Quiz 7
Math 252
Thursday, May 28
Name:
Recitation:
Unjustified answers will receive no credit. Justification includes mathematical work as
well as verbal explanations. Verbal explanations should be complete sentences. For multiple
choice questions you get 1pt for the correct answer and up to 4pts for the explanation.
(5pts) Problem 1:
Z
Consider the integral
2x3 ln(x) dx. Using integration by parts, suppose we let
u = ln(x), dv = x3 dx. Which of the following is equivalent to our original integral?
Z
2 1
D. This choice of u, dv will not work.
A. 3x x − 3x2 ln(x) dx
E. None of these is correct.
Z
1 4
1 4 1
B. 4 x ln(x) − 4 x x dx
Z
C.
3x2 ln(x) dx
Justify your choice and then finish evaluating the integral.
Solution: Rubric: 1pt for correct choice, 2pt for justification, 2pts for evaluating
R
R
Recall that the integration by parts formula is u dv = uv − v du. With the given
choice of u, dv, we must find v, du. I like to use a little table:
u = ln(x)
↓
du =
1
x
v = 14 x4
↑
dx dv = x3 dx
Plugging this into the formula gives:
Z
Z
1 4
1 4 1
3
x ln(x) dx = x ln(x) −
x
dx
4
4
x
Thus option B. is the correct choice.
Then we evaluate:
1 4
x ln(x) −
4
Z
1 4
x
4
1
1
dx = x4 ln(x) −
x
4
1 4
= x ln(x) −
4
Z
1
x3 dx
4
1 4
x +C
16
OVER→
(6pts) Problem 2:
An aquarium has a triangular viewing window (pictured below). The window is 1.1
meters wide and 1.5 meters tall, and its top sits 1 meter below the surface of the water.
Calculate the total force the water exerts against the window. Explain your reasoning
and be sure to include units (ρ = 1, 000kg/m3 ).
w
h
Solution: Rubric: 1pt for units, 1pt for calculation, 1pt for explanation, 3pts for
perfect setup (1.5pts if errors, but close)
Force equals pressure times area and pressure is proportional to depth. So we will
approximate the force by considering small horizontal strips of the window, where the
depth is approximately constant.
We calculate the force on a strip: Fi = Pi · Ai = (ρgdi )Ai . Then we need to find the
depth of each strip (di ), as well as the area of each strip, Ai . Let’s have yi measure
depth, so that part is easy. For the area of each strip, we will need to know the width
(wi ) at a given depth (yi ). Let’s use similar triangles. Then
wi
11
1.1
=
→ wi = (2.5 − yi )
1.5
2.5 − yi
15
Then we can calculate the total force in the usual way:
11
Fi = (ρgyi )
(2.5 − yi ) ∆y
15
↓ limit of sum as ∆y → 0
Z 2.5
11
F = ρg
y (2.5 − y) dy
15
1
24, 255
=
N
2
= 12, 127.5N
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