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(MATH014)[2010](s)midterm~jtu^_10396.pdf
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Math014 Calculus II Midterm Exam Solution, Spring 2010
1. [10points]Thegiven .gureshowsarectangleinscribedinthesemi-circular regionbetweenthex-axis
and the graph of y =9.x2, with a vertex P:(x,y)lying in the .rst quadrant.
(a) Express the area of the rectangle as a function of x, and indicate the domain of your area function. [2 points]
Solution
area = A(x)=2x 9.x2 , 0 x 3 .
(b) Find the average value of the areas of all such inscribed rectangles. [5 points]
Solution
3 .. .3
12 2)3/2
average area = average value of A(x)= 2x 9.x2 dx = . (9.x =6 .
3.090
0
(c) How many such rectangles canhavetheir areas equal tothe average value obtainedinpart(b)? [3 points]
22)
Solution Two. In fact, 2x 9.x=6 .. x2(9.x=3, hence
9 45 9 45
2
x 4 .9x 2 +9 =0 .. x = .. x = .
22
(For other exercises in average values, see Ex. 6.5.)
2. ([10 points]) The base ofa solid is the area enclosed by the curves y = x2 and y =8.x2, and the cross-sections of the solidperpendiculartothe x-aixs are semi-circular discs. Find the volume of the solid.
2
Solution When the two curves intersect, 8.x= x2 .. x = 2.
A cross-section of the solid perpendicular to the x-aixs is a semi-circular disk with diameter (8 .
2
x2).x=8.2x2, hence its area is 1(4.x2)2 .
2
2
8
volume = 1 (4.x 2)2dx
2
.2
6
2
2
4
=1 (16.8x + x 4)dx
2
.2
2
..2
1 83 15
0
= 16x . x+ x
-2-1 0 1 2
235
.2
256
=
15
(For other exercises in cross-sections andvolumes, see Ex. 6.2, #56-62.)
2 .4x
3. ([10 points]) Find the area under the graph of the function y = 16(x+ x)eover the in.nite interval[0,).
Solution Integration by parts:
22
16(x + x)e .4xdx = .4(x + x)de.4x 2
00
.
2 .4x
= .4(x + x)e + 4(2x +1)e .4xdx
00 1
=0+ .(2x +1)de.4x
0
. 0
0 2 4
.4x
= .(2x +1)e +2e .4xdx
00
1 . 13
.4x
=1. e =1+ =
2 22
0 (For relatedexercises in integration by parts and improper integrals, see 7.1(eg., Examle3,p.455), 7.8(eg., Ex7.8,#60).
4. ([12 points]) Suppose that the region under the graphof y = f(x)shown in the following .gure is rotated about the y-axis to form a solid of revolution.
(a) Express the volume of the solid in the form of an integral. [4 points] Solution By the method of cylindrical shells,
88
volume = 2xf(x)dx =2 xf(x)dx .
00
(b) According to the integral in part (a) and the given graph, apply Simpsons rule with n =8 subintervals to estimate the volume. [8 points]
.
(Recall that for any polynomialp(x)of degree 2, p(x)dx =[p()+4p(+)+p()].)
62
Solution By the Simpsons rule with n =8 subintervals of length 1,
8
volume =2 xf(x)dx
0
2
0f(0)+41f(1)+22f(2)+43f(3)+24f(4)+45f(5)+26f(6)+47f(7)+8f(8)
3 2 1124
= 0+4+6+24+48+140+96+196+48 =
33
(Similar to7.7, # 37.)
5. ([16 points]) Evaluate t