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(MATH024)[2010](s)final~1944^_10406.pdf
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Math 024 -Honors Calculus II
Final Examination, Spring Semester, 2010

Time Allowed: 3 Hours Total Marks: 100
Note: Calculators are not allowed. Please show all necessary details. Only the following integrals can be used:
n+1
x
x
x n dx =+ C, (n .= .1); e x dx = e + C;
n +1

dx dx
= ln |x| + C; = arctan x + C;
xx2 +1

dx
sin x dx = . cos x + C; = arcsin x + C;
1 . x2
cos x dx = sin x + C; (.1)dx

= arccos x + C.
1 . x2
1. (24 marks) For each of the following series, determine if it is condition-ally or absolutely convergent.
(2)2 (4 2)2 (6 4 2)2 (8 6 4 2)2
(a) . + . + ;
20 (1)2 22 (3 1)2 23 (5 3 1)2 24 (7 5 3 1)2

()()()()
1+ 1 . ln 1+ 1 +ln 1+ 1 . ln 1+ 1
1234

(b) ln

+ ;

()n2
1
1 .
(c)

;

n


n=2
( )

nk +1 . nk
sin2
(d)

,k is an integer.

n=1
1
2. (10 Marks) Expand the function f(x) = as its power series at
2x . x2
x = 1.


3. (10 Marks) Let f(x)= anx n for . <x< +. For the function
n=0
f(x)
F (x) = , calculate F (100)(0) in terms of the sequence {an}.
1 . x2
4. (10 Marks) Show that for x> 0,
()n ()
1 . x 1+ x
= ln
1+ x2x
.



n
n=1
1

5. (10 Marks) A thin shell can be obtained by rotating the smooth curve

x = x(t),y = y(t),t [a, b]
about x-axis. Assume that mass is uniformly distributed on the shell. Find the center of mass.
6. (14 Marks) A 3D region is obtained by rotating the 2D region de.ned
by:
x 0,y 1,x 2 +(y . 1)2 4,

about x-axis.
(a)
Find the total surface area of the 3D region.

(b)
If mass is uniformly distributed in the 3D region, .nd the center of mass.


7. (12 Marks) Establish the identity: for .. x .,
2
(.1)n.1 cos nx .2 x
= . .
2
n12 4
n=1
11
8. (10 Marks) Let xn = 1+ + + . 2 n. De.ne the series an
2 n
by
a1 = x1,an = xn . xn.1, (n> 1).

Prove that the limit
[]
11
lim 1+ + + . 2 n
n
2 n

exists by showing an is convergent.
1. (24 marks) For each of the following series, determine if it is condition-ally or absolutely convergent.
(2)2 (4 2)2 (6 4 2)2 (8 6 4 2)2
(a) . + . + ;
20 (1)2 22 (3 1)2 23 (5 3 1)2 24 (7 5 3 1)2
()()()()
(b) ln 1+ 11 . ln 1+ 12 +ln 1+ 31 . ln 1+ 14 + ;
()n2
1
(c) 1 . ;
n
n=2
( )

k
(d) sin2 nk +1 . n,k is an integer.
n=1

Solution

(a) This series is , with
n=1 an
[(2n)!!]2
an =(.1)n.1 .
2n.1[(2n . 1)!!]2
Since
. [2n+2!!]2
an+1 2n[(2n+1)!!]2
lim = lim
n n [(2n)!!]2
an
2n.1[(2n.1)!!]2
(2n + 2)2 1
= lim = < 1,
n 2(2n + 1)2 2
by the Ratio Test, the series is absolutely convergent.

(b) The series is , with
n=1 an
()
an =(.1)n.1 ln 1+ 1 .
n
()
Let f(x)=ln 1+ x 1 . Since for x> 0,
. 1