Midterm Exam #2—Math 101, Section 207
March 13, 2015
Duration: 50 minutes
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Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be given
for the answer without the correct accompanying work. Problems 5–7 are long-answer: give complete
arguments and explanations for all your calculations—answers without justifications will not be marked.
Continue on the back of the page if you run out of space.
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Problem Out of Score Problem Out of Score
1
6
5
8
2
6
6
8
3
6
7
8
4
3
Total
45
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 2 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
1a. [3 pts] Find the average value of f (x) = √xx2 +4 on the interval [−1, 1].
We recall that the average value of the function f (x) on the interval [a, b] is given by
Z b
1
f (x) dx.
favg =
b−a a
Thus in our case we have
Z 1
1
x
√
favg =
dx;
1 − (−1) −1 x2 + 4
using the substitution u = x2 + 4, du = 2x dx we have
Z
1 5 (1/2)du
favg =
= [u1/2 ]5u=5 = 0,
2 5
u1/2
so the average value of the function on [−1, 1] is 0.
Alternatively, we could have shown that the function f (x) is odd, and therefore its integral
over the symmetric interval [−1, 1] which is centered on zero must be zero.
Grading scheme:
• 1pt for the right favg formula.
• 1pt for making a valid substitution.
• 1pt for finding the answer 0.
Z
1b. [3 pts] Find
arctan x dx.
1
We use integration by parts: with u = arctan x, dv = dx, we have du = 1+x
2 , v = x,
hence
Z
Z
x
arctan x dx = x arctan x −
dx.
1 + x2
This integral is similar to that from part (a); using the analagous substitution u = x2 +
1, du = 2x dx, we have
Z
Z
x
(1/2)du
1
1
=
= ln u + C = ln(1 + x2 ) + C,
2
1+x
u
2
2
and so
Z
1
arctan x dx = x arctan x − ln(1 + x2 ) + C.
2
Grading scheme:
• 1pt for a valid integration by parts.
• 1pt for correctly solving the second integral.
• 1pt for finding the correct antiderivative.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 3 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
Z √
4 − x2
2a. [3 pts] Find
dx.
x2
√
Recognizing 4 − x2 , we use the trigonometric substitution x = 2 sin θ, so that dx =
2 cos θ dθ and thus
Z √
Z p
Z
Z
4 cos2 θ
4 − x2
4 − 4 sin2 θ
dx =
(2 cos θ) dθ =
dθ = cot2 θ dθ.
x2
4 sin2 θ
4 sin2 θ
By the identity cot2 θ + 1 = csc2 θ, we have
Z
Z
2
cot θ dθ = (csc2 θ − 1) dθ = − cot θ − θ + C.
Finally, considering the right triangle whose side opposite angle θ has length√x and whose
2
adjacent
= 4−x
, and so
hypotenuse has length 2, we find that θ = arcsin x2 and cot θ = opposite
x
√
Z √
2
2
x
4−x
4−x
dx
=
−
−
arcsin
+ C.
x2
x
2
Grading scheme:
• 1pt for making a valid substitution.
• 1pt for integrating csc2 θ.
• 1pt for putting the answer in terms of x.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 4 of 8
2b. [3 pts] Determine the partial fraction decomponision of f (x) =
need to integrate the resulting expression.
2x−1
.
x3 −x2 +x−1
You do not
We begin by factoring the denominator: we notice that x = 1 is a root so (x − 1) divides
the denominator, and performing the factorization gives
x3 − x2 + x − 1 = (x − 1)(x2 + 1),
and since x2 + 1 is irreducible (that is, it has no linear factors, since b2 − 4ac = −4 < 0)
we know this factorization is complete. Therefore the partial fraction decomposition takes
the form
2x − 1
A
Bx + C
A(x2 + 1) + (Bx + C)(x − 1)
=
+
=
.
x 3 − x2 + x − 1
x−1
x2 + 1
(x − 1)(x2 + 1)
Regrouping the numerator on the right hand-side, we have the equality of numerators
2x − 1 = (A + B)x2 + (C − B)x + (A − C),
which gives us the system of equations
A+B =0
−B + C = 2
A − C = −1
Adding the third equation to the second gives A − B = 1, and adding this equation to the
first gives 2A = 0 hence A = 21 . It follows that B = − 12 and therefore that C = 32 . Thus
x3
3/2
(−3/2)x + 1/2
3−x
1
2x − 1
=
+
=
+
.
2
2
2
−x +x−1
x−1
x +1
2(x + 1) 2(x − 1)
Grading scheme:
• 1pt for factoring the denominator.
• 1pt for getting the form of coefficients correct (e.g. linear factors over irreducible
quadratics).
• 1pt for finding the correct values of A, B, C.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 5 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
Z ∞ √
5x x
3a. [3 pts] Compute
dx.
x5
1
First, we simplify the integrand:
√
5x x
5x3/2
5
=
= 7/2 = 5x−7/2 .
5
5
x
x
x
Thus, by definiton we have
Z A
Z ∞ √
5 −5/2 A
5x x
5x−7/2 dx = lim −5/2
dx = lim
x
x=1
5
A→∞
A→∞ 1
x
1
= lim [−2A−5/2 − (−2)] = lim 2 −
A→∞
A→∞
2
A5/2
= 2.
Grading scheme:
• 1pt for simplifying the integrand.
• 1pt for turning the improper integral into a limit.
• 1pt for evaluating the limit.
3b. [3 pts] Solve the initial value problem
dy
xy sin x
=
,
y(0) = 1.
dx
y+1
You may leave your answer as an implicit function of x; that is, you do not need to solve
the resulting expression for y.
We begin by separating the equation:
dy
xy sin x
y+1
=
⇐⇒
dy = x sin x dx.
dx
y+1
y
We write 1 +
1
y
for
y+1
y
and integrate to obtain
Z
Z
y + ln |y| = x sin x dx = −x cos x + cos x dx = sin x − x cos x + C,
where we have used integration by parts with u = x, dv = sin x dx. Subbing in the initial
values, we have
1 + ln 1 = sin 0 − (0) cos 0 + C ⇐⇒ C = 1,
and we have that the solution to our IVP is given by
y + ln y = sin x − x cos x + 1.
Note that we may ignore the absolute value signs since y(0) > 0 and y is never zero (since
then ln |y| is not defined), and so y > 0 everywhere.
Grading scheme:
• 1pt for separating the equation.
• 1pt for correctly performing the integration.
• 1pt for solving for the value of C.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 6 of 8
Problems 1–4 are short-answer questions: put a box around your final answer, but no credit will be
given for the answer without the correct accompanying work.
n
4. [3 pts] Show that the sequence an = (−1)n 2 2−1
diverges.
n
n
We first write 2 2−1
= 1 − 21n ; now it is clear that 12 ≤ |an | < 1 for every n. The limit
n
does not exist, therefore, because there are infinitely many terms greater than 21 and also
infinitely many terms less than − 12 . There are several ways to show this:
• The difference of successive terms is always at least 1 in absolute value:
1
|an+1 − an | = (−1)n+1 1 − 2n+1
− (−1)n 1 − 21n 1
1
= |(−1)n | 2n+1
− 1 − 1 − 21n = 2 − 21n − 2n+1
≥ 2 − 12 − 41 > 1.
If the sequence converged we would have limn→∞ |an+1 − an | = 0.
• Given any N > 0, there exists some n > N such that an is negative and some
n0 > N for which an is positive. Thus the limit of an can be neither positive nor
negative, so if it exists it must equal zero. But our work above shows |an | ≥ 21 for
every n ∈ N, so the limit cannot be 0, and we deduce that the limit does not exist.
• Suppose limn→∞ an exists – call it L – and let bn = 1 − 21n . Then limn→∞ bn
clearly exists and equals 1, and 12 ≤ bn < 1 for every n; in particular, bn 6= 0 and
−1
so cn = b−1
= 1. So, since
n is a well-defined sequence with limit limn→∞ bn
n
an = (−1) bn , one has
lim an cn = lim an
lim cn = (L)(1) = L;
n→∞
n→∞
n→∞
but then
n
lim an cn = lim an b−1
n = lim (−1) = L,
n→∞
n→∞
n→∞
a contradiction because this limit does not exist.
Because of the varied ways of solving this problem, the grading scheme is necessarily somewhat
vague.
Grading scheme:
• 1pt for indicating the possible limits of the sequence (i.e. ±1).
• 1pt for indicating divergence of the sequence (−1)n .
• 1pt for a logically consistent argument.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 7 of 8
Problems 5–7 are long-answer: give complete arguments and explanations for all your calculations—
answers without justifications will not be marked.
Z ∞
dx
√ is divergent.
5a. [4 pts] Without computing any limits, explain why the integral
1+ x
1
√
√
√
For any x ≥ 1 we have that 1 ≤ x and so in particular 1 + x ≤ 2 x. Thus 1+1√x ≥
Z ∞
dx
1
√ , and so by the comparision test it suffices to show that
√ is divergent. But
2 x
2 x
1
Z ∞
Z
dx
1 ∞ dx
√ =
,
2 1 x1/2
2 x
1
Z ∞
dx
√ dominates this integral, we deduce
which is divergent by the p-test. Since
1+ x
1
by the comparison test that it too is divergent.
Grading scheme:
• 1pt for citing the comparison test.
• 1pt for comparing the integral in the correct way (i.e. showing this is greater than a
divergent integral).
• 1pt for showing that the integral to which we compare, diverges (by p-test, or referring to
lecture, etc.)
• 1pt for a logically consistent argument.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
Z
∞
page 8 of 8
2
5b. [4 pts] Compute
−∞
x
dx.
1 + x6
We begin by writing
Z
∞
−∞
x2
dx =
1 + x6
Z
0
−∞
x2
dx +
1 + x6
Z
0
∞
x2
dx.
1 + x6
Consider first the integral over the positive reals: using the substitution u = x3 , du =
3x2 dx, we have
Z ∞
Z ∞
du/3
x2
dx =
du,
6
1+x
1 + u2
0
0
where the upper limit is ∞ because limx→∞ u = ∞. Thus
Z ∞
∞
π
1
1 π
x2
arctan
u
dx
=
=
−
0
= .
2
u=0
1 + x6
3
3
6
0
Using the same substitution we obtain
Z 0
Z
0
x2
du
π
1 0
1
dx =
dx =
arctan u u=−∞ = ,
6
2
3 −∞ 1 + u
3
6
−∞ 1 + x
(of course, we could also have used symmetry, since the integrand is even), and so by
definition
Z ∞
x2
π π
π
dx = + = .
6
6
6
3
−∞ 1 + x
Grading scheme:
• 1pt for writing the integral as the sum of improper intgrals (from −∞ to a, and from a to
∞).
• 1pt each for computing both integrals, or computing one and citing symmetry.
• 1pt for finding the answer π3
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 9 of 8
Z
1
x3 − x + 1 dx. Simplify your answer
6a. [4 pts] Use Simpson’s rule with n = 4 to estimate
−1
completely.
With n = 4 we have that the interval length is
1−(−1)
4
= 12 , and so our points are
x0 = −1, x1 = − 12 , x2 = 0, x3 = 12 , x4 = 1.
Simpson’s rule gives
S4 =
∆x
[f (x0 )
3
+ 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )],
and so calculating
f (−1) = 1, f (− 12 ) = 74 , f (0) = 1, f ( 21 ) = 58 , f (1) = 1,
we have
Z
1
x3 − x + 1 dx ≈ S4 = 61 [1 + 4
−1
Z 1
So our estimate for
x3 − x + 1 dx is 2.
−1
Grading scheme:
• 1pt for the right value of ∆x
• 1pt for the correct form of Simpson’s rule.
• 1pt for evaluating at the correct xi points.
• 1pt for simplifying the answer.
11
8
+ 2(1) + 4
5
8
+ 1] = 2.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 10 of 8
6b. [4 pts] Give an upper bound for the error
Z 1resulting from the estimate used in part (a). Explain
x3 − x + 1 dx.
why this allows us to find the true value of
−1
The error ES involved in using Simpson’s rule with n subintervals is bounded by
K(b − a)5
,
180n4
where K is a constant such that |f (4) (x)| ≤ K for all x ∈ [a, b]. We compute the
derivatives of f (x) = x3 − x + 1:
|ES | ≤
f 0 (x) = 3x2 − 1,
f 00 (x) = 6x,
f 000 (x) = 6,
f (4) (x) = 0.
We have that f (4) (x) = 0 for all x, and therefore we may take K = 0 and so
0 · 25
|ES | ≤
= 0.
180 · 44
Z
1
Thus ES = 0, and since ES =
x3 − x + 1 dx − S4 it follows that S4 = 2 is the true
−1
value of the integral.
Grading scheme:
• 1pt for recalling the error bound for Simpson’s rule.
• 1pt for computing derivatives to find K.
• 1pt for finding the error to be zero.
• 1pt for stating that the estimate from part (a) is exact.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 11 of 8
7a. [4 pts] Determine whether or not the sequence an = ln(n + 1) − ln n converges. If it does,
find its limit, fully justifying your answer.
We observe that
an = ln(n + 1) − ln n = ln n+1
= ln 1 + n1 .
n
Since ln x is continuous on its domain, and 1 + n1 lies in the domain of ln x for every
n ≥ 1, we have that
1
1
lim an = lim ln 1 + n = ln lim 1 + n ,
n→∞
provided the limit limn→∞ 1 +
and it follows that
n→∞
1
n
n→∞
exists. Of course, this limit clearly exists and equals 1,
lim an = ln 1 = 0.
n→∞
Grading scheme:
• 1pt for using logarithm laws.
• 1pt for citing continuity of ln x on the interval [1, ∞).
• 1pt for passing the limit inside the argument of ln(1 + n1 ).
• 1pt for computing the limit 0.
Midterm Exam #2—Math 101, Section 207—March 13, 2015
page 12 of 8
7b. [4 pts] Find the centroid of the region bounded by the x-axis and the graph of y = 1 − x2 .
By definition, the x- and y-co-ordinates of the centroid of a region bounded above by the
graph of f (x) and below by the x-axis, between a and b, are given respectively by
Z
Z
1 b
1 b1
x̄ =
xf (x) dx and ȳ =
[f (x)]2 dx,
A a
A a 2
where
Z
b
f (x) dx.
A=
a
In our case we have 1 − x2 = 0 ⇐⇒ x = ±1, so a = −1, b = 1, f (x) = 1 − x2 , so
Z 1
4
3 1
(1 − x2 ) dx = x − x3 x=−1 = .
A=
3
−1
Hence
Z
3 2
3 1
4 1
x − x3 dx = x2 − x4 −1 = 0
x̄ =
4 −1
4
and
Z
Z
3
3 1
3 1 1
2
3
5 1
2 2
1 − 2x2 + x4 dx = x − 2x3 + x5 x=−1 = .
(1 − x ) dx =
ȳ =
2
4 −1
8 −1
8
5
So the centroid of the region is (x̄, ȳ) = (0, 52 ).
Grading scheme:
• 1pt for recalling the formulas for x̄, ȳ.
• 1pt for calculating A.
• 1pt each for calculating x̄, ȳ.