Math 101—Midterm Exam #2, Practice Midterm B

Duration: 50 minutes

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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
Math 101—Midterm Exam #2, Practice Midterm B 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] Determine whether the following sequences converge or diverge. If they converge,
find the limit
(a) an = √sin n
n+1
(b) bn = ln(n + 1) − ln n
(c) cn = cos( πn2
)
Solution
(a) We remember that −1 ≤ sin n ≤ 1, for all n, hence √−1 n+1
≤ √sin n
n+1
≤ √n+11
. However,
−1 1
both dn = √n+1 and en = √n+1 are easily seen to converge to zero, and by the
Squeeze Theorem for sequences, an → 0, as n → ∞.
(b) Notice that if we were to take the limits of each component we would get ∞ − ∞
which is a type of indeterminate form, so we have to algebraically manipulate the
expression before we take the limit. ln(n + 1) − ln(n) = ln n+1 n
and given that the
n+1
limit of n is 1, and that ln x is a continuous function at 1, we get that
lim bn = ln 1 = 0
n→∞

(c) This limit does not exist. Check that when n is an odd integer, cos( nπ2
) = 0 while
nπ
when n is even cos( 2 ) is either 1 or −1. The sequence never gets arbitrarily close to
any value, so it diverges.

1b. [3 pts] Determine whether the improper integral

Z ∞
4 + sin x
p , dx
1 x − 1/2
converges or not. If it converges, find its value.
Solution R∞
We will compare the given integral with 1 √ 3 dx.
x− 12
Indeed, since sin x ≥ −1, 4 + sin x ≥ 3. But
Z ∞ Z t
3 3
q dx = lim q dx
t→∞ 1
1 x − 12 x − 12
Rt q q
and it is straighforward to verify that 1 √ 3 1 dx = 6[ x − 12 ]t1 = 6( t − 1
2
− 21 ) and
x− 2
thus Z t
3
lim q dx = ∞,
t→∞ 1
1 x− 2
R∞
so the integral 1
√
4+sin x
dx diverges by the comparison test.
x−1/2
Math 101—Midterm Exam #2, Practice Midterm B 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.
2a. [3 pts] Use Simpson’s Rule with n = 4 to approximate the value of the integral
Z 5
(2x2 + 3x) dx.
−3

Solution
∆x = b−a n
= 5−(−3)
4
= 2, so the x-coordinates for the relevant for Simpson’s Rule
approximation are: −3, −1, 1, 3, 5.
S4 = [f (−3) + 4f (−1) + 2f (1) + 4f (3) + f (5)] ∆x
3
= 23 [9 − 4 + 10 + 108 + 65] = 376
3

Z π/4
2b. [3 pts] Evaluate tan3 x sec2 x dx. Simplify your answer fully.
0

Solution
Making the substitution u = tan x: du = sec2 xdx, when x = 0 : u = 0, when
Z π/4 Z 1
3 2 u4 1
π
x = 4 : u = 1 we see tan x sec x dx = u3 du = [ ]10 = .
0 0 4 4
Math 101—Midterm Exam #2, Practice Midterm B page 4 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.
3a. [3 pts] Find the general antiderivative of f (x) = x2 sin(πx).
Solution Z
We are asking for the indefinite integral I = x2 sin(πx) dx. Integrating by parts we get
2 2
I = − xπ cos(πx) − (− cos(πx)·2x dx. Integrating by parts once more: I = − xπ cos(πx) +
R
π
2x sin(πx) 2 2
− π22 sin(πx) dx = (2−π x ) cos(πx)+2πx sin(πx)
R
π2 π3
+ C.
Z 2
x2
3b. [3 pts] Evaluate √ dx.
1 x2 − 1
Solution
Z 2 Z 2
x2 x2
This is an improper integral due the singularity at x = 1 ; √ dx = lim+ √ dx.
1 x2 − 1 t→1 t x2 − 1
The last integral is rather complicated to compute, so, for simplicity, we will first evaluate
the indefinite integral and plug in the endpoints at the end (before taking the limit).

We make the (“inverse” ) substitution x = sec s, 0 ≤ s < π/2 and dx ds

= sec s tan s and in
sec2 s sec s tan s
Z
differential form dx = sec s tan s ds which results in the new integral ds =
Z Z tan s
sec3 s ds. For I = sec3 s ds we will perform integration by parts with u = sec s, dv =
sec2 sds (then du = sec s tan sds, v = tan s), so
Z Z
I = sec s tan s − tan s sec s ds = sec s tan s − (sec2 s − 1) sec s ds
2

Z Z
= sec s tan s + sec s − sec3 s ds = sec s tan s + ln | sec s + tan s| − I + C
because we observed that the integral we started with, appeared on the RHS. Rearranging:
1
2I = sec s tan s + ln | sec s + tan s| + C ⇒ I = (sec s tan s + ln | sec s + tan s|) + C
2
Since we didn’t change the endpoints after the substitution (we were only treating the in-
definite integral) we need to go back to the original variable x, i.e.,
√ we need to√express
sec s, tan s in terms of x. But sec s = x and to find tan s : tan s = sec2 −1 = x2 − 1.
Two remarks: (i) we picked the positive branch of the square root and that was because of
the range of the original substitution (0 ≤ s < π/2 : tan s ≥ 0.) (ii) We could have arrived
at the same conclusion by drawing a right triangle with hypotenuse x and one side with
length 1 (as explained in section 7.3 in the book) .
√ √
In the original variable x the integral is = 21 (x x2 − 1 + ln |x + x2 − 1|) + C and
plugging in the endpoints t and 2 we get:
1 √ √ √ √
Z 2
x2
√ dx = [2 3 + ln(2 + 3) − t t2 − 1 − ln |t + t2 − 1|.
t x2 − 1 2
Math 101—Midterm Exam #2, Practice Midterm B page 5 of 8
But
√ √
2
x2 1 √ √ √ √
Z
2 3 + ln(2 + 3)
√ dx = lim+ [2 3 + ln(2 + 3) − t t2 − 1 − ln |t + t2 − 1| =
1
2
x −1 t→1 2 2
Math 101—Midterm Exam #2, Practice Midterm B 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.
Z 4
x−1
4. [3 pts] Evaluate I = 2
dx.
0 x − 4x − 5
Solution
The numerator is of lower degree than the denominator so there is no need for long division. To
factor the denominator we use the quadratic formula to get that the solutions of the quadratic
equation x2 − 4x − 5 = 0 are x = 5 and x = −1.
x−1 A B
= +
x2 − 4x − 5 x−5 x+1
Multiplying both sides with (x − 5)(x + 1) and matching coefficients, we get A − 5B = −1 and
A + B = 1. The solution to this system is (A, B) = ( 32 , 13 ).
Z 4
x−1 2 4 1 1 4 1
Z Z
2 1
I= 2
dx = dx + dx = [ln |x − 5|]40 + [ln |x + 1|]40
0 x − 4x − 5 3 0 x−5 3 0 x+1 3 3
− ln 5
= 23 [ln 1 − ln 5] + 13 [ln 5 + ln 1] = 3
Math 101—Midterm Exam #2, Practice Midterm B 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.
5.
(a) [4 pts] A lamina with density ρ = 3 is in the shape of the region under the graph of
y = e2x between x = −1 and x = 1. Find My , the moment with respect to the y-axis, of
this lamina.
Solution
R1 2x R1 3xe2x 2x 3e2
My = ρ −1 xe2x dx = 3(x e2 |1−1 − 12 −1 e2x dx) = 2
|1−1 − 3e4 |1−1 = 4
+ 9
4e2
.

(b) [4 pts] Evaluate the integral

Z √
(x + 1) x2 + 2x + 2 dx.

Solution
R √ R p
I = (x + 1) x2 + 2x + 2 dx = (x + 1) (x + 1)2 + 1 dx.

Making the (“inverse”) substitution x + 1 = tan θ, π2 < θ < π2 , dx

dθ
= sec2 θ :
Z p Z √ Z
(x + 1) (x + 1) + 1 dx = tan θ sec θ sec θ dθ = tan θ sec θ sec2 θdθ
2 2 2

Now set u = sec θ, and thus du = sec θ tan θ dθ to get

u3 sec3 θ (x2 + 2x + 2)3/2
Z
I = u2 du = +C = +C = + C.
3 3 3
Math 101—Midterm Exam #2, Practice Midterm B page 8 of 8
6.
(a) [4 pts] Use integration to calculate the area of the ellipse
x2 y 2
+ =1
4 9
Hint: exploit the symmetry of the ellipse with respect to both axes.
Solution
q
2
Solving the equation of the ellipse for y we get y = ± 9 − 9x4 . Because the ellipse is
symmetric to both axes, the total area A is four
q times the area in the first quadrant which
9x2
the area of the region below the graph of y = 9 − 4
between x = 0 and x = 2.
R2q 2 R2p
A = 4 0 9 − 9x4 dx = 12 0 1 − ( x2 )2 dx.

Set x2 = sin θ, −π
2
≤ θ ≤ π2 , so x = 2 sin θ and dx
dθ
= 2 cos θ. When x = 0, θ = 0 and
when x = 2, θ = π/2 (because of the range of the substitution).
R π/2 p R π/2
Then A = 12 0 2 cos θ 1 − sin2 θ dθ = 24 0 cos2 θ dθ, which by the half-angle
R π/2 π/2
formula becomes A = 12 0 (1 + cos(2θ)) dθ = 12[θ + 21 sin(2θ)]0 = 6π. (“sanity”
check: A = a · b · π = 3 · 2 · π (where a and b are the lengths of the axes of the ellipse))

6b. [4 pts] Determine whether the integral

Z e3
ln x dx
0
converges or diverges. If it converges, find its value.
Solution

Z e3 Z e3
ln x dx = lim+ ln x dx
0 t→0 t
Integrating by parts with u = ln x, dv = dx to get
Z e3
e3 1
= lim+ [x ln x |t − x dx] = e3 ln(e3 ) − lim+ (t ln t + e3 − t)
t→0 t x t→0

= 3e3 − e3 + lim+ (t − t ln t)
t→0
But
ln t
lim+ t ln t = lim+
t→0 t→0 1/t
1/t
which by L’Hospital’s Rule is the same as lim+ = − lim+ t = 0 thus the limit (and
t→0 −1/t2 t→0
3
consequently the integral) equals 2e .
Math 101—Midterm Exam #2, Practice Midterm B page 9 of 8
7. [8 pts] Find the solution to the differential equation
dy y
= 4
dx x + x2
which satisfies the initial condition y(1) = 1/e.
Solution
dy
We first consider the case y = 0. Then dx
= 0 ⇒ y = constant = 0. However this cannot happen
since y(1) = 1e .
dy dx
So we can divide, and writing in differential form: y
= x4 +x2
. Integrating:
R 1 R 1
ln |y| = x4 +x 2 dx = x2 (x2 +1)
dx.

To calculate the last integral, you can either use the trigonometric substitution x = tan θ (but in the
process you will have to remember the anti-derivative of csc2 x) or you resort to a partial fraction
decomposition, which is what we will do here:
1
x2 (x2 +1)
= Ax + xB2 + Cx+D
x2 +1
since there is a repeated and an irreducible quadratic term. Multiplying
2 2
both sides with x (x + 1) and matching coefficients, it is easy to see that A = C = 0, B = 1, D =
−1
1 −1
dx = ( x12 − 1
R R
Consequently, x2 (x2 +1) x2 +1
) dx = x
− arctan x + C.
1
Putting everything together and “exponentiating”: |y| = e−( x +arctan x ) · C. To find the constant C
we use the information that y(1) = 1/e to get C = eπ/4
π 1
y = ±e 4 − x −arctan x and we can exclude the negative case by another use of y(1) = 1/e.

In conclusion, the solution of the differential equation with the extra condition y(1) = 1/e is
π 1
y = e 4 − x −arctan x