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(MATH024)024_final_2007.pdf
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Math 024 -Calculus II
Final Examination, Spring Semester, 2007

Time Allowed: 3 Hours Total Marks: 100
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. (10 Marks) For each of the following improper integrals, determine if it converges:
. 1+
11
(a) dx; (b) sin dx.
0 x 1 x
+
x
2.
(10 Marks) Evaluate the improper integral 3 dx.

0 1+ x

3.
(25 Marks) For each of the following series, determine if it is converges.


..n


(a) ; (d)(.1)n 1+ ;
2
1+ nn
n=1 n=1
ln n

1



(b) ; (e) sinn 1+ .
n ln2(n + 1)n
n=1 n=1
1

1


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

;


. 2
4. (5 Marks) Determine the region where the power series x nis con-
n=1
vergent.
5. (10 Marks) Find the Talor expansion of f(x) = (1+ x)3 about
(a) x = 0; (b) x = 1.
6. (15 Marks)
(a) Derive the expansion:
1

2 =1 . x 2 + x 4 . x 6 + ,
1+ x
for |x| < 1.

(b) Use (a) to derive the following identity:
111

arctan x = x . x 3 + x 5 . x 7 + .
357
for |x| < 1.

(c) Prove that

3.1 3.3
3 30 3.2
= . + . + .
6 1357
7. (20 Marks) Consider f(x) = sin(2x) de.ned on (0,).
(a)
Expand it as a sine series;

(b)
Expand it as a cosine series;

(c)
Show that for x [0,],



8 . cos(2k + 1)x
sin 2x = . .
k=0 (2k + 1)2 . 4
(d) Show that

2 1
111111 1 1
. =+ .. ++ .. + .
153759711 913 1115 1317 1519
83
8. (5 Marks) Show that for |x| < 1, the identity
..2

x n =(n + 1)x n n=0 n=0
holds.
1. (10 Marks) For each of the following improper integrals, determine if it converges:
. 1+
11
(a)



dx;

(b) sin dx.

0 x 1 x
Solution
(a) Since
. 1 . 1
11
dx = lim

dx



x

x .0+
0 . 1




lim 2

lim (2 . 2 .)=2,

=

=

x

.0+ x=..0+
the existence of the limit means that the improper integral is conver-gent, by the de.nition.
(b) Since ..
sin 1

x
lim =1,
1
x+
x
hence, for su.ciently large x, we have
sin 1
x 1
<,
1 2
x
which implies ..
11

< sin
2xx
for su.ciently large x. Since
+ 1 dx
1 2x
+ 1is divergent, by the Comparison Test, sin dx is divergent.
x
1
+
x
2. (10 Marks) Evaluate the improper integral dx.
3
0 1+ x
Solution Since
x 1 (1/3)(x . 1/2) + 1/2
dx =(.1/3)+ dx
1+ x3 x +1 x2 . x +1
11 x . 12 1dx
= . ln(x +1)+ dx +
3 3(x . 1 )2 + 3 2(x . 1 )2 + 3
24 24
1 1131 dx
= . ln(x +1)+ ln (x . )2 ++ .. .
.2
3 6 242 2
(3 ) (x . 1 ) +1
42
3

1 1 143 21
= . ln(x + 1) + ln(x 2 . x + 1) + arctan (x . )+ C
3 6 232 32
6 x2 . x +1 1 2x . 1