Math 142
Midterm 2 a Solutions
Part I. (20 points) There are 2 problems, do both of them.
1. Circle T for true or F for false, or fill in the blank line. (Work on scratch paper.)
Z
1
√
dx can be evaluated with the substitution x = 2 tan(θ).
a) The integral
x 4 − x2
T
F
The correct substitution is x = 2 sin(θ).
b) The correct form of the partial fraction decomposition of the rational function
x2 + x + 1
=
x2 (x2 + 4)
Cx + D
A
B
+ 2+ 2
x
x
x +4
x2 + x + 1
is
x2 (x2 + 4)
.
c) Using the form of the partial fraction decomposition in part b, what is the form of the integral
Z
x2 + x + 1
dx =
x2 (x2 + 4)
A ln(|x|) − B/x + 12 C ln(x2 + 4) + 12 D arctan(x/2) + E
Z
d) Using the Midpoint Rule, n = 6,
0
Z
e) Using Simpson’s Rule, n = 6,
0
Z
3
f) Using your calculator,
0
3
3
1
dx ≈
x3 + 1
1
dx ≈
x3 + 1
1
dx ≈
x3 + 1
1.1551
(round to 4 decimals).
1.1614
1.1544
.
(round to 4 decimals).
(round to 4 decimals).
2. Find the centroid of the region in the xy-plane that lies between the graphs of y = x and y = x2 for 0 ≤ x ≤ 1.
Show all of your work in the space below. (Exact answer, no calculator)
Z
The area of the region is A =
0
1
1
x − x2 dx = 12 x2 − 13 x3 =
1
2
−
1
3
=
1
6
. Therefore,
0
1
Z
Z 1
1 1
1 3 1 4 2
2
3
x=
x − x x(x − x ) dx = 6
x − x dx = 6
A 0
3
4
0
0
1 1
1
1
=6
−
= ,
=6·
3 4
12
2
and
1
Z
Z 1
1 11 2
1 3 1 5 4
2
4
y=
(x − x ) dx = 3
x − x dx = 3
x − x A 0 2
3
5
0
0
1 1
2
2
=3
−
=3·
= .
3 5
15
5
The centroid is (x, y) = 12 , 25 .
Part II. (30 minutes, 30 points) There are 4 problems, do problem 1 on this page and 2 more for a total of 3.
1. (You must do this problem.) Evaluate three the following four integrals, showing all of the steps in your
calculation. Neatness counts. (No calculator)
Z
a)
Z
x+1
dx
x2 + 2x
x3
dx
x2 + 4
Let x = 2 tan(θ), dx = 2 sec2 (θ) dθ .
Z
Z
8 tan3 (θ)
x3
√
p
dx =
· 2 sec2 (θ) dθ
x2 + 4
4 tan2 (θ) + 4
Z
= 8 tan3 (θ) sec(θ) dθ
Observe that the numerator is almost the
derivative of the denominator.
Z
Z
x+1
1
2x + 2
dx
=
dx
2
x + 2x
2
x2 + 2x
=
√
b)
1
ln |x2 + 2x| + C
2
Z
This can also be done using partial fractions.
Z
=8
(sec2 (θ) − 1) sec(θ) tan(θ) dθ
=8
=
Z √
c)
x−2
dx
x+2
√
Let u = x − 2, u2 = x − 2, 2u du = dx .
Z √
Z
x−2
u
dx =
· 2u du
x+2
u2 + 4
Z
Z 2
u2
u +4−4
=2
du = 2
du
u2 + 4
u2 + 4
Z
4
=2 1− 2
du
u +4
Now let u = 2v, du = 2 dv .
Z √
Z x−2
4
dx = 2
1− 2
· 2 dv
x+2
4v + 4
Z
1
=4 1− 2
dv
v +1
= 4 (v − arctan(v)) + C
= 4 (u/2 − arctan(u/2)) + C
√
√
= 2 x − 2 − 4 arctan( x − 2/2) + C
tan2 (θ) sec(θ) tan(θ) dθ
=8
∞
Z
√
d)
2
1
3
sec3 (θ) − sec(θ) + C
p
(x2 + 4)3/2
− 4 x2 + 4 + C
3
1
dx
x+3
This is an improper integral.
Z
2
∞
1
√
dx = lim
b→∞
x+3
Z
b
(x + 3)−1/2 dx
2
b
= lim 2(x + 3)1/2 b→∞
= lim 2
b→∞
√
0
√ b + 3 − 3 = ∞.
The integral diverges.
2. Find the exact length of the curve y = 1 − x2/3 from the point A(1, 0) to the point B(8, −3). (No calculator)
Solve for x, x2/3 = 1 − y, x = (1 − y)3/2 . Therefore, dx/dy = 23 (1 − y)1/2 · (−1) and
r
Z 0r
9
13 9
1 + (1 − y) dy =
L=
− y dy
4
4
4
−3
−3
Z 0p
Z 0
p
1
1
−9 13 − 9y dy
13 − 9y dy = −
=
2 −3
18 −3
0
2
1
= − (13 − 9y)3/2 · 18
3 −3
1 3/2
=
40 − 133/2 .
27
Z
0
3. Find the exact area of the surface obtained by rotating about the x-axis the curve y 2 = 4x + 4, 0 ≤ x ≤ 8. (No
calculator)
√
1
The curve that is rotated is y = 2 x + 1 . Since dy/dx = √
,
x+1
r
Z 8
√
1
dx
S=
2π · 2 x + 1 · 1 +
x
+
1
0
r
Z 8
√
x+2
=
dx
2π · 2 x + 1 ·
x+1
0
Z 8
√
= 4π
x + 2 dx
0
8
2 = 4π(x + 2) · 3 0
8 = π 103/2 − 23/2 .
3
3/2
4. The end of a tank containing water is vertical and has the indicated shape. Express the force on the end as
an integral and then evaluate it. Give your answer in exact form in terms of δ = weight density of water in
pounds per cubic foot. (No calculator)
Put the origin of an xy-coordinate system at the midpoint of the base of the triangle. Then consider a thin
horizontal strip of width dy at a point y between 0 and 3 as indicated in the figure above. The length of the
strip is `.
6−y
6
2
By similar triangles,
= so ` = (6 − y) and the area of the strip is
`
4
3
dA = ` dy = 32 (6 − y) dy .
Since the depth of the strip is 3 − y, the pressure is approximately P = (3 − y) · δ, and the force on the strip is
approximately
dF = P · dA = (3 − y) · δ · 32 (6 − y) dy .
Consequently, the total force on the end of the tank is
Z 3
F =
(3 − y) · δ · 32 (6 − y) dy
0
=
2
δ
3
Z
3
18 − 9y + y 2 dy
0
3
9 2 1 3 2
= δ 18y − y + y 3
2
3
0
2
81
= δ 54 −
+9
3
2
2
81
= δ 63 −
3
2
= δ (42 − 27) = 15 δ .