Math 126
Name:
Fall 2016
Score:
/45
Show all your work
Dr. Lily Yen
No Calculator permitted in this part. Read the questions carefully. Show all your work
and clearly indicate your final answer. Use proper notation.
Problem 1: Evaluate the following integrals analytically.
Z
a.
xe−2x dx
Test 2
Using integration by parts,
Z
Z
−2x
−2x
1
xe
dx = − 2 xe
− − 12 e−2x dx = − 12 xe−2x − 14 e−2x + C
Score:
Z
b.
/3
cos3 (θ) sin8 (θ) dθ
If
dθ, so
R u =3 sin(θ),8 then du R= cos(θ)
R
8
2
cos
(θ)
sin
(θ)
dθ
=
cos
(θ)
sin
(θ)
cos(θ)
dθ
=
(1 − sin2 (θ)) sin8 (θ) cos(θ) dθ =
R
R 8
8
10
1 1
1 9
1
u 1+C =
sin (θ) − sin (θ) cos(θ) dθ = u − u 0 du = 9 u − 11
9
11
1
1
sin (θ) − 11 sin (θ) + C
9
Alternatively,
using integration
by parts,
R
R
2
cos3 (θ) sin8 (θ) dθ
=
cos
(θ)
cos(θ) sin8 (θ) dθ =
R
cos2 (θ) 91 sin9 (θ) −R (−2 cos(θ) sin(θ)) 91 sin9 (θ) dθ =
1
2
cos2 (θ) sin9 (θ) + 29 cos(θ) sin10 (θ) dθ = 91 cos2 (θ) sin9 (θ) + 99
sin11 (θ) + C (and these
9
two results are actually equal).
Score:
/3
Z
c.
x+1
dx
− 2x − 15
By the method of partial fractions, constants A and B exist such that
x2
x2
x+1
x+1
A
B
=
=
+
,
− 2x − 15
(x + 3)(x − 5)
x+3 x−5
so x + 1 = A(x − 5) + B(x + 3). Substituting x = 5 yields that 6 = 8B, so B = 34 ,
and substituting x = −3 yields that −2 = −8A, so A = 14 .
Hence
Z
Z
x+1
1/4
3/4
dx
=
+
dx = 14 ln|x + 3| + 43 ln|x − 5| + C
x2 − 2x − 15
x+3 x−5
Score:
/4
Z
d.
e−3x cos(5x) dx
Using integration by parts twice,
Z
Z
−3x
1 −3x
e
cos(5x) dx = − 3 e
cos(5x) − − 13 e−3x (−5) sin(5x) dx
Z
1 −3x
5
= −3e
cos(5x) − 3 e−3x sin(5x) dx
Z
−3x
−3x
−3x
cos(5x) − 53 − 31 e
sin(5x) − − 13 e 5 cos(5x) dx
= − 31 e
Z
1 −3x
5 −3x
25
= −3e
cos(5x) + 9 e
sin(5x) − 9
e−3x cos(5x) dx,
R −3x
)
e
cos(5x) dx = − 31 e−3x cos(5x) + 59 e−3x sin(5x), so
so R(1 + 25
9
34
−3x
e
cos(5x) dx = − 31 e−3x cos(5x) + 59 e−3x sin(5x), so
9
Z
3 −3x
5 −3x
e
cos(5x) + 34
e
sin(5x) + C
e−3x cos(5x) dx = − 34
Score:
/4
Problem 2: Find the exact area of the region bounded between y = 3x2 − 3 and the x-axis
for x in the interval [0, 3].
y
x
1
Z
2
3
1
2
Z
0 − (3x − 3) dx +
0
1
3
2
Z
1
Z
3
(3x − 3) − 0 dx =
−3x + 3 dx +
3x2 − 3 dx
1
1 0
3
= −x3 + 3x + x3 − 3x = (2 − 0) + (18 − (−2)) = 22
0
2
1
Score:
Page 2
/3
Math 126
Math 126
Name:
Fall 2016
Show all your work
Dr. Lily Yen
Calculators permitted from here on.
Problem 3: Determine analytically the convergence of the following integral.
Z 1 x
e
dx
2
0 x
Test 2
If x ≥ 0, then ex ≥ 1. Therefore
R 1 ex
R1 1
R1
+
dx
≥
dx
=
lim
2
2
b→0
0 x
0 x
b
R 1 ex
dx = ∞.
0 x2
1
x2
dx = limb→0+
1
−1 x b
= limb→0+ (−1 + 1b ) = ∞, so
Score:
/4
Problem 4: The graph of y1 = 1/x, y2 = 1/x , and the functions f , g, h, and k are shown
below.
2
y
y = k(x)
y = h(x)
y = 1/x
y = g(x)
y = f (x)
y = 1/x2
x
1
a. Is the area between y1 and y2 on the interval [1, ∞] finite or infinite? Explain.
∞
R∞ 1
The area under y = 1/x is 1 x dx = ln(x) = ∞ while the area under y = 1/x2 is
0
∞
R∞ 1
1
dx
=
−
=
1.
x
1 x2
0
The area between 1/x and 1/x2 is therefore infinite (as in ∞ − 1 = ∞).
Score:
/1
b. Using the graph, decide whether the integral of each of the functions f , g, h, and k on
the interval [1, ∞] converges, diverges, or whether it is impossible to tell. Reason must
be provided for each correct answer.
The area under f is smaller than the finite area under 1/x2 , so finite.
The areas under h or k are larger than the infinite area under 1/x so infinite.
Since all you know about the area under g is that it is larger than the finite area
under 1/x2 but smaller than the infinite area under 1/x, you cannot tell whether the
area under g is finite or infinite.
Score:
Page 3
/4
Math 126
Problem 5: A 20 m tall parabolic arch spans 50 m. Suppose a supporting beam is to be
constructed across the arch at its average height. How high is the cross beam placed?
The parabola has the form y = 20 − ax2 and passes through (±25, 0). Filling in (25, 0)
4
.
yields that 0 = 20 − 625a, so a = 125
The average height is then
25
R 25
1
1
1
2
3 20
−
ax
dx
=
(20x
−
ax
)
= 20 − 625
a = 40
= 13.3 m.
50 −25
50
3
3
3
−25
Score:
/3
Problem 6: Draw y = e
and y = (x−1) +e −1 on the grid. Shade the region bounded
between these two graphs in the interval [0, 2].
2−3x
2
2
y
7
6
5
4
3
2
1
x
1
−1
−2
2
3
Use integrals to express the following. Do not evaluate your integrals. Draw a
cross-sectional strip for each solid of rotation.
a. The area of the shaded region.
Z
2
2
2
(x − 1) + e − 1 − e
0
2−3x
dx =
1
(x
3
3
2
− 1) + e x − x +
2
1 2−3x e
3
0
= − 43 + 53 e2 + 31 e−4
Score:
/2
b. The volume of a solid that has the shaded region as its base, and cross-sections perpendicular to the x-axis are semi-circles.
Z
0
2
π
2
(x − 1)2 + e2 − 1 − e2−3x
2
2
dx =
π
8
Z
2
(x − 1)2 + e2 − 1 − e2−3x
2
dx
0
Score:
/2
c. The volume of the solid obtained by rotating the region around y = −2.
Using washers, the outer radius is 2 + (x − 1)2 + e2 − 1 = (x − 1)2 + e2 + 1 and the
inner radius is 2 + e2−3x , so the volume is
Z 2
π((x − 1)2 + e2 + 1)2 − π(2 + e2−3x )2 dx
Score:
/2
0
d. The volume of the solid obtained by rotating the region around x = 3.
Using cylindrical shells, the height of each shell is (x − 1)2 + e2 − 1 − e2−3x , so the
volume is
Z 2
2π
(3 − x) (x − 1)2 + e2 − 1 − e2−3x dx
0
Score:
/2
Page 4
Math 126
Problem 7: Suppose the force, F , required to compress a spring by 2 cm is 0.8 N. Find the
work done in stretching the spring from equilibrium by 13 cm.
By Hooke’s Law, F = kx, so 0.8 N= k × 0.02 m, so k = 40 N/m. The required work is
R 0.13
0.13
therefore W = 0 40x dx = 20x2 = 0.338 J.
0
Score:
/3
Problem 8: A tank is in the shape of an inverted circular cone with height 5 m and diameter
4 m. The tank is filled with water to 3 m deep. Find work done to pump all the water out a
valve 2 m above the top of the tank.
A slice of water at height h from the bottom of the tank has radius 25 h and needs to be
R3
lifted 7 − h meters. Hence the required work is 0 (7 − h)9.8 · 1000 · π( 52 h)2 dh =
3
R3 2
7 3
1 4 3
1568π 0 7h − h dh = 1568π( 3 h − 4 h ) = 67032π = 211 000 J = 211 kJ
0
Score:
Page 5
/5
Math 126