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(MATH1024)[2012](s)midterm~=vfemuek^_76164.pdf
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Time Allowed: 2 Hours Total Marks: 100 Note: Calculators are not allowed. Please show all necessary details in your work. The following integrals are given to you:
n+1 .
x
x n dx =+ C, .x
(n = .1);
e x dx = e + C;
n +1
dx
dx
= ln |x| + C;
= arctan x + C;
. x
x2 +1
dx
sin x dx = . cos x + C;
= arcsin x + C;
. 1 . x2
(.1)dx
cos x dx = sin x + C;
= arccos x + C.
1 . x2
1. (60 Marks) Evaluate the following integrals:

(a) x 2 log(x + 1 + x2) dx;
.
1
(b) (x + 1)2(x2 + 1) dx;
. /2 sin x
(c) x dx;
0 . 1 (1 + sin x) tan . x 2
(d) 0 x . 1 dx.
. + e.x2
2. (10 Marks) Show that the improper integral . . |x| dx converges.
3. (10 Marks) Compute d dt . b+t a+t e .x2 dx,

where a and b are some constants. Please justify your work.
1 4. (10 Marks) Suppose f is Riemann integrable on [a, b], and f(x) m> 1
0 for any x [a, b]. Show that is Riemann integrable on [a, b].
f(x)
.
2
5. (10 Marks) Show that sin(x 2)dx> 0.
0
1. (60 Marks) Evaluate the following integrals:

(a) x 2 log(x + 1+ x2)dx;
Solution
x 2 log(x + 1+ x2)dx
1 3)
= log(x + 1+ x2) d(x
3
x
. 1+
3 . 3 2
xx
1+ x
= log(x + 1+ x2) . dx
33 x + 1+ x2
3 . 3
xx
= log(x + 1+ x2) . dx
2
3 31+ x
3 . 2
xx
= log(x + 1+ x2) . d(1 + x 2)
2
3 61+ x
3 .
x1+ x2 . 1
= log(x + 1+ x2) . d(1 + x 2)
2
3 61+ x
3 .. .
x1 2).1/2
= log(x + 1+ x2) . (1 + x 2)1/2 . (1 + x d(1 + x 2)
36
3
x. 1 . 2 .
2)1/2
= log(x + 1+ x2) . (1 + x 2)3/2 . 2(1 + x + C
3 63
3 .. = xlog(x + 1+ x2) . 1(x 2 . 2) 1+ x2 + C.
39
1
(b) dx;
(x + 1)2(x2 + 1)
Solution Write

1 A BCx + D
=+ + .
(x + 1)2(1 + x2) x +1 (x + 1)2 x2 +1
Then
1= A(x + 1)(x 2 + 1) + B(x 2 + 1) + (Cx + D)(x + 1)2 ,
which gives
x = .1: 1=2B
x 3 : 0= A + C,
x =0: 1= A + B + D,

d .
. : 0=2A . 2B.
dx x=.1
Thus
11 1
A = ,B = ,C = . ,D =0,
222
so that
1 1111 1 x

= + . .
(x + 1)2(1 + x2)2 x +1 2(x + 1)2 2 x2 +1
Hence,

1 1111 1 x
dx = + . dx
(x + 1)2(x2 +1) 2 x +1 2(x + 1)2 2 x2 +1 1 11
= log |x +1|. . log(x 2 + 1) + C.
2 2(x +1) 4
. /2
sin x
(c) dx;
x
0 (1 + sin x) tan 2
x
Solution By y = tan , we have
2 2y 2
sin x = , dx =dy.
1+ y2 1+ y2
Hence
2y

2
sin x 1+ y12
dx = dy
x 2yy 1+ y2
(1 + sin x) tan
1+
2 2
1+ y
1
=4 dy.
(1 + y)2(1 + y2)
The last integral is given in the last problem. We have

. /2 . .. y=1
sin x 1 11 .
dx = 4 log |y +1|. . log(y 2 + 1) .
x
2 2(y +1) 4 y=0
0
(1 + sin x) tan 2 1 11 1
= 4 log2 .. log2+ =1+log2.
2 44 2
4
. 1 .
x
(d) dx;
1 . x
0
Solution
. 1 .
x
dx
0 . x . 1 . + 2y xy2 2y
= y dy, = y, x = , dx =dy
2)2 22)2
(1 + y1 . x 1+ y(1 + y
0
+

= . 2(1 + y2 . 1) dy
(1 + y2)2

0
+ +
2dy =dy . 2
22)2
1+ y(1 + y
00
. y=+ + dy
= 2 arctan y. . 2 y=0 (1 + y2)2
0
. /2 1

22
=2 . 2 sec u