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(MATH024)[2008](s)final~=8hrzkjdb^_26706.pdf
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Math 024 -Calculus II
Final Examination, Spring Semester, 2008

Time Allowed: 2.5 Hours Total Marks: 105
Note: Calculators are not allowed. Please show all necessary details in your work. 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. (20 marks) Evaluate the following integrals.
+
dx
(a) ;
. x2 +4
. 1

dx
(b) . ;
0 x(1 . x)

. b
dx
(c) , (a< 1 <b);
3
a x . 1
. 1
f.(x) x2 . 1
(d) dx, where f(x)= .
.1 1+ f2(x) x
2. (20 marks) For each of the following series, determine if it is convergent.

1
(a) ;
n=2 n(ln n)1/n

11
(b) . ln1+ ;
nn
n=1

. (.1)n
(c) ;
n=1 ln(n + 1)

1
(d) sin , (1 <q<p)
pn . qn
n=1
3.
(10 marks) The curve y = 2 ln(4 . x2) crosses the x-axis at A and B. Find the length of the arc AB.

4.
(20 marks) Consider the .nite region R bounded by the curve y2 = a(1 . x) in the .rst quadrant, where a> 0.

(a)
Find the volume of the 3D region obtained by rotating R about the x-axis;

(b)
For what value of a, the volume of the region obtained by rotating R about the y-axis is the same as that obtained by rotating about the x-axis?




. 1
5. (15 marks) For k> 0, let Ik,n = x k lnn x dx.
0
n
(a) Show that Ik,n = . Ik,n.1;
k +1
n!
(b) Show that Ik,n =(.1)n ;
(k + 1)n+1
(c) Without giving detailed justi.cation for every step of the calcula-tion, please show that
. 1
1dx
=
nn xx
0 n=1
by using
. 1 . 1
dx
.x ln x dx
= e
xx
00
and the power series of the exponential function.
6. (10 marks) Suppose that the function f(x) satis.es the following con-ditions:
(i)
f is an odd continuous function in (., +);

(ii)
lim x 2f(x) = 0.


x
Prove that
+
f(x)dx =0.
.
7. (10 marks) By manipulating the power series

. n 1 x = , |x| < 1,
n=0 1 . x
show that
..2 ..4 ..6
1315171 4 3
++++ = +2ln .
2426282 3 4
Solution
1. (20 marks) Evaluate the following integrals.
+
dx
(a) ;
. x2 +4
. 1

dx
(b) . ;
0 x(1 . x)

. b
dx

(c) 3 , (a< 1 <b);
a x . 1
. 1
f.(x) x2 . 1
(d) dx, where f(x)= .
.1 1+ f2(x) x
Solution
(a)
+ dx u dx
= lim
. x2 +4 u+,v. v x2 +4
1 x.u
= lim arctan .
u+,v. 22 v 1 1 .
= . = .
2222 2
(b)
. 1 .
v
dx dx
. = lim .
0 x(1 . x) u0+,v1. u x(1 . x)
v
1 x
= lim dx.
u0+,v1. u x 1 . x
If we substitute
x = t2 ,
1 . x
then

t2 2t(1 + t2) . t2 2t 2t
x = , dx =dt =dt.
1+ t2 (1 + t2)2 (1 + t2)2
Thus,
1 x 12t
dx = t dt
x 1 . x t2 (1 + t2)2
(1+t2)
2
=dt
1+ t2
= 2 arctan t + C = 2 arctan x + C.
1 . x
Hence,

. x(1 . x)0 = = = + C x x=v lim 2 arc