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(MATH014)[2009](s)final~1748^_10393.pdf
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Math014 Final Exam Solution, Spring 09
1. ([8pts]) Considertheregionenclosedbythecurveand the x-axisasshowninthefollowing .gure.
y y
66
4
2 2
x
24 6810

24 6810
(a) (b)
(a) Use the trapezoidal rule with n =8 subintervals to estimate the area of the enclosed region. [3 pts]
Solution Just a matter of calculating 8 trapezoids of base length 1: 1+0 3+1 5+3 6+5 5+6 3+5 1+3 1+0
area +++++++
22222222 or 1
= (0+2(1)+2(3)+2(5)+2(6)+2(5)+2(3)+2(1)+0)=24
2
(b) Usethediskmethod(washermethod) and midpoint rule with n =3 subintervals to estimate the volume of the solid obtained by rotating the region about the y-axis. [5 pts]
Solution Note that the disk method says
. 6
volume = (x2 left)dy
right .x2
0
Divide the y interval[0,6] into three subintervals of length 2 and get by the midpoint rule : volume 2[(92 .32)+(82 .42)+(72 .52)]=288
2. ([8 pts]) Use integration by parts, or otherwise, to show the following.

(a) If afunction f satis.es f(1)= f(1)=0, f
(x)|4,
is continuous on theinterval[0,1] and |f

show that

. 1
0
f(x)dx

2

. [4 pts]
3
Solution

Hence

. 1
0
(x)
1 1
. 1 . 1
2 2
x

x

(x)dx
=

(x)dx
f(x)dx = xf(x)

. xf

. f

2

+ f

2

0 00 0
. 1
0
. 1 . 1
12
|f (x)|x 2dx 2 x 2dx =
2
00
f(x)dx



3

. 1
(b) Let In = (1.x 2)ndx, for n =0,1,2,3,... . Show that In+1 = CnIn for some constant Cn
0
22n(n!)2

depending on n, and then show In = . [4 pts]
(2n +1)!
Solution
. 1
In+1 = (1.
0
x

2)n+1dx = x(1.x 2)n
1
. 1
x

2)n
x 2dx
+ 2(n +1)(1 .

00
. 1 . 1
=2(n +1) (1.x 2)n(x 2 .1)dx+2(n +1) (1.x 2)ndx
00
2n +2
(2n +3)In+1 =2(n +1)In .. In+1 = In
2n +3 Repeatedly using the relation,
2n 2n(2n .2)
In = In.1 = In.2 =
2n +1 (2n +1)(2n .1) 2n(2n .2)(2n .4) 2 [2n(2n .2)(2n .4) 2]2 22n(n!)2
= I0 ==
(2n +1)(2n .1)(2n .3) 3 (2n +1)! (2n +1)!
3. ([8 pts]) b is a constant in each of the following improper integrals. Determine the value of the constant b for which the corresponding improper integral converges. Show your reasoning for full credit.
. 3
x
(a) bdx [4 pts]
1+x
Solution For all su.ciently large x,

33
xx1
=
bbb.3
1+xxx
and hence the integral converges if b> 4, and diverges if b 4 by the integral test. Or by more detailed comparison:
. 3 .
x1
dx< dx< for b> 4
bb.3
1+xx
11
and
. 3 . 3
xx
dx dx = for b 0
b
1+x2
11 . 3 . 3 .
xx11
dx dx =

dx = for 0 <b 4
bb + bb.3
1+xxx2 x
11 1
. . bx 1 .
(b) . dx [4 pts]
1 x2 +1 2x
Solution Similarly, for all su.ciently large x,

bx 1 b 1 11
. . =(b.

)
x2 +1 2xx 2x 2x
1
and hence it is a divergent integral, unless b = 2
.
Or by more detailed calculation like the following:

.
. bx 1 ..b 1 . 1(L2 +1)b b
. dx =

ln(x 2 +1) .

lnx = lim

ln .

ln