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Exam 2 with Answer Key for Integral Calculus | M 408L, Exams of Mathematics

Material Type: Exam; Class: INTEGRAL CALCULUS; Subject: Mathematics; University: University of Texas - Austin; Term: Fall 2010;

Typology: Exams

2009/2010

Uploaded on 12/07/2010

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Version 023 L EXAM 2 meth (54960) 1
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page find all choices
before answering.
001 10.0 points
Evaluate the definite integral
I=Z2
2
6
xx21dx .
1. I=1
2
2. I=π
3. I=3
4π
4. I=1
2πcorrect
5. I= 1
6. I=3
4
Explanation:
Set x= sec u. Then
dx = sec utan u du , x21 = tan2u ,
while
x=2 =u=π
4,
x= 2 =u=π
3.
In this case,
I= 6 Zπ/3
π/4
sec utan u
sec utan udu =Zπ/3
π/4
6du .
Consequently,
I= 6π
3π
4=1
2π.
002 10.0 points
Determine the integral
I=Zx(ln x)2dx .
1. I=x2(ln x)2ln x+1
2+C
2. I=1
2x2(ln x)2ln x+1
2+Ccorrect
3. I=1
2x2(ln x)2+ ln x+1
2+C
4. I=x2(ln x)2+ ln x+1
2+C
5. I=x2(ln x)2+ ln x1
2+C
6. I=1
2x2(ln x)2ln x1
2+C
Explanation:
After integration by parts,
Zx(ln x)2dx =1
2x2(ln x)2Zx21
xln x dx
=1
2x2(ln x)2Zxln x dx.
But after integration by parts once again,
Zxln x dx =1
2x2ln x1
2Zx21
xdx
=1
2x2ln x1
2Zx dx
=1
2x2ln x1
4x2+C.
Thus
Zx(ln x)2dx
=1
2x2(ln x)21
2x2ln x+1
4x2+C.
Consequently,
I=1
2x2(ln x)2ln x+1
2+C .
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This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

001 10.0 points

Evaluate the definite integral

I =

√ 2

x

x^2 − 1

dx.

1. I =

  1. I = π

3. I =

π

4. I =

π correct

5. I = 1

6. I =

Explanation: Set x = sec u. Then

dx = sec u tan u du , x^2 − 1 = tan^2 u ,

while

x =

2 =⇒ u =

π 4

x = 2 =⇒ u =

π 3

In this case,

I = 6

∫ (^) π/ 3

π/ 4

sec u tan u sec u tan u

du =

∫ (^) π/ 3

π/ 4

6 du.

Consequently,

I = 6

( (^) π

3

π 4

π.

002 10.0 points

Determine the integral

I =

x(ln x)^2 dx.

  1. I = x^2

(ln x)^2 − ln x +

+ C

2. I =

x^2

(ln x)^2 −ln x+

+C correct

3. I =

x^2

(ln x)^2 + ln x +

+ C

  1. I = x^2

(ln x)^2 + ln x +

+ C

  1. I = x^2

(ln x)^2 + ln x −

+ C

6. I =

x^2

(ln x)^2 − ln x −

+ C

Explanation: After integration by parts, ∫ x(ln x)^2 dx =

x^2 (ln x)^2 −

x^2

x

ln x dx

x^2 (ln x)^2 −

x ln x dx.

But after integration by parts once again, ∫ x ln x dx =

x^2 ln x −

x^2

x

dx

x^2 ln x −

x dx

x^2 ln x −

x^2 + C.

Thus ∫ x(ln x)^2 dx

x^2 (ln x)^2 −

x^2 ln x +

x^2 + C.

Consequently,

I =

x^2

(ln x)^2 − ln x +

+ C.

keywords: integration by parts, log function

003 10.0 points

Evaluate the integral

I =

∫ (^) π/ 4

0

2 sec^4 x dx.

1. I = 2

2. I =

3. I =

4. I =

correct

5. I = 3

Explanation: Since

(tan x)′^ = sec^2 x , sec^2 x = 1 + tan^2 x ,

use of the substitution u = tan x is suggested. For then du = sec^2 x dx ,

while x = 0 =⇒ u = 0 ,

x =

π 4

=⇒ u = 1.

Thus

I = 2

∫ (^) π/ 4

0

(tan^2 x + 1) sec^2 x dx

0

(u^2 + 1) du = 2

[ 1

u^3 + u

] 1

0

Consequently,

I =

004 10.0 points Evaluate the integral

I =

x^3 + x

dx.

  1. I = 2 ln

correct

  1. I = ln
  1. I = ln
  1. I = 4 ln
  1. I = 4 ln
  1. I = 2 ln

Explanation: By partial fractions,

4 x^3 + x

A

x

Bx + C x^2 + 1

To determine A, B, and C multiply through by x^3 + x: for then

4 = A(x^2 + 1) + x(Bx + C)

= (A + B)x^2 + Cx + A ,

which after comparing coefficients gives

A = −B , C = 0 , A = 4.

Thus

I = 4

1

x

x x^2 + 1

dx

[

ln x −

ln(x^2 + 1)

] 2

1

= 2

[

ln

( (^) x 2 x^2 + 1

) ] 2

1

Consequently,

I = 2 ln

1. I =

π 2

  • ln 2

2. I =

π 2

− 2 ln 2

  1. I = π + 2 ln 2
  2. I = π − 2 ln 2

5. I =

π 2

− ln 2 correct

  1. I = π + ln 2

Explanation: First set x = 2u. Then dx = 2du, while

x = 0 =⇒ u = 0,

x = 2 =⇒ u = 1.

In this case

I = 2

0

tan−^1 u du.

We now use integration by parts. For then

I = 2

[

u tan−^1 u

] 1

0

0

u 1 + u^2

du

π −

[

ln(1 + u^2 )

] 1

0

Consequently,

I =

π 2

− ln 2.

008 10.0 points

Determine the indefinite integral

I =

5 + 4x − x^2

dx.

  1. I = tan−^1

x + 3 2

+ C

  1. I = 3 tan−^1

x − 3 2

+ C

  1. I = 3 sin−^1

x − 3 2

+ C

  1. I = 3 sin−^1

x − 2 3

  • C correct
  1. I = sin−^1

x + 3 2

+ C

  1. I = sin−^1

x + 2 3

+ C

  1. I = tan−^1

x + 2 3

+ C

  1. I = 3 tan−^1

x − 2 3

+ C

Explanation: After completion of the square we see that

5 + 4x − x^2 = 9 − (x − 2)^2.

Thus

I =

9 − (x − 2)^2

dx.

Now set x − 2 = 3 sin u. Then

dx = 3 cos u du

and

9 − (x − 2)^2 = 9

1 − sin^2 u

= 9 cos^2 u.

This shows that

I =

9 cos u 3 cos u

du =

3 du = 3u + C.

Consequently,

I = 3 sin−^1

x − 2 3

+ C

with C an arbitrary constant.

009 10.0 points Evaluate the definite integral

I =

e

√ t (^) dt.

  1. I = 4e^3 correct
  2. I = 6e^9
  3. I = 4e^3 − 2 e
  4. I = 6e^3
  5. I = 4e^3 + 2e
  6. I = 6e^9 + 2e

Explanation: Let w =

t, so that

t = w^2 , dt = 2w dw.

Then

I =

1

2 w ew^ dw.

To evaluate this last integral we use now use integration by parts:

I =

[

2 w ew^

] 3

1

1

ew^ dw

= 6e^3 − 2 e − 2(e^3 − e).

Consequently,

I = 4e^3.

010 10.0 points Determine the indefinite integral

I =

x^2 x + 6

dx.

1. I =

x^2 2

  • 6x + 6 ln |x + 6| + C
  1. I = x^2 − 6 x + 36 ln |x + 6| + C
  2. I = x^2 + 6x + 36 ln |x + 6| + C

4. I =

x^2 − 6 x + 36 ln |x + 6| + C correct

  1. I = x^2 − 6 x + 6 ln |x + 6| + C

6. I =

x^2 2

  • 6x + 36 ln |x + 6| + C

Explanation: After division we see that

x^2 x + 6

x^2 − 36 x + 6

x + 6

= x − 6 +

x + 6

Thus

I =

x − 6 +

x + 6

dx.

Consequently,

I =

x^2 2

− 6 x + 36 ln |x + 6| + C

with C an arbitrary constant.

011 10.0 points Determine if the improper integral

I =

3

2 xe−^3 x

2 dx

converges, and if it does, find its value.

  1. I does not converge

2. I =

e−^27

  1. I = 2e−^27

4. I =

e^27

5. I =

e^27

6. I =

e−^27 correct

Explanation:

P

x

z

y

  1. fx > 0 , fy = 0
  2. fx > 0 , fy > 0 correct
  3. fx = 0 , fy > 0
  4. fx < 0 , fy = 0
  5. fx = 0 , fy = 0
  6. fx = 0 , fy < 0
  7. fx < 0 , fy > 0
  8. fx < 0 , fy < 0

Explanation: The value of fx at P is the slope of the tangent line to graph of f at P in the x- direction, while fy is the slope of the tangent line in the y-direction. Thus the sign of fx indicates whether f is increasing or decreasing in the x-direction, or whether the tangent line in that direction at P is horizontal.

Similarly, the value of fy at P is the slope of the tangent line at P in the y-direction, and so the sign of fy indicates whether f is increasing or decreasing in the y-direction, or whether the tangent line in that direction at P is horizontal.

From the graph it thus follows that at P

fx > 0 , fy > 0.

keywords: surface, partial derivative, first or- der partial derivative, graphical interpreta- tion

014 10.0 points

Determine fxxfyy − (fxy)^2 when

f (x, y) =

x^3 + 3y^2 + 8x + 5y + 2xy.

  1. fxxfyy − (fxy)^2 = 6x − 4
  2. fxxfyy − (fxy)^2 = 6x + 4
  3. fxxfyy − (fxy)^2 = 12x − 4 correct
  4. fxxfyy − (fxy)^2 = 12x − 2
  5. fxxfyy − (fxy)^2 = 12x + 4

Explanation: After differentiation once

fx = x^2 + 8 + 2y, fy = 6y + 5 + 2x.

After differentiating a second time therefore, we see that

fxx = 2x, fxy = 2, fyy = 6.

Consequently,

fxxfyy − (fxy)^2 = 12x − 4.

015 10.0 points Determine the value of the iterated integral

I =

0

{∫^2

1

(7 + 2xy) dx

dy.

1. I =

2. I =

3. I =

4. I =

5. I =

correct

Explanation: Integrating with respect to x and holding y fixed, we see that

∫ (^2)

1

(7 + 2xy) dx =

[

7 x + x^2 y

]x= x=

Thus

I =

0

7 + 3y

dy =

[

7 y +

y^2

] 3

0

Consequently,

I =

keywords:

016 10.0 points

Evaluate the double integral

I =

A

5 + x^2 1 + y^2

dxdy

when

A =

(x, y) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1

1. I =

π

2. I =

π

  1. I = 3π

4. I =

π correct

5. I =

π

Explanation: Since

A =

(x, y) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1

is a rectangle with sides parallel to the coor- dinate axes, the double integral can be repre- sented as the iterated integral

I =

0

0

5 + x^2 1 + y^2

dx

dy.

Now ∫ (^2)

0

5 + x^2 1 + y^2

dx =

1 + y^2

[

5 x +

x^3

] 2

0

Thus

I =

0

1 + y^2

dy =

[

tan−^1 y

] 1

0

Consequently,

I =

π.

017 10.0 points

Evaluate the integral

I =

0

∫ (^) x^2

0

(x − 3 y) dydx.

1. I = −

2. I =

3. I = −

4. I = −

5. I = −

correct