Math 231 DL1
Final Exam
I. (A) (6 points) Give the precise definition of limx→a f (x) = L.
(B) (6 points) Carefully state the Fundamental Theorem of Calculus.
(A) For all ǫ > 0 there is a δ > 0 such that if 0 < |x − a| < δ, then
|f (x) − L| < ǫ.
(B) If f is continuous on the interval [a, b] and F is any antiderivative of f ,
then
Z b
f (x) dx = F (b) − F (a).
a
1
II. (12 points) Calculate
Z
2x − 1
dx.
x2 − 5x + 6
The denominator factors as (x − 2)(x − 3), so we can use partial fractions to
write the integrand as −3/(x − 2) + 5/(x − 3). Thus
Z
Z
5
−3
2x − 1
dx =
+
dx = −3 ln |x − 2| + 5 ln |x − 3| + C.
x2 − 5x + 6
x−2 x−3
2
III. (12 points) Calculate
Z
x+1
√
dx.
4 − x2
Here we make the trigonometric substitution x = 2 sin(θ), so dx = 2 cos(θ) dθ.
Then
Z
Z
2 sin(θ) + 1
x+1
√
q
2 cos(θ) dθ
dx =
2
4−x
4 − 4 sin2 (θ)
Z
Z
2 sin(θ) + 1
=
2 cos(θ) dθ = 2 sin(θ) + 1 dθ = −2 cos(θ) + θ + C.
2 cos(θ)
q
By definition, since sin(θ) = x/2, θ = sin−1 (x/2) and cos(θ) = 1 − sin2 (θ) =
p
1 − x2 /4, so our integral is
p
p
−2 1 − x2 /4 + sin−1 (x/2) + C = − 4 − x2 + sin−1 (x/2) + C.
3
IV. (12 points) Calculate
Z
2
∞
ln(x)
dx.
x3/2
First we use integration by parts to find the antiderivative of the integrand.
Let
u = ln(x)
Then
Z
dv =
dx
x3/2
du =
ln(x)
−2 ln(x)
√
dx =
−
x
x3/2
=
Z
−2
v=√ .
x
dx
x
−2
√ dx
x x
4
−2 ln(x)
√
− √ + C.
x
x
Since this integral is improper, we need to take a limit when we evaluate.
Z ∞
Z t
ln(x)
ln(x)
dx = lim
dx
3/2
3/2
t→∞
x
2
2 x
µ
¶ ¯t
−2 ln(x)
4 ¯¯
√
= lim
−√
¯
t→∞
x
x ¯
2
4
2 ln(2) + 4
4
2 ln(2)
−2 ln(t)
√
√
+√ =
−√ + √
= lim
t→∞
t
t
2
2
2
using L’Hôpital’s Rule.
4
V. (12 points) Find the Taylor polynomial of degree three for the function
arctan(x) around the point 0.
Recall the third degree Taylor polynomial is
3
X
f (k) (a)
k=0
k!
(x − a)k = f (a) + f ′ (a)(x − a) +
f ′′ (a)
f ′′′ (a)
(x − a)2 +
(x − a)3 .
2!
3!
The derivatives of f (x) = arctan(x) are
f (0) (x) = f (x) = arctan(x)
1
1 + x2
−2x
f (2) (x) =
(1 + x2 )2
f (1) (x) =
f (3) (x) =
−2(1 + x2 )2 + 8x2 (1 + x2 )
−2(1 + x2 ) + 8x2
=
.
2
4
(1 + x )
(1 + x2 )3
When we evaluate these at a = 0 we get 0, 1, 0, −2 respectively, so when we plug
in we have
0
−2 3
1
P3 (x) = 0 + x + x2 +
x = x − x3 .
2!
3!
3
5
VI. (12 points) Explain why the following series converges or diverges.
∞
X
(4 − 3 cos2 (3n))
(n3/2 + 2)
n=1
Note that each term of our series is positive since the numerator is at least
1 and the denominator is at least 3. Also, since 0 ≤ cos2 (x) ≤ 1, we have
4
4
(4 − 3 cos2 (3n))
≤ 3/2
≤ 3/2 .
(n3/2 + 2)
(n + 2)
n
The series
∞
X
4
3/2
n
n=1
converges since it is a multiple of the p-series with p = 3/2 > 1.
By the Comparison Test, the above series converges as well.
6
VII. (A) (8 points) Find the open interval of convergence of the power series
∞
X
2n (x − 2)n
.
3n
n=1
(B) (4 points) Give the reason why the series does or does not converge at
the largest endpoint of its interval of convergence.
(A) Using the Ratio Test, we have
¯ 2n+1 (x − 2)n+1 ¯
¯
¯
¯
¯
3n + 3
ρ = lim ¯
¯
n→∞ ¯
2n (x − 2)n ¯
3n
¯
¯
¯ 2n+1 (x − 2)n+1 3n ¯
¯
¯
= lim ¯ n
¯ = 2|x − 2|.
n→∞ ¯ 2
(x − 2)n 3n + 3 ¯
The open interval of convergence is where ρ < 1, so we solve 2|x − 2| < 1 to
get |x − 2| < 1/2 or 3/2 < x < 5/2. Hence the open interval of convergence is
(3/2, 5/2).
(B) When we plug in x = 5/2, our series becomes
inf
Xty
n=1
∞
X
2n (1/2)n
1
=
3n
3n
n=1
which diverges since it is a multiple of the harmonic series
diverges.
7
P
1
n,
which we know
VIII. (12 points) Consider the parametric curve C defined by
x=
1 2
t ,
2
y=
1 3
t ,
3
0 ≤ t ≤ 1.
Calculate the length of the curve C.
The integral which gives arc length is
Z 1p
x′ (t)2 + y ′ (t)2 dt.
0
Differentiation shows x′ (t) = t and y ′ (t) = t2 , so the arc length is
Z
1
0
Z
p
t2 + t4 dt =
0
1
t
p
1 + t2 dt.
Using the substitution u = 1 + t2 and du = 2tdt, we have
Z p
Z
1
1
1√
2
t 1 + t dt =
u du = u3/2 + C = (1 + t2 )3/2 + C.
2
3
3
Then the arc length is
µ
1
(1 + t2 )3/2
3
√
¶ ¯¯1
2 2−1
¯
.
¯ =
¯
3
0
8
IX. (A) (6 points) Again consider the parametric curve C of the previous
problem defined by
x=
1 2
t ,
2
y=
1 3
t ,
3
0 ≤ t ≤ 1.
Let P be the plane region bounded by C, the y-axis, and the line y = 1/3.
Calculate the area of P .
(B) (6 points) Write down the integral giving the volume of the solid of
revolution obtained by revolving the region P around the line x = −4. Do NOT
evaluate the integral.
R
(A) If we find the area by computing ydx, then since our region lies above
C and below y = 1/3, we have y = 1/3 − t3 /3 and dx = tdt. Then the area is
Z
0
1
¯1
¯
¯
t/3 − t4 /3 dt = (t2 /6 − t5 /15)¯ = 1/6 − 1/15 = 1/10.
¯
0
If we find the area by computing xdy then x = t2 /2 and dy = t2 dt, so the
area is
¯1
Z 1
¯
¯
t4 /2 dt = t5 /10¯ = 1/10.
¯
0
R
0
(B) If we use the washer method, we integrate dy since the axis of rotation is
vertical. Then the outer radius of the washer will be 4 + x and the inner radius
is 0. Also, dy = t2 dt. Then the integral giving volume is
Z 1
Z 1
2
π(4 + x) dy =
π(4 + t2 /2)2 t2 dt.
0
0
If we use the shell method, the height of the shells are 1/3 − y and the radius
is 4 + x. Since dx = t dt, the integral giving volume is
Z 1
Z 1
2π(1/3 − y)(4 + x) dx =
2π(1/3 − t3 /3)(4 + t2 /2)t dt.
0
0
9
X. (A) (4 points) Explain what it means for the complex exponential function ex to be an exponential function.
(B) (8 points) DERIVE the half-angle formula
sin2 (θ) =
1 − cos(2θ)
2
from the fact that the complex exponential function is an exponential function.
(A) For all complex numbers a and b, ea eb = ea+b .
(B) Start with eiθ eiθ = ei2θ . Since
eiα = cos(α) + i sin(α),
we have
(cos(θ) + i sin(θ))2 = cos(2θ) + i sin(2θ).
Expanding, we get
(cos2 (θ) − sin2 (θ)) + i(2 sin(θ) cos(θ)) = cos(2θ) + i sin(θ).
Equating real parts, we have
cos(2θ) = cos2 (θ) − sin2 (θ) = 1 − 2 sin2 (θ),
so
sin2 (θ) =
1 − cos(2θ)
.
2
10