Math 126
Name:
Summer 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.
x5 ln(x) dx
Test 2
Using
by parts, R
R
R 5 integration
1 6
x + C.
x ln(x) dx = 16 x6 ln(x) − 16 x6 · x1 dx = 16 x6 ln(x) − 16 x5 dx = 16 x6 ln(x) − 36
Score:
/3
Z
b.
1
dx
tan(6x)
R 1
1
If u = sin(6x), then du
=
6
cos(6x),
so
dx
=
du.
Therefore,
dx =
6 cos(6x)
tan(6x)
R cos(6x)dx
R
R cos(6x)
1
dx =
· 6 cos(6x)
du = 61 · u1 du = 16 ln|u| + C = 16 ln|sin(6x)| + C.
sin(6x)
u
Score:
Z
c.
/3
e2x sin(3x) dx
Using
parts twice,R
R 2x integration by
R
1 2x
e sin(3x) dx = 2 e sin(3x)− R 21 e2x 3 cos(3x) dx = 12e2x sin(3x)− 32 e2x cos(3x) dx =
1 2x
e sin(3x) − 32 21 e2x cos(3x) −R 21 e2x (−3) sin(3x) dx =
2
1 2x
e sin(3x) − 34 e2x cos(3x) − 49 e2x sin(3x) dx + C. Therefore
2 R
13
e2x sin(3x) dx = 12 e2x sin(3x) − 34 e2x cos(3x) + C, so
4
Z
2
3
e2x sin(3x) dx = e2x sin(3x) − e2x cos(3x) + C2
13
13
Score:
/4
Z
d.
4+x
dx
x2 + x
By the method of partial fractions, constants A and B exist such that
4+x
B
4+x
= x(x+1)
= Ax + x+1
, so 4 + x = A(x + 1) + Bx. Substituting x = 0 yields that
x2 +x
4 = A, and substituting x = −1 yields that 3 = −B, so B = −3. Hence
Z
Z
4+x
4
3
dx
=
−
dx = 4 ln|x| − 3 ln|x + 1| + C
x2 + x
x x+1
Score:
/4
Problem 2: Evaluate if possible, otherwise state why the integral does not exist.
Z 7
1
dx
2/3
0 (x − 5)
R5
1
The integrand is undefined at x = 5, so split the interval of integration: 0 (x−5)
2/3 dx =
z
√
Rz
√
√
3
1
1/3 limz→5− 0 (x−5)
= limz→5 3 3 z − 5 − 3 3 −5 = 0 + 3 5.
2/3 dx = limz→5− 3(x − 5)
0
p
R7
1
3
Similarly, 5 (x−5)
dx
=
3
(2).
Hence
2/3
Z
0
7
√
√
1
3
3
dx
=
3
5
+
3
2
2/3
(x − 5)
Score:
Page 2
/4
Math 126
Math 126
Name:
Summer 2016
Show all your work
Dr. Lily Yen
Calculators permitted from here on.
Problem 3: Determine analytically the convergence of the following integral.
Z ∞
sin(x) + 2
√
dx
x
1
Test 2
√
≥ √1x for all x. Therefore
Note that sin(x)+2
x
R ∞ sin(x)+2
R∞ 1
Rz 1
√ z
√
√
dx ≥ 1 √x dx = limz→∞ 1 √x dx = limz→∞ 2 x = limz→∞ 2 z − 2 = ∞, so
1
x
1
Z
1
∞
sin(x) + 2
√
dx = ∞
x
Score:
/3
Problem 4: Suppose −1 h(z) dz = 7, and that h(z) is an even function. Calculate the
following:
R1
a.
R1
0
h(z) dz =
7/2
Since h is even, h(−x) = h(x), so
b.
R −2
−4
h(z) dz =
−1
R1
0
h(z) dz.
35
5h(z + 3) dz =
Substituting z = u − 3,
R0
R −2
−4
5h(z + 3) dz =
R1
−1
5h(u) du = 5
R1
−1
h(u) du = 5 × 7 = 35.
Score:
/3
2
Problem 5: Consider the area bounded by the y-axis, x + y = 6, and y = x in the first
quadrant. Find the average vertical height of this area.
y
6
4
x
2
6
The height at x is (6 − x) − x2 , so the average
height is
2
R
2
1
6 − x − x2 dx = 12 (6x − 21 x2 − 13 x3 ) = 12 (12 − 2 − 38 ) =
2−0 0
0
11
.
3
Score:
Page 3
/4
Math 126
Problem 6: Draw y = 6 − 3x and y = x2 − 4 on the grid. Shade the region bounded by
these two graphs and the y-axis for non-negative values of x.
y
6
4
2
1
2
3
4
5
6
7
8
x
9
−2
−4
−6
−8
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
Z 2
2
(6 − 3x) − (x − 4) dx =
0
0
2
3
1 2 34
10 − 3x − x2 dx = 10x − x2 − x3 =
2
3 0
3
Score:
/2
b. The volume of a solid that has the shaded region as its base, and cross-sections perpendicular to the y-axis are equilateral triangles.
√
√
√
For x ≤ 0, the base of each triangle is x = y + 4, so the area is√ 43 x2 = 43 (y + 4).
For x ≥ 0, the side of each triangle is x = 2 − 31 y, so the area is 43 (2 − 31 y)2 . Hence
the volume is
Z 0 √
Z 6√
√
3
3
(y + 4) dy +
(2 − 31 y)2 dy = · · · = 4 3 ≈ 6.928
4
4
−4
0
Score:
/2
c. The volume of the solid obtained by rotating the region around x = 5.
Using cylindrical shells, the height of each shell is (6 − 3x) − (x2 − 4) = 10 − 3x − x2 ,
so the volume is
Z 2
292
2π(5 − x)(10 − 3x − x2 ) dx = · · · =
π ≈ 305.8
3
0
Score:
/2
d. The volume of the solid obtained by rotating the region around y = −7.
Using washers, the outer radius is y + 7 = 13 − 3x and the inner radius is
y + 7 = 3 + x2 , so the volume is
Z 2
828
π(13 − 3x)2 − π(3 + x2 )2 dx = · · · =
π ≈ 520.2
5
0
Score:
Page 4
/2
Math 126
Problem 7: For your summer job, you work on a scaffolding 75 ft above the ground. Suppose
you need to lift a 500 lb bucket of cement from the ground to a point 30 ft above the ground
by pulling on a rope weighing 0.5 lb/ft. How much work is required?
Lifting just the bucket of cement requires 30 ft × 500 lb = 15 000 ft · lb of work.
Similarly, lifting the bottom 45 ft of rope requires 30 ft × 0.5 lb/ft × 45 ft = 675 ft · lb.
Lifting the top 30 ft of the rope—the part that ends up on the scaffolding
with
30
R 30
R 30
2
you—requires 0 (30 − x)0.5 dx = 0 15 − 0.5x dx = 15x − 0.25x = 225 ft · lb.
0
Total 15 000 ft · lb + 675 ft · lb + 225 ft · lb = 15 900 ft · lb.
Score:
/4
Problem 8: Calculate the work done in pumping oil from a full (lower) hemispherical tank
of radius 10 m out of a spout 2 m above the top of the hemisphere. Assume that the oil has
density 800 kg/m3 .
2
10
10 − x
r
x
p
√
The slice at height x is a disk with radius r = 102 − (10 − x)2 = 20x − x2 . Therefore
the volume of the disk is πr2 dx = π(20x − x2 ) dx. Since the density is 800 kg/m3 , the mass
of the disk is 800π(20x − x2 ) dx, so the force of gravity on the disk is
9.8 × 800π(20x − x2 ) dx = 7840π(20x − x2 ) dx. The disk has to be lifted 12 − x metres,
which then requires a work of 7840π(20x − x2 )(12 − x) dx = 7840π(240x − 32x2 + x3 ) dx.
The total amount of work required is thus
Z 10
90160000
π ≈ 94.4 MJ
7840π(240x − 32x2 + x3 ) dx = · · · =
3
0
Score:
Page 5
/5
Math 126