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(MATH024)[2010](s)midterm~1292^_10407.pdf
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Math 024 -Honors Calculus II
Midterm Test, Spring Semester, 2010

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
n dx =(n = .1);
x + C, .e x dx = e x + C;
n +1

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

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

= arccos x + C.
1 . x2
1. (30 Marks) Evaluate the following de.nite integrals:
1
(a) f(x)dx, where
.1
.
.
1 . e.x3 , if x> 12 ,
.
f(x)= 43 .|x|, if |x| 1 ,
2
.

. ex3 , if x< .12 ;

1
(arcsin x)2
(b) dx;
0 x(1 . x)


1
x ln(x + 1+ x2)
(c) dx.
2
0 1+ x
2. (30 Marks) Evaluate the following inde.nite integrals:

dx
(a) ;
(a2 . x2)3


x
(b) dx;
x4 + x2 +1 1

sin x cos x
(c) dx.
(sin x + cos x)4
3. (10 Marks) Find the limit
[] 2
x
1
.(x+t)2
lim e dt.
2
x.0
x0
4. (10 Marks) For 0 <. and m 1, prove
2

dx. .. (x2 . 2x cos . + 1)m sin2m.1 ..
5. (10 Marks) For a<b, evaluate
b
dx
.
a
(b . x)(x . a)
6. (10 Marks) Assume that f is monotonic and continuous in [a, +),

with lim f(x) = 0. Prove that f(x) sin2 x dx is convergent if
x.+.
a
and only if f(x)dx is convergent.
a
Solution
1. (30 Marks) Evaluate the following de.nite integrals:
1
(a) f(x)dx, where
.1
.
3
1
.

1 . e.x
, if x>

,

2
.
3
if |x| 1
2
.|x|, , if x< .12 .
3
f(x)=

,

4
.
.
x
e
Solution

()
3
.|x|
4
1 .
1
1
1
1
()
2
2
3 d+x
3
.x
x
1 . e

f(x)dx

dx +

dx

= e

1
.1 .1 .
22
() 1
11
3

4

. x

1

22
.t3
3
.x
(.1) dt +2

.

dx +

dx

= e

e

2

1
2
10
(()2)
31 11
.
42 22
1

+ =1

2

=

2


1
(arcsin x)2
(b) dx. 0 x(1 . x)

Solution Let x = t, then x = t2 and dx =2t dt. Thus,

1 1
(arcsin x)2 (arcsin t)2
dx = 2t dt 0 x(1 . x) 0 t 1 . t2
t=1 .3 .3
= 2(arcsin t)3. =2 = .
3t=0 38 12

1
x ln(x + 1+ x2)
(c) dx.
2
0 1+ x
Solution

1
x ln(x + 1+ x2)
dx
2
0 1+ x
1
= ln(x + 1+ x2)d( 1+ x2)
0

x
1+
. x=1 1 2
.1+ x
= 1+ x2 ln(x + 1+ x2). . 1+ x2 dx
x=0 2
0 x + 1+ x 1 =2 ln(1 + 2) . 1dx 0 =2 ln(1 + 2) . 1.
2. (30 Marks) Evaluate the following inde.nite integrals:

dx
(a) .
(a2 . x2)3

Solution Let x = a sin t. Then

dxa cos t dt 1dt
== .
33 t
acosa2 cos2 t
(a2 . x2)3
We substitute u = tan t, then du = sec2 t dt. Thus,

dx 1 u
=du =+ C
22
aa
(a2 . x2)3
tan t
=+ C
a2
1 x
= a + C
a2 1 . (x )2 a
x
= + C.
a2 a2 . x2

x
(b) dx.
x4 + x2 +1
Solution Since
xx
=
x4 + x2 +1 (x2 + 1)2 . x2 x
=