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This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points
Determine if
I =
0
f (x) dx
is convergent or divergent when
f (x) =
x−^1 /^3 , x ≤ 1 , x , 1 ≤ x ≤ 2 ,
and find its value if convergent.
- I = 4
2. I =
- I not convergent
- I = 3 correct
5. I =
6. I =
Explanation: The integral is improper because f (x) → ∞ as x → 0 +. Thus
I = lim t → 0 +^
It , It =
t
f (x) dx.
But for t < 1,
It =
t
x^1 /^3
dx +
1
x dx
[ 3
x^2 /^3
] 1
t
[ 1
x^2
] 2
1
=
t^2 /^3 +
On the other hand,
lim t → 0 +^
t^2 /^3 = 0.
Consequently, I is convergent and
I = 3.
002 10.0 points Determine if
I =
3
x √ (^3) x (^2) − 7 dx
converges, and if it does, compute its value.
1. I = −
3 · 22 /^3
2. I = −
3 · 22 /^3
- I does not converge correct
4. I =
3 · 22 /^3
5. I =
22 /^3
6. I = 2^2 /^3
Explanation: The integral I is improper because of the infinite interval of integration. Thus I will converge if
lim t → ∞
∫ (^) t
3
x √ (^3) x (^2) − 7 dx
exists. To evaluate this last integral, we use substitution, setting u = x^2 − 7. For then
du = 2x dx ,
while
x = 3 =⇒ u = 2 ,
x = t =⇒ u = t^2 − 7.
In this case
∫ (^) t
3
x √ (^3) x (^2) − 7 dx =
∫ (^) t^2 − 7
2
u^1 /^3
du
[
u^2 /^3
]t^2 − 7 2
(t^2 − 7)^2 /^3 − 22 /^3
However,
lim t → ∞ (t^2 − 7)^2 /^3 = ∞.
Consequently,
I does not converge.
003 10.0 points Determine if the improper integral
I =
3
4 x (9 + x^2 )^2
dx
converges, and if it does, compute its value.
1. I =
- integral doesn’t converge
3. I =
4. I =
correct
5. I =
Explanation: The integral
I =
3
4 x (9 + x^2 )^2
dx
is improper because of the infinite interval of integration. To overcome this, we truncate and consider the limit
lim t → ∞ It, It =
∫ (^) t
3
4 x (9 + x^2 )^2
dx.
To evaluate It, set u = 9 + x^2. Then
du = 2x dx ,
in which case ∫ 4 x (9 + x^2 )^2
dx = 2
u^2
du.
Thus
It =
∫ (^) t
3
4 x (9 + x^2 )^2
dx
[
9 + x^2
]t
3
9 + t^2
Consequently, since
lim t → ∞
9 + t^2
we see that I converges and that
I = lim t → ∞
∫ (^) t
3
4 x (9 + x^2 )^2
dx =
004 10.0 points
Determine if the improper integral
I =
−∞
3 x e−^2 x
2 dx
is convergent or divergent, and if it is conver- gent, find its value.
1. I =
- I = 0 correct
- I is divergent
4. I =
5. I =
Explanation: The integral is improper because the range if integration is infinite in both directions. Thus we have to check if each of the limits
I 1 = lim s → ∞
−s
3 x e−^2 x
2 dx
Which (if any) of the following integrals is convergent?
− 6
x^2
dx
- none of them
0
x
dx correct
− 6
8 x^2 dx
0
x
x
dx
6
7 x
x dx
Explanation: Each of the improper integrals
I =
0
xp^
dx, I =
− 6
xp^
dx
is convergent if and only if p < 1, while each of
I =
6
xp^
dx, I =
− 6
xp^
dx
is convergent if and only if p > 1. Of the given integrals, therefore, only
I =
0
x
dx
is convergent.
007 10.0 points
Determine if the integral
I =
0
(x − 1)^2 /^3
dx
is convergent or divergent; and if convergent, find its value.
- I is divergent
2. I = 2
- I = 6 correct
4. I = 0
5. I = 3
Explanation: The integral is improper because the func- tion f (x) =
(x − 1)^2 /^3 has a vertical asymptote at x = 1 in the inter- val of integration. Thus we have to consider each of the improper integrals
I 1 =
0
(x − 1)^2 /^3
dx,
I 2 =
1
(x − 1)^2 /^3
dx
separately. Now for t < 1, ∫ (^) t
0
(x − 1)^2 /^3
dx =
[
3(x − 1)^1 /^3
]t
0
= 3(t − 1)^1 /^3 + 3.
But lim t → 1 −
(t − 1)^1 /^3 = 0,
so I 1 is convergent and
I 1 = lim t → 1 −
(t − 1)^1 /^3 + 1
On the other hand, for t > 1, ∫ (^2)
t
(x − 1)^2 /^3
dx =
[
3(x − 1)^1 /^3
] 2
t
= 3 − 3(t − 1)^1 /^3.
But lim t → 1+ (t − 1)^1 /^3 = 0,
so I 2 is convergent and
I 2 = lim t → 1+
1 − (t − 1)^1 /^3
Consequently, I is convergent and
I = I 1 + I 2 = 6.
008 10.0 points
Determine if the improper integral
I =
0
x (3 +
x)^4
dx
exists, and if it does, find its value.
1. I =
correct
2. I =
3. I =
4. I =
5. I =
Explanation: The integral I is improper because the in- terval of integration is infinite and because
lim x → 0
x
Thus we have to truncate both at x = 0 and at x = ∞, so it is convenient to split the integral by setting I = I 1 + I 2 where
I 1 =
0
x (3 +
x)^4
dx
and
I 2 =
1
x (3 +
x)^4
dx.
But first let’s compute the indefinite integral ∫ 7 √ x (3 +
x)^4
dx.
Set u =
x. Then
du dx
x
so ∫ 7 √ x (3 +
x)^4
dx =
(3 + u)^4
du
(3 + u)^3
+ C = −
x)^3
+ C
with C an arbitrary constant. Now
I 1 = lim t → 0+
t
x (3 +
x)^4
dx
= lim t → 0+
[
x)^3
} ] 1
t
On the other hand,
I 2 = lim s → ∞
∫ (^) s
1
x (3 +
x)^4
dx
= lim s → ∞
[
x)^3
} ]s
1
Consequently,
I = I 1 + I 2 =
009 10.0 points
A manufacturer of lightbulbs wants to pro- duce bulbs that last about 350 hours but, of course, some bulbs burn out faster than others. Having taken calculus, he thinks im- proper integrals might help him. Let F (t) be the fraction of the company’s bulbs that burn out in the first t hours of use and let r(t) be the derivative F ′(t) of F. (Note that F (t) always satisfies the inequalities 0 ≤ F (t) ≤ 1.)
Determine if the improper integral
I =
2
7 xe−^4 x (4x − 1)^2
dx
exists, and if it does, find its value.
1. I =
e−^8
2. I =
e−^8
- I does not exist
4. I =
e−^8 correct
5. I =
Explanation: The integral
I =
2
7 xe−^4 x (4x − 1)^2
dx
is improper because of the infinite range of integration. To overcome this, we restrict to a finite interval of integration and consider the limit
lim t → ∞
It, It =
∫ (^) t
2
7 xe−^4 x (4x − 1)^2
dx.
To minimize the algebra involved in comput- ing the value of In, it’s simplest to determine the indefinite integral ∫ 7 xe−^4 x (4x − 1)^2
dx,
and then find the value of the definite integral. We use integration by parts beginning with the fact that
d dx
4 x − 1
(4x − 1)^2
Then ∫ 7 x e−^4 x (4x − 1)^2
dx =
xe−^4 x 4 x − 1
4 x − 1
) (^) d dx
(xe−^4 x) dx
But
d dx
(xe−^4 x) = e−^4 x(1 − 4 x),
so the integral on the right becomes ∫ 1 4 x − 1
e−^4 x(1 − 4 x) dx
e−^4 x^ dx =
e−^4 x^ + C
with C an arbitrary constant. Thus ∫ 7 x e−^4 x (4x − 1)^2
dx = −
7 e−^4 x 16(4x − 1)
+ C.
But then by the Fundamental Theorem of Calculus,
It =
∫ (^) t
2
7 x (4x − 1)^2
e−^4 x^ dx
[
7 e−^4 x 16(4x − 1)
]t
2
( (^) e− 8
8 − 1
e−^4 t 4 n − 1
Since
lim t → ∞
e−^4 t 4 t − 1
the limit lim t → ∞
It exists and
I =
e−^8.
012 10.0 points
Determine fx when
f (x, y) = (2xy − 1)e−xy^.
- fx = y(3 − 2 xy)e−xy^ correct
- fx = y(2xy − 3)e−xy
- fx = y(1 − 2 xy)e−xy
- fx = x(xy − 3)e−xy
- fx = x(3 − xy)e−xy
- fx = y(1 + 2xy)e−xy
- fx = x(1 − xy)e−xy
- fx = x(1 − 2 xy)e−xy
Explanation: From the Product Rule we see that
fx = 2ye−xy^ − y(2xy − 1)e−xy^.
Consequently,
fx = y(3 − 2 xy)e−xy^.
013 10.0 points Determine ∂z ∂y
when z = ex/^3 y.
∂z ∂y
x 3 y^2
ex/^3 y^ correct
∂z ∂y
3 y^2
ex/^3 y
∂z ∂y
x 3 y
ex/^3 y
∂z ∂y
x 3 y
ex/^3 y
∂z ∂y
x 3 y^2
ex/^3 y
∂z ∂y
3 y^2
ex/^3 y
Explanation: Differentiating z with respect to y keeping x fixed we see that
∂z ∂y
= ex/^3 y^ ·
∂(x/ 3 y) ∂y
Consequently,
∂z ∂y
x 3 y^2
ex/^3 y^.
014 10.0 points
Determine h = h(x, y) so that
∂f ∂x
h(x, y) (3 x^2 + 4 y^2 )^2
when
f (x, y) =
5 x^2 y 3 x^2 + 4 y^2
- h(x, y) = 40xy^2
- h(x, y) = 20x^3 y
- h(x, y) = 20xy^2
- h(x, y) = 40x^3 y
- h(x, y) = 20xy^3
- h(x, y) = 40xy^3 correct
Explanation: Differentiating with respect to x using the quotient rule we obtain
∂f ∂x
10 xy(3x^2 + 4 y^2 ) − 30 x^3 y (3x^2 + 4y^2 )^2
Consequently,
h(x, y) = 40xy^3.
015 10.0 points
Find the value of fx and fy at (1, −1) when
f (x, y) =
xy
018 10.0 points
Determine ∂z/∂y when z = z(x, y) is de- fined by
6 x^2 + 3xy + 2yz + 5z^2 = 5.
∂z ∂y
3 x − 2 z 10 z
∂z ∂y
3 x + 2z 10 z
∂z ∂y
3 x + 2z 2 y + 10z
correct
∂z ∂y
3 y − 2 x 2 x + 10y
∂z ∂y
3 x − 2 z 2 x + 10y
Explanation: Differentiating z implicitly with respect to y, we see that
3 x + 2z + 2y
∂z ∂y
∂z ∂y
Consequently,
∂z ∂y
3 x + 2z 2 y + 10z
019 10.0 points
Determine fxx + fyx when
f (x, y) = 2x^2 + 3xy^3 − 4 y^2 + 8.
- fxx + fyx = 2 + 3y
- fxx + fyx = 4x + 4 + y^3
- fxx + fyx = 4 + 9y^2 correct
- fxx + fyx = 4x + 3y^2 − 8 y
- fxx + fyx = 2xy − 8 + 2y^2
Explanation: Using the law for differentiation of mono- mials we see that
fx = 4x + 3y^3 , fxx = 4,
while
fy = 9xy^2 − 8 y, fyx = 9y^2.
Consequently,
fxx + fyx = 4 + 9y^2.
020 10.0 points
Determine the value of fxy at (1, 0) when
f (x, y) = x^2 ye^7 x+4y^.
- fxy(1, 0) = 9
- fxy(1, 0) = 7
- fxy(1, 0) = 7e^9
- fxy(1, 0) = 7 + e^9
- fxy(1, 0) = 9e^7 correct
Explanation: Differentiating with respect to x, we see that
fx = 2xye^7 x+4y^ + 7x^2 ye^7 x+4y,
and hence that
fxy = (2x + 8xy + 7x^2 + 28x^2 y)e^7 x+4y
after differentiating again, but now with re- spect to y. The value of fxy at (1, 0) is thus given by
fxy(1, 0) = 9e^7.
