MAT 266: Test 2 Solutions
1. AlgebraicallyZdetermine whether the given integral converges or diverges, and if it converges, to
2
dx
√
what value:
3
x−2
1
A) Diverges
B) Converges to −2/3
C) Converges to −3/2
D) Converges to −1.488
E) None of the above
The integral is improper at 2. We compute:
Z 2
Z t
dx
dx
√
√
= lim−
3
3
x − 2 t→2 1
x−2
1
t
3
2/3 = lim− (x − 2) t→2 2
1
3
2/3
= lim− ((t − 2) − (−1)2/3 )
t→2 2
3
= (0 − 1)
2
3
=−
2
So the answer is C.
2. Find the area of the region bounded by the curves y = x2 + 2x − 1 and y = 3x + 5.
A)
B)
C)
D)
95
3
95
6
125
3
125
6
E) None of the above
First we find where the curves intersect:
x2 + 2x − 1 = 3x + 5 ⇐⇒ x2 − x − 5 = 0
⇐⇒ (x − 3)(x + 2) = 0
⇐⇒ x = 3 or x = −2
In this region, y = 3x + 5 is on top. So the area is
Z 3
Z
2
(3x + 5) − (x + 2x − 1) dx =
3
x + 6 − x2 dx
−2
3
x3 x2
+ 6x − =
2
3 −2
9
27
4
−8
= + 18 −
− ( − 12 −
)
2
3
2
3
125
=
6
−2
So the answer is D.
3. Set up the integral using the method of cylindrical shells to find the volume of the solid
obtained by rotating the region bounded by the curves y = 4x − x2 and y = x around the y-axis.
R3
A) 0 2πx(3x − x2 ) dx
R3
B) 0 2πx(x2 − 3x) dx
R3
C) 0 2πx(5x − x2 ) dx
R3
D) 0 2πx(x2 − 5x) dx
E) None of the above
First, we find the intersection of the curves:
4x − x2 = x ⇐⇒ x2 − 3x = 0
⇐⇒ (x − 3)x = 0
⇐⇒ x = 0 or x = 3
In this range, 4x − x2 is on top. We are chopping up the x-axis. The distance from a shell to
the axis of rotation is x, so we get
Z 3
Z 3
2
2πx(3x − x2 ) dx
2πx(4x − x − x) dx =
0
0
So, the answer is A.
4. Determine whether the sequence converges or diverges, and if it converges, find the limit:
2n+3
an = n+1
5
A) Diverges
B) Converges to 0
C) Converges to 1
D) Converges to 2/5
E) None of the above
2n+3
8 2n
=
lim
n→∞ 5n+1
n→∞ 5 5n
n
8 2
= lim
n→∞ 5
5
n
8
2
= lim
5 n→∞ 5
8
= ·0=0
5
lim
The key point is that 0 < 2/5 < 1 for the final limit. So, the answer is B.
5. Determine whether the series converges or diverges, and if it converges, find the sum:
∞
X
(3n + 5)n
(4n − 2)2
n=1
A) Diverges
B) Converges to 0
C) Converges to 3/4
D) Converges to 3/16
E) None of the above
The point here is that if a series converges, then the terms go to 0 and
3n2 + 5n
(3n + 5)n
=
lim
n→∞ 16n − 16n + 4
n→∞ (4n − 2)2
3
=
16
lim
The terms converge to 3/16, which is not 0, so the series diverges. The answer is A.
6. Determine whether the series converges or diverges, and if it converges, find the sum:
∞
X
3 + 2n
n=1
3n
A) Diverges
B) Converges to 5/2
C) Converges to 7/2
D) Converges to 11/2
E) Converges to 13/2
The terms being summed can be split as the sum of two parts, each of which is a geometric
series:
∞
X
3
1
3
=
1 =
n
3
2
1− 3
n=1
and
∞
X
2n
n=1
So
3n
=
∞ n
X
2
3
n=1
∞
X
3 + 2n
n=1
3n
=
=
2/3
=2
1 − 32
7
3
+2=
2
2
and the answer is C.
7. Set up and simplify, but do not evaluate, an integral for the length of the curve y = ln(sec(x))
for 0 ≤ x ≤ π4 .
R π/4
A) 0 sec(x) dx
R π/4
B) 0 sec2 (x) dx
R π/4
C) 0 csc(x) dx
R π/4
D) 0 csc2 (x) dx
E) None of the above
d
1
ln(sec(x)) = sec(x)
· (sec(x) tan(x)) = tan(x). So, we want
dx
Z π/4 p
Z π/4 p
Z π/4
2
2
1 + (tan(x)) dx =
sec(x) dx =
sec(x) dx
First, the derivative:
0
0
0
and sec(x) is positive in this interval (so we took the right square root). So, the answer is A.
√
8. Find the exact volume of the solid obtained by rotating the region bounded by y = x+1, y = 2,
x = 0, about the line y = 2.
√
The curves y = 2 and y = x + 1 meet at x = 1. If we do this by disk/washer, we get
Z 1
Z 1
√
√
2
1 − 2 x + x dx
π(2 − ( x + 1)) dx = π
0
0
!
3/2
2 1
2x
x
=π x−2
+ 3
2 0
4 1
π
=π 1− + −0 =
3 2
6
√
If we do it by shells, we need that y = x + 1 becomes x = (y − 1)2 . Note, the curve hits the
y-axis at y = 1. Then, the volume is
Z 2
Z 2
2
2π(2 − y)(y − 1) dy = 2π
2 − 5y + 4y 2 − y 3 dy
1
1
2 !
5 2 4 3 y 4 = 2π 2y − y + y − 2
3
4 1
2
7
2π
π
= 2π
−
=
=
3 12
12
6
9. A tank in the shape of a hemisphere of radius 3 meters is full of water. Given that a cubic
meter of water weighs 9800 Newtons, set up, but do not evaluate the integral for the work
required to pump the water out of a spout which is 1 meter above the top of the tank. Note, the
acceleration due to gravity is 9.8m/s2 .
We will measure distance from the top of the tank with h, so h = 0 is the top of the tank and
h = 3 is the bottom of the tank. At height h, the cross-section has radius r where r2 + h2 = 32 ,
so a horizontal slice has volume ≈ πr2 δh = π(9 − h2 )δh. Multipling by 9800, we get the weight
of a slice. Multiply by the distance traveled (h + 1) to get the work for our slice, add them up
and take a limit to get
Z
3
9800π(9 − h2 )(h + 1) dh
0
10. Set up, but do not evaluate
√ an integral to find the volume of the solid generated by revolving
the region bounded by y = 25 − x2 and y = 4 about the x-axis.
Intesection points are when
√
4 = 25 − x2 ⇐⇒ x2 = 25 − 16 = 9 ⇐⇒ x = ±3 .
By disk/washer:
Z
3
Z
√
2
2
2
π(( 25 − x ) − 4 ) dx =
−3
3
π(9 − x2 ) dx
−3
√
and by shells, it helps to recognize that y = 25 − x2 is the top of a circle of radius 5, so the
upper bound will be 5. Then we get:
Z 5
Z 5 p
p
p
2
2
2πy( 25 − y − (− 25 − y )) dy = 4π
y 25 − y 2 dy
4
For the record, both evaluate to 36π.
4