series. One has
6
1
1
=3
−
(2k − 1)(2k + 1)
2k − 1 2k + 1
thus
n
X
6
(2k − 1)(2k + 1)
k=1
n X
1
1
1 1 1
1
1
= 3
−
= 3 1 − + − ± ··· +
−
2k − 1 2k + 1
3 3 5
2n − 1 2n + 1
k=1
1
= 3 1−
→ 3 as n → ∞.
2n + 1
sn =
Thus
∞
X
n=1
6
= 3.
(2n − 1)(2n + 1)
[10] Determine which of the following series:
i. Converge absolutely.
ii. Converge conditionally.
iii. Diverge.
Justify your answer.
(a)
∞
X
(−1)n+1
n=1
Solution.
1
n3/2
.
It converges absolutely because
∞ ∞
X
X
1
1
n+1
(−1)
=
3/2
3/2
n
n
n=1
n=1
is a p-series with p = 3/2 > 1, thus converges.
(b)
∞
X
1
(−1)n+1 √ .
n
n=1
Solution. The series converges by the alternate series test. However, the series
of absolute values is a p-series with p = 1/2 < 1, hence diverges.
The series converges conditionally .
8
[11] Find the radii and intervals of convergence of the following power series
(a)
∞
X
x2n+1
n=1
n!
.
The radius is r = ∞, the interval is (−∞, ∞).
Solution.
(b)
∞
X
n=1
xn
√ .
3n n n
Solution.
The radius is r = 3, the interval is [−3, 3].
∞
X
2n
(x − 3)n .
(c)
n
+
1
n=0
1
5
The radius is r = , the interval is
, f rac72 .
2
2
Solution.
(d)
∞
X
(−1)n
n=0
3n
(x + 1)n .
Solution.
The radius is r = 3, the interval is (−4, 2).
[13] Find the Taylor polynomials of order 3 generated by f at x = 0 when
ex + e−x
.
2
Solution. Method 1. We know (or have no excuse for not knowing) that ex =
2
3
4
2
3
4
1+x+ x2 + x6 + x24 +· · · ; replacing x by −x one gets e−x = 1−x+ x2 − x6 + x24 ∓· · · ;
2
4
adding and dividing by two, (ex + e−x )/2 = 1 + x2 + x24 + · · · . Keeping only the
terms of degree ≤ 3 we get that the polynomial is:
(a) f (x) =
1+
x2
.
2
Method 2. We compute three derivatives:
f 0 (x) =
ex − e−x
,
2
f 00 (x) =
ex + e−x
,
2
f 000 (x) =
ex − e−x
.
2
Setting x = 0 (also in f (x)) we get that f (0) = 1, f 0 (0) = 0, f 00 (0) = 1, f 000 (0) = 0.
The third order Taylor polynomials at x = 0 is:
1
1
x2
T3 (x) = f (0) + f 0 (0)x + f 00 (0)x2 + f 000 (0)x3 = 1 + .
2
6
2
(b) f (x) = sin(x).
9
Solution. Method 1. We know (or have no excuse for not knowing) that
3
x5
sin x = x − x6 + 120
± · · · . Keeping only the terms of degree ≤ 3 we get that the
polynomial is:
x3
x− .
6
Method 2. We compute three derivatives:
f 0 (x) = cos x,
f 00 (x) = − sin x,
f 000 (x) = − cos x.
Setting x = 0 (also in f (x)) we get that f (0) = 0, f 0 (0) = 1, f 00 (0) = 0, f 000 (0) = −1.
The third order Taylor polynomials at x = 0 is:
1
x3
1
T3 (x) = f (0) + f 0 (0)x + f 00 (0)x2 + f 000 (0)x3 = x − .
2
6
6
[14] Write out a Maclaurin series for the following functions f (x). In each case give the
interval of validity (in what interval is f (x) equal to the sum of its series)
(a) f (x) = x2 ex .
Remember to state the interval where the Maclaurin series converges to f (x).
P
xn
Solution. Since ex = ∞
n=0 n! , and this is valid for −∞ < x < ∞, we get by
the simple expedient of multiplying by x2 ,
2 x
xe =
∞
X
xn+2
n=0
n!
= x2 + x 3 +
x4 x5
+
+ · · ·.
2
6
The interval of validity is, of course, still −∞ < x < ∞.
(b) f (x) =
1
.
1−4x
Remember to state the interval where the Maclaurin series converges to f (x).
∞
X
1
Solution. We saw zillions of time that
=
xn if (and only if) |x| < 1.
1 − x n=0
Thus (replacing x by 4x)
∞
X
1
=
4n xn = 1 + 4x + 16x2 + 64x3 + · · ·.
1 − 4x n=0
10
for |4x| < 1. Because if a power series centered at 0 converges to a function in an
interval around zero then the power series is the Maclaurin series, the series in the
frame above is the Maclaurin series for 1/(1 − 4x). Because it converges to the
1 1
.
function if and only if |4x| < 1, the interval of validity is − ,
4 4
(c) f (x) = arctan x
Solution.
Done in class