Calculus I : Midterm Exam SolutionApril 22, 2014 7:00 PM ∼ 10:00 PM

Spring 2014 MAS 101

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Do not write in this table.Problem Score Problem Score Problem Score

1 / 48 5(a) / 4 6(b) / 10

2 / 20 5(b) / 4 7 / 20

3 / 20 5(c) / 4 8 / 20

4 / 20 6(a) / 10 9 / 20

Total Score:

– 1 –

2014 Calculus 1 Mid-term Solution TA Jonghun Yoon

1

48 points

(a) If∑

(an + bn) converges and∑

an diverges, then∑

bn diverges.

(b) If all an are positive and an+1

an< 1 for all n, then

∑an converges.

(c) If all an are positive and n√an < 1 for all n, then

∑an converges.

(d) If 0 < an+1 < an for all n, then∑

(−1)nan converges.

(e) If∑

an converges, then∑|an| converges.

(f) If∑

an converges, then∑

(−1)nan converges.

(g) If∑

an converges, then∑

a2n converges.

(h) If the an are positive and∑

an converges, then∑

a2n converges.

(i) (x+ sinx) · coshx = o (xx) as x→∞.

(j) 2(x+ (lnx)2) tan−1 (ex) = O (x− lnx) as x→∞.

(k)∫∞1

ln xx2 dx converges.

(l) The domain of coth−1 x is (−∞,−1) ∪ (1,∞)

Solution.

(a) T, If∑

bn converges, then∑

an =∑{(an + bn)− bn} converges.

(b) F, The counter example is an = 1/n.

(c) F, The counter example is an = 1/(n+ 1).

(d) F, The counter example is an = 1 + 1/n.

(e) F, The counter example is an = (−1)n/n.(f) F, The counter example is an = (−1)n/n.(g) F, The counter example is an = (−1)n/

√n.

(h) T, Since∑

an converges, limn→∞a2nan

= 0. Hence, for all large n, 0 < a2n ≤ Man for some

M > 0.

(i) T, Use limx→∞( ex )x = 0. This is true since e

x < 12 for all large x.

(j) T, By L’ohospital’s rule, limx→∞x+(ln x)2

x−ln x = 1. And | tan−1 x| ≤ π2 .

(k) T, By Integral by parts,∫ a1

ln xx2 dx =

[− ln x

x −1x

]a1= − ln a

a −1a + 1. Hence

∫∞1

ln xx2 dx = 1

(l) T, See the text book 439 page.

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2014 Calculus 1 Mid-term Solution TA Jongbaek Song

2

20 points

Decide whether the infinite series

∞∑n=1

(−1)n(√

n5 + n2 −√n5),

converges absolutely, converges conditionally, or diverges. You should justify your

answer.

Solution. Let an =√n5 + n2 −

√n5 = n2

√n5+n2+n5

= 1√n+ 1

n2 +√n.

• [9 points]∑∞

n=1 an diverges.

Indeed, use the sequence bn = 1√nto apply the limit comparison test,

limn→∞

anbn

=1

2.

Since∑∞

i=11√ndiverges by p-series, so does

∑∞n=1 an by limit comparison test.

• [9 points]∑∞

n=1(−1)nan converges. Use the following alternating series test (Leibniz’s test).

1. an ≥ 0, for all n, obviously. · · · · · · (3 points)

2. an = 1√n+ 1

n2 +√n≥ 1√

n+1+√n+1

≥ 1√n+1+ 1

(n+1)2+√n+1

= an+1

Hence, an is a non-increasing sequence. · · · · · · (3 points)

3. an = 1√n+ 1

n2 +√n→ 0 as n → ∞. · · · · · · (3 points)

• [2 points] Therefore,∑∞

n=1(−1)nan converges conditionally.

Grading Policy

• You may use other methods to show that∑∞

n=1 an diverges or an is an decreasing sequence.

• If you just state something without any explanation, you wouldn’t get the associated points.

Especially, you have to explain why the sequence an is decreasing.

• 2 points assigned in the final conclusion would follow only when your solution explain that.

In other words, If your solution has nothing to do with conditionally convergence, then 2

points might not be given, even though you conclude correctly.

■

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Calculus I Midterm Solution and Scoring Criteria TA. Giwon Suh

3. Let a > 1. We define

f(x) =

∫ x

1

loga tdt +

∫ loga x

0

atdt,

for x > 1. Find f(x) and f ′(x).

Solution.

∫ x

1

loga tdt =

∫ln t

ln adt =

1

ln a

∫ x

1

ln tdt =1

ln a[t ln t− t]x1 =

1

ln a(x lnx− x + 1)

(+6 points)

∫ loga x

0

atdt =

∫ loga x

0

et ln adt =1

ln a[et ln a]

ln xln a0 =

1

ln a(eln x − 1) =

1

ln a(x− 1)

(+6 points)

Therefore, f(x) = 1ln ax lnx = x loga x, (+3 points)

and f ′(x) = 1ln a (1 + lnx) = 1

ln a + loga x (+5 points)

Solution 2.

- t

6

yy = at

loga x

x

1

A

B

In the figure, Area of A and B denotes the value of integration∫ x

1loga tdt and

∫ loga x

0atdt, respec-

tively. Therefore, f(x) = x loga x. (+15 points)

and f ′(x) = 1ln a (1 + lnx) = 1

ln a + loga x (+5 points)

Solution 3.

f ′(x) = loga x +aloga x

ln a× 1

lnx=

1

ln a+ loga x

(+5 points)∫f ′(x)dx =

∫(

1

ln a+ loga x)dx =

x

ln a+

x lnx− x

ln a=

x lnx

ln a

f(x) =x lnx

ln a+ C

April. 24, 2014 Typeset by LATEX

Calculus I Midterm Solution and Scoring Criteria TA. Giwon Suh

(+12 points)

Since f(1) = 0, C = 0. Therefore, f(x) = x ln xln a (+3 points)

Grading Policy

For each step, computation mistake -3 points.

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2014 Calculus I Midterm Solution and Scoring Criteria TA. Minsu Ko

4. You made a cup of coffee with the original temperature 80◦C. You went out to the

balcony with it and the outside was 80◦C. After 5 minutes, the temperature of the

coffee dropped to A◦C. What will be the temperature of the coffee after 20 minutes.

Solution.

According to Newton’s Law of Cooling, (+5 points)

dH

dt= −k (H −Hs) ,

where H is the temperature of coffee at time t and Hs is the constant surrounding temperature.

If we substitute y for (H −Hs), then

dy

dt=

d

dt(H −Hs) =

dH

dt− d

dt(Hs)

=dH

dt= −k (H −Hs) = −ky.

Now we have to solve differential equation

dy

dt= −ky.

We know that solution of previous differential equation is y = y0e−kt, thus (+5 points)

H −Hs = (H(0) −Hs) e−kt

Since H(0) = 80 and Hs = 10 given,

H(t) = 10 + 70e−kt.

From the condition that H(5) = A , (+5 points)

A = H(5) = 10 + 70e−5k, i.e. e−kt =A− 10

70.

Therefore we can conclude that (+5 points)

H(20) = 10 + 70(e−5k

)4= 10 + 70

(A− 10

70

)4

.

Scoring Criteria.

1. Each simple mistake counts -2 points.

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2014 Calculus 1 Mid-term Solution TA. Giung Bae (배기웅)

5

10 points

(a) Decide whether the following series converge or diverge. You should justify your

answer.∞∑

n=1

3√n+ lnn

n

n2 − (−1)nn+ 7√n.

Solution.

limn→∞

3√n+ lnn

n

n2 − (−1)nn+ 7√n÷ 1

n32

= limn→∞

3 + lnnn√n

1− (−1)nn + 7√

n

= 3.

Therefore, by the Limit Comparison Test,∑∞

n=13√n+ lnn

n

n2−(−1)nn+7√nand

∑∞n=1

1

n32are both convergent

or divergent. Since∑∞

n=11

n32are convergent p-series,

∑∞n=1

3√n+ lnn

n

n2−(−1)nn+7√nare also convergent.

Grading Policy

• You get 4 points if everything are correct.

• You get 2 points if your idea and logic are right and there is only a trivial mistake.

• You get 0 points if your logic is wrong, even if you performed some suitable test, regardless

of your answer.

• You get 0 points if you argued that

limn→∞

an = 0 implies

∞∑n=1

an <∞.

�

5

10 points

(b) Decide whether the following series converge or diverge. You should justify your

answer.∞∑

n=1

(9− 1n )

n

8nn.

Solution.

limn→∞

(an)1n = lim

n→∞

9− 1n

8n1n

=9

8> 1,

since limn→∞1n = 0 and limn→∞ n

1n = 1. Therefore, the series is divergent by the Root Test.

Grading Policy

• You get 4 points if everything are correct.

• You get 2 points if your idea and logic are right and there is only a trivial mistake.

• You get 0 points if your logic is wrong, even if you performed a suitable test, regardless of

your answer.

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2014 Calculus 1 Mid-term Solution TA Yewon Jeong

5

4 points

(c)

∞∑n=2

(n− 1)!

(n− 1)n(Determine whether the series converges or diverges.)

Solution. This series converges. Let’s prove this. Let an = (n−1)!(n−1)n for n ≥ 2.

(Method 1 : Ratio Test)

Enough to show that (1) an > 0 for any n > 2 and (2) limn→∞an+1

anexists and the value is smaller

than 1. (1) is obvious when n ≥ 2.

(2) limn→∞

an+1

an= lim

n→∞n!

nn+1 · (n−1)n

(n−1)! = limn→∞

(n−1n )n = lim

n→∞(1− 1

n )n. (+2 points)

This is e−1 because limn→∞

(1− 1n )n = lim

n→∞{(1− 1

n )−n}−1 = limh=−1

n →0{(1 + h)1/h}−1 = e−1 < 1.

So the series converges by the Ratio Test. (+2 points)

(Method 2 : Comparison Test)

Since an > 0 for any n ≥ 2, enough to find a converging series∑∞

n=2 bn satisfying 0 < an ≤ bn for

any n ≥ 2. Let bn = ( 1n−1 )2, then an = (n−1

n−1 ) · (n−2n−1 ) · · · · · ( 2

n−1 ) · ( 1n−1 )2 ≤ bn. (+2 points)

Furthermore, the series∑∞

n=2 bn converges by the p-series test. So the series∑∞

n=2 an also con-

verges by the Comparison Test. (+2 points)

(Method 3 : Integration and Root Test)

(step 1) ln(n− 1)! = ln 1 + ln 2 + · · ·+ ln(n− 1). Since y = lnx is an increasing function,∫ n−1

1

lnx dx ≤ ln 1 + ln 2 + · · ·+ ln(n− 1) ≤∫ n

1

lnx dx.

Then we get (n− 1)n−1 · e2−n ≤ (n− 1)! ≤ nn · e1−n and e2−n

n−1 ≤ an ≤ nn·e1−n

(n−1)n . (+2 points)

(step 2) From step 1, we also get (n− 1)−1/n · e(2−n)/n ≤ n√an ≤ n

n−1 · e(1−n)/n.

Since limn→∞

(n− 1)−1/n · e(2−n)/n = e−1 and limn→∞

nn−1 · e

(1−n)/n = e−1, limn→∞

n√an also converges to

e−1 < 1 by the Sandwich Theorem. So∑∞

n=2 an converges by the Root Test. (+2 points)

Typical wrong answers

• limn→∞

an = 0 does not imply the convergence of∑∞

n=2 an. (For example, limn→∞

1n = 0 , but∑∞

n=21n diverges.)

• limn→∞

an = 0 and an+1

an< 1 for any n ≥ 2 also do not imply the convergence of

∑∞n=2 an.

This series is not Alternating !

• When you show the existence of limn→∞

n√an , the boundedness of n

√an is not enough, and so

we need the help of the Sandwich Theorem.

• (1− 1n )n < 1 for any n ≥ 2 does not imply lim

n→∞(1− 1

n )n < 1. (For example, 1− 1n < 1 for

any n ≥ 2, but limn→∞

(1− 1n ) ≮ 1.)

• By the same reason, n√an < 1 for any n ≥ 2 also do not imply lim

n→∞n√an < 1.

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2014 Spring Calculus 1 Midterm Solution TA. Jeong-seop Kim

6

20 points

Let 0 ≤ x < π2 .

(a) (10 points) Find a function y = f(x) that satisfies

cosxdy

dx= −y

√1− y2 and y(0) = 1.

(b) (10 points) Let y = f(x) be the solution of (a). Find the value of (f−1)′(0.8).

Solution.(a)-1.

From the given formula,∫−1

y√

1− y2dy =

∫1

cosxdx. (+1 point)

By integrating each side,

(LHS) = sech−1y + C1, (+3 points)

(RHS) = ln(secx+ tanx) + C2, (∵ 0 ≤ x < π/2) (+2 points)

we have sech−1y = ln(secx+ tanx) + C. Since y(0) = 1,

0 = ln(1 + 0) + C =⇒ C = 0. (+2 points)

Now,

y = sech(ln(secx+ tanx)) =2eln(sec x+tan x)

e2 ln(sec x+tan x) + 1

=2(secx+ tanx)

sec2 x+ tan2 x+ 2 secx tanx+ 1=

2(secx+ tanx)

2 secx(secx+ tanx)

= cosx (+2 points)

Solution.(a)-2.

From the given formula,∫−1

y√

1− y2dy =

∫1

cosxdx. (+1 point)

By integrating each side with a subtitution y = cos θ,

(LHS) =

∫sin θ

cos θ sin θdθ =

∫1

cos θdθ = ln | sec θ + tan θ|+ C1, (+3 points)

(RHS) = ln(secx+ tanx) + C2. (∵ 0 ≤ x < π/2) (+2 points)

So we have ln | sec θ + tan θ| = ln(secx+ tanx) + C. Since y(0) = 1,

ln |1 + 0| = ln(1 + 0) = C =⇒ C = 0. (+2 points)

Comparing both sides, we obtain a solution θ = x. That is,

y = cosx. (+2 points)

Grading Policy

• -1 point for a minor miss in computation.

• For the second solution, if there is a gap when deriving y = cosx, at most -3 points.

• Though one uses the equation y = cosx in (b), no point is given for other answer in (a).

•∫

1

y√

1−y2= sech−1y + C is not correct unless stating a range of sech−1y otherwise.

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2014 Spring Calculus 1 Midterm Solution TA. Jeong-seop Kim

Solution.(b)-1.

We know

(f−1)′(y) =1

f ′(f−1(y)). (+2 points)

Thus,

f ′(0.8) =1

− sin(cos−1(0.8))=

−1√1− cos2(cos−1(0.8))

=−1√

1− 0.82(+5 points)

= − 1

0.6= −5

3. (+3 points)

Solution.(b)-2. — starting from y = sech(ln(secx+ tanx))

We know

(f−1)′(y) =1

f ′(f−1(y)). (+2 points)

Here,

f ′(x) = −sech(lnT ) · tanh(lnT ) · secx tanx+ sec2 x

secx+ tanx

= −sech(lnT ) · tanh(lnT ) · secx (+2 points)

where T = secx+ tanx. From f−1(0.8) = x0 ⇔ f(x0) = 0.8,

0.8 = sech(lnT ) =2elnT

e2 lnT + 1=

2T

T 2 + 1

=⇒ T = 2 or T =1

2=⇒ T = 2 (∵ 0 < x ≤ π/2),

and hence secx0 + tanx0 = secx0 +√

sec20 x− 1 = 2. Thus, secx0 = 1.25. (+3 points)

By plugging in secx = 1.25 and sech(lnT ) = y = 0.8, we get

(f−1)′(0.8) =1

−0.8×√

1− 0.82 × 1.25= − 1

0.6= −5

3. (+3 points)

Solution.(b)-3. — using given formula in (a)

When y = 0.8, cosx = 0.8(needs some calculation!). Thus, (+3 points)

(f−1)′(0.8) =1

dy/dx

∣∣∣∣y=0.8

=cosx

−y√

1− y2(+4 points)

= − 0.8

0.8×√

1− 0.82= − 1

0.6= −5

3. (+3 points)

Grading Policy

• -2 points for a computation mistake.

• On the second line of the third solution, one can get only +2 points if he or she does not

take the reciprocal of dydx .

• There is no partial point in the last numerical calculation if one has a critical miss in the

above steps.

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Calculus I Midterm TA : Jeongseok Oh

7

20 points

Decide whether ∫ π2

0

x cosx− sinx

x sinxdx

converges or not. If it converges, find its value.

Solution. It converges. By definition,∫ π2

0

x cosx− sinx

x sinxdx = lim

a→0

∫ π2

a

x cosx− sinx

x sinxdx.

On the open interval (0, π2 + ε) for sufficiently small ε > 0 with coordinate x, it is easy to see

thatx cosx− sinx

x sinxdx = d(ln(

sinx

x)).(+15 points)

So, for 0 < a < π2 ,∫ π

2

a

x cosx− sinx

x sinxdx =

∫ x=π2

x=a

d(ln(sinx

x)) =

[ln(

sinx

x)

]x=π2

x=a

= ln2

π− ln

sin a

a.

Since ln : R>0 → R is continuous,

lima→0

lnsin a

a= ln lim

a→0

sin a

a= ln 1 = 0.(+3 points)

Therefore, the convergence is∫ π2

0

x cosx− sinx

x sinxdx = lim

a→0

∫ π2

a

x cosx− sinx

x sinxdx = ln

2

π.(+2 points)

• The answer without justifications gets no points.

• You can use any differentials to calculate the integral∫ π2

a

x cosx− sinx

x sinxdx.

If you calculate this correctly, then you can get 15 points.

• If all your methods are right, but the answer is wrong, then the penalty is -2.

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2014 Calculus 1 Mid-term Solution TA. Huisang Noh

8

20 points

Evaluate ∫sec2 x

(secx+ tanx)3/2

Solution.

Let

I =

∫sec2 x

(secx+ tanx)3/2

Using integration by parts,

I = tanx(secx+ tanx)−3/2 +3

2

∫tanx secx(secx+ tanx)−3/2dx

and ∫tanx secx(secx+ tanx)−3/2dx = secx(secx+ tanx)−3/2

+3

2

∫sec2 x(secx+ tanx)−3/2dx

That is,

I = tanx(secx+ tanx)−3/2 +3

2secx(secx+ tanx)−3/2 +

9

4I

Hence,

I = − 4 tanx+ 6 secx

5(secx+ tanx)3/2+ C

Grading Policy

• Simple mistakes: -3pt (for each)

• integration constant: -2pt

• A wrong answer: -5pt

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Calculus I Midterm Solution and Scoring Criteria TA. Yunbae Kim

9. Determine all positive values of p and q for which the following integral converges:∫ ∞0

x−p(1 + x)−qdx.

Solution: ∫ ∞0

x−p(1 + x)−qdx =

∫ 1

0

x−p(1 + x)−qdx+

∫ ∞1

x−p(1 + x)−qdx.

(+4 points)

i) On 0 < x < 1, since

0 < x−p2−q ≤ x−p(1 + x)−q ≤ x−p,

the integrals of x−p(1 + x)−q and x−p are both converge or both diverge.

(Or, let x = 1/t, then ∫ 1

0

x−p(1 + x)−qdx =

∫ ∞1

tp+q−2(1 + t)−qdt.

Also, we have

limt→∞

tp+q−2(1 + t)−q

tp−2= 1.

Hence, the integrals of tp+q−2(1 + t)−q and tp−2 are both converge or both diverge on

1 < t <∞. ) (+4 points)

Since∫ 1

0x−pdx converges only when p < 1,

∫ 1

0x−p(1 + x)−qdx also converges only when

p < 1.

(Or, since∫∞1

tp−2dt converges only when p− 2 < −1, i.e., p < 1,∫∞1

tp+q−2(1 + t)−qdt also

converges only when p < 1. )

(+4 points)

ii) On 1 ≤ x <∞, since

0 < (1 + x)−(p+q) ≤ x−p(1 + x)−q ≤ x−(p+q)

and ∫ ∞2

x−(p+q)dx =

∫ ∞1

(1 + x)−(p+q)dx ≤∫ ∞1

x−p(1 + x)−qdx ≤∫ ∞1

x−(p+q)dx,

the integrals of x−p(1 + x)−q and x−(p+q) are both converge or both diverge.

(Or, since

limx→∞

x−p(1 + x)−q

x−(p+q)= 1,

the integrals of x−p(1 + x)−q and x−(p+q) are both converge or both diverge. )

(+4 points)

Since∫∞1

x−(p+q)dx converges only when p+ q > 1,∫∞1

x−p(1 + x)−qdx also converges only

when p+ q > 1. (+4 points)

By combining i) and ii),∫∞0

x−p(1 + x)−qdx converges only when 0 < p < 1, p + q > 1, and

q > 0.

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