quest hw1 solutions

lui (el22284) Quest HW1 perutz (56210) 1

This print-out should have 10 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.Each question receives a score between +10

and -10 depending on how many tries youneed. The scores are arranged so that a ran-dom guesser will on average score zero. Yourscore for the whole assignment will never benegative, and you get bonus participationcredit for attempting all questions. Thisscoring system is designed so that studentsgrades dont get too bunched up (its veryhard to assign As, Bs etc. fairly at the end ofthe semester if everyone has almost the samescore).

001 10.0 points

Determine if the improper integral

I =

3

7

(x+ 6)2dx

converges, and if it does, compute its value.

1. I =7

9correct

2. I =7

10

3. I doesnt converge

4. I = 2

5. I = 79

Explanation:

The integral is improper because of the infi-nite interval of integration. To overcome thiswe set

I = limt

t3

7

(x+ 6)2dx

whenever the limit on the right hand sideexists. Now t

3

7

(x+ 6)2dx =

[ 7x+ 6

]t3

= 7t+ 6

+7

9.

Since

limt

7

t+ 6= 0,

it follows that the improper integral convergesand that

I =

3

7

(x+ 6)2dx =

7

9.

002 10.0 points


I =

0

2

7x 3 dx

converges, and if it does, find its value.

1. I =2

7

2. I =2

3

3. I =7

2

4. I =3

2

5. I is not convergent correct

Explanation:

The integral is improper because the in-terval of integration is infinite. To test forconvergence, therefore, we have to check ifthe limit

limt

0t

2

7x 3 dx

exists. But

0t

2

7x 3 dx =[ 27ln (|7x 3|)

]0t

=( 27ln(3) 2

7ln (|7t 3|)

).


On the other hand,

limt

ln (|7t 3|) = .

Consequently, the integral

I is not convergent .

003 10.0 points

Determine if

I =

3

x5x2 4 dx


1. I =5 54/5

8

2. I does not converge correct

3. I =54/5

8

4. I = 5 54/5

4

5. I = 5 54/5

8

6. I = 54/5

Explanation:

The integral I is improper because of theinfinite interval of integration. Thus I willconverge if

limt

t3

x5x2 4 dx

exists. To evaluate this last integral, we usesubstitution, setting u = x2 4. For then

du = 2x dx ,

while

x = 3 = u = 5 ,x = t = u = t2 4 .

In this case

t3

x5x2 4 dx =

1

2

t245

1

u1/5du

=5

8

[u4/5

]t245

=5

8

((t2 4)4/5 54/5

).

However,

limt

(t2 4)4/5 = .

Consequently,

I does not converge .

004 10.0 points


I =

12xe2x

2

dx

converges, and if it does, find its value.

1. I =1

2e2

2. I does not converge

3. I = 2e2

4. I = e2

5. I = e2

6. I =1

2e2 correct

Explanation:

The integral

I =

12xe2x

2

dx

is improper because of the infinite range ofintegration. To overcome this we restrict to afinite interval of integration and consider thelimit

I = limt

It, It =

t1

2xe2x2

dx.


To evaluate It we use the substitution u = x2.

For then

It =

t21

e2u du = 12

[e2u

]t21

=1

2

(e2 e2t2

).

But,

limx

eax2

= 0,

for any a > 0, so

limt

e2t2

= 0.

Consequently, I converges and

I =1

2e2 .

005 10.0 points


I =

e

1

t ln(4t)dt


1. I = 4

2. I = 4e

3. I =4

e

4. I =1

4

5. I does not converge correct

Explanation:

The integral is improper because of the in-finite interval of integration, so we set

I = limn

In, In =

ne

1

t ln(4t)dt,

whenever the limit exists. Now with the sub-stitution u = ln(4t) we see that

1

t ln(4t)dt =

1

udu = ln(u) + C.

Consequently,

In =[ln(ln(4t))

]ne=(ln(ln(4n))ln(ln(4e))

).

But

limn

ln(ln(4n)) = ,

so limn

In does not exist. Hence

I does not converge .

006 10.0 points


I =

10

4 sin1(x)1 x2 dx

is convergent or divergent, and if convergent,find its value.

1. I is divergent

2. I =1

4

3. I =1

2pi2 correct

4. I =1

4pi2

5. I =1

2

Explanation:

The integral is improper because

limx 1

sin1(x)1 x2 = .

Thus we have to check if

limt 1

t0

4 sin1(x)1 x2 dx


is convergent or divergent. To determine

4 sin1(x)

1 x2 dx

set u = sin1(x). Then

du =1

1 x2 dx ,

and so

4 sin1(x)1 x2 dx = 4

u du = 2u2 + C ,

from which it follows that t0

4 sin1(x)1 x2 dx

= 2[(sin1(x))2

]t0= 2(sin1(t))2 .

Now

limt 1

(sin1(t))2 =pi2

4.

Consequently, I is convergent and

I =1

2pi2 .

007 10.0 points

When f, g, F and G are functions such that

limx 1

f(x) = 0, limx 1

g(x) = 0,

limx 1

F (x) = 2, limx 1

G(x) = ,

which, if any, of

A. limx 1

f(x)g(x) ,

B. limx 1

F (x)g(x) ,

C. limx 1

g(x)

G(x).

are indeterminate forms?

1. C only

2. A and C only

3. A only

4. all of them

5. B only

6. A and B only

7. B and C only

8. none of them correct

Explanation:

A. By properties of limits

limx 1

f(x)g(x) = 0 0 = 0 ,

so this limit is not an indeterminate form.

B. By properties of limits

limx 1

F (x)g(x) = 20 = 1 ,


C. By properties of limits

limx 1

g(x)

G(x)=

0

= 0 ,


008 10.0 points

Determine if the limit

limx 1

x4 1x3 1

exists, and if it does, find its value.

1. limit =4

3correct

2. limit =

3. limit =3

4

4. none of the other answers

5. limit = Explanation:


Set

f(x) = x4 1, g(x) = x3 1 .Then

limx 1

f(x) = 0, limx 1

g(x) = 0 ,

so LHospitals rule applies. Thus

limx 1

f(x)

g(x)= lim

x 1

f (x)

g(x).

But

f (x) = 4x3, g(x) = 3x2 .

Consequently,

limit =4

3.

009 10.0 points

Use LHospitals Rule to determine which ofthe inequalitites

A.100

x< ex,

B. ex < x2 + 100,

C. e2x > xex + 100,

holds for all large x.

1. none of them

2. B and C only

3. C only correct

4. all of them

5. A and C only

6. A and B only

7. B only

8. A only

Explanation:

The notion of limit at infinity tells us thatif

limx

f(x)

g(x)= ,

thenf(x)

g(x)> 1

for all large x. But then

f(x) > g(x)

holds for all large x so long as g(x) > 0 forlarge x, allowing us to multiply through theinequality by g(x).Similarly, if

limx

f(x)

g(x)= 0 ,

thenf(x)

g(x)< 1

for all large x, so if g(x) > 0 for all large x,then

f(x) < g(x)

for all large x.On the other hand, if

limx

f(x) = = limx

g(x) ,

then we can use LHospitals Rule to deter-mine

limx

f(x)

g(x).

Similarly, if

limx

f(x) = 0 = limx

g(x) ,

then we can use LHospitals Rule to deter-mine

limx

f(x)

g(x).

In this way, we can use LHospitals Rule tocompare the rates of growth or decay of f(x)and g(x) when x.For the three given inequalities, therefore,

we have to choose appropriatef and g andmake sure that g(x) > 0 for all large x.


A. FALSE: set

f(x) = ex , g(x) =100

x.

Then

limx

f(x) = 0 = limx

g(x) ,

and by applying LHospitals Rule we see that

limx

f(x)

g(x)= 0 .

Thus the inequality

100

x> ex ,

not

100

x< ex ,


B. FALSE: set

f(x) = ex , g(x) = x2 + 100 .

Then

limx

f(x) = = limx

g(x) ,

and by applying LHospitals Rule twice wesee that

limx

f(x)

g(x)= .

Thus the inequality

ex > x2 + 100 ,

not

ex < x2 + 100 ,


C. TRUE: set

f(x) = e2x , g(x) = xex + 100 .

Then

limx

f(x) = = limx

g(x) ,

and by applying LHospitals Rule twice wesee that

limx

f(x)

g(x)= .

Thus the inequality

e2x > xex + 100


keywords:

010 10.0 points

Find the value of

limx 0

1 cos(3x)4 sin2(2x)

.

1. limit does not exist

2. limit =11

32

3. limit =5

16

4. limit =9

32correct

5. limit =3

8

Explanation:

Set

f(x) = 1 cos(3x), g(x) = 4 sin2(2x) .Then f, g are differentiable functions suchthat

limx 0

f(x) = limx 0

g(x) = 0 .

Thus LHospitals Rule can be applied:

limx 0

f(x)

g(x)= lim

x 0

f (x)

g(x)

= limx 0

3 sin(3x)

16 sin(2x) cos(2x).

To compute this last limit we can either applyLHospitals Rule again or use the fact that

limx 0

sin(3x)

x= 3, lim

x 0

sin(2x)

x= 2 .

Consequently,

limx 0

1 cos(3x)4 sin2(2x)

=9

32.

quest hw1 solutions

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