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The answers to Test 2 of MAT 266 course at ASU. The test consists of 14 questions worth a total of 60 points, with questions 1-10 being multiple choice and questions 11-14 being free response. The topics covered in the test include finding areas and volumes of regions, evaluating integrals, and calculating work. The document also includes an honor statement that confirms the student has not given or received any unauthorized assistance on the exam.
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By signing below you confirm that you have neither given nor received any unauthorized assistance on this exam. This includes any use of a graphing calculator beyond those uses specifically authorized by the Mathematics Department and your instructor. Furthermore, you agree not to discuss this exam with anyone until the exam testing period is over. In addition, your calculator’s program memory and menus may be checked at any time and cleared by any testing center proctor or Mathematics Department instructor.
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b. 25 3 c. 20 d.
e. None of these
6 π (^) b. 10
3 π (^) c. 5
(^2) d. 5
π (^) e. None of these
a.
1
0
(^1) dx ∫ (^) x b.
1
0
(^1) dx ∫ (^) x c.
1
0
(^1) dx ∫ (^) x d.
1 2 0
(^1) dx ∫ (^) x e. none of these
n
n
∑ Select the correct answer
a.
b.
c.
d.
e. None of these
Answer: d
Answer: b
Answer: c
Answer: c
Answer: a
Answer: a
Answer: b
Solution: Volume of the ith^ element =
Weight of the ith^ element = 62.5 (25 )
Work done to pump out the ith
Work done to pump out all the water
Solution:
First observe
∫ (^) x (ln x ) ∫
So,
2 2 = lim[ ]
= lim[ ]
x x x x
∞ ∫ ∫
The tank shown is full of water. Given that water weighs 62.5 lb/ft^3 work required to pump the water out of the tank.
element = π (25 − xi^2 )∆ x 62.5 (25 2 ) π − xi ∆ x element = 62.5 π (25 − xi^2 ) xi ∆ x
Work done to pump out all the water = ith^ element =
5 2 0
∫^ 62.5^ π^ (25^ −^ x^ )^ xdx^ =^ 9765.625π ft^ − lb
or show that it is divergent: (^4) 2
(ln )
dx x x
∞ ∫.
4 4
3
(^1) ( ln ) (ln ) 1 3[ln ]
dx u du u x x x
x
∫ ∫
4 4 2 2
3 2
3 3 3
(^1) lim 1 (ln ) (ln ) 1 = lim[ ] 3(ln )
= lim[ 1 1 ]^1 3(ln ) 3(ln 2) 3(ln 2)
t t x t t x
t
dx dx x x x x
x
t
→∞ =
→∞
∫ ∫
62.5 lb/ft 3 and R = 5 ft, find the
62.5 π (25 − x ) xdx = 9765.625π ft − lb
Solution: Outer radius of the cross-section = y 1/3^ + 1 Inner radius of the cross-section = y^3 + 1
Volume =
1 1/3 2 3 2 0
π (^) ∫ [( y + 1) − ( y +1) ] dy
Solution:
The height of an equilateral triangle with side-length equal to s units =
s
So, area of the equilateral triangle = 1 3 3 2 2 2 4
⋅ s ⋅ s = s
For our problem, area of the equilateral triangle = 3 2 64
x
Hence, volume of the monument =
20 2 0
∫^ x^ dx^ =