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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 =
- 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.
- 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
- I = x^2
(ln x)^2 + ln x +
+ C
- 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.
- I = 2 ln
correct
- I = ln
- I = ln
- I = 4 ln
- I = 4 ln
- 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
2. I =
π 2
− 2 ln 2
- I = π + 2 ln 2
- I = π − 2 ln 2
5. I =
π 2
− ln 2 correct
- 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.
- I = tan−^1
x + 3 2
+ C
- I = 3 tan−^1
x − 3 2
+ C
- I = 3 sin−^1
x − 3 2
+ C
- I = 3 sin−^1
x − 2 3
- I = sin−^1
x + 3 2
+ C
- I = sin−^1
x + 2 3
+ C
- I = tan−^1
x + 2 3
+ C
- 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.
- I = 4e^3 correct
- I = 6e^9
- I = 4e^3 − 2 e
- I = 6e^3
- I = 4e^3 + 2e
- 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
- I = x^2 − 6 x + 36 ln |x + 6| + C
- I = x^2 + 6x + 36 ln |x + 6| + C
4. I =
x^2 − 6 x + 36 ln |x + 6| + C correct
- I = x^2 − 6 x + 6 ln |x + 6| + C
6. I =
x^2 2
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.
- I does not converge
2. I =
e−^27
- I = 2e−^27
4. I =
e^27
5. I =
e^27
6. I =
e−^27 correct
Explanation:
P
x
z
y
- fx > 0 , fy = 0
- fx > 0 , fy > 0 correct
- fx = 0 , fy > 0
- fx < 0 , fy = 0
- fx = 0 , fy = 0
- fx = 0 , fy < 0
- fx < 0 , fy > 0
- 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.
- fxxfyy − (fxy)^2 = 6x − 4
- fxxfyy − (fxy)^2 = 6x + 4
- fxxfyy − (fxy)^2 = 12x − 4 correct
- fxxfyy − (fxy)^2 = 12x − 2
- 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
- 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
