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(MATH024)[2008](s)midterm~=8hrzkjdb^_80058.pdf
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Math 024 -Calculus II
Midterm Test, Spring Semester, 2008

Time Allowed: 2 Hours Total Marks: 110
Note: Calculators are not allowed. Please show all necessary details in your work. The following integrals are given to you:
. 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
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:
. 2
(a) x e|x.1| dx.
0
. 2
(b) f(x) dx, where
.2
f(x) = . 2 . 2x2 , 2 . 2|x|, if |x| 1 if |x| > 1
.
2
(c) sin x ln(sin x) dx.
4

2. (30 Marks) Evaluate the following inde.nite integrals:
. x3 +1
(a) dx.
(x . 1)4 1
(b) dx.
x(x3 + x2 + x + 1)
x
(c) dx.
b . x
3. (30 Marks)
(a) Let m and n be non-negative integers. Show that
. 1
m! n!
x m(1 . x)n dx = . (0! = 1)
0 (m + n + 1)!
. 1
x4(1 . x)4
(b) Evaluate 2 dx.
0 1+ x
(c) Use (a) and (b) to prove
122 1

< . < .
1260 7 630
..1
n
1 (2n)!
4. (10 Marks) Compute lim .
n
nn!
5. (10 Marks) Find the limit:
. h
1
lim [f(x + h) . f(x)] dx,
h0 h2
0
where f is di.erentiable function in (., +).
1. (30 Marks) Evaluate the following de.nite integrals:
. 2
(a) x e|x.1| dx.
0
. 2
(b) f(x) dx, where
.2

2 . 2x2 , if |x| 1
f(x)=
2 . 2|x|,
if |x| > 1

(c)

2

4

sin x ln(sin x)dx.

Solution
(a)
. 2
.1
. 2
x e|x.1| dx = x e .(x.1) dx + x e x.1 dx
0 0 . 1 1 . 2
= x e .(x.1) dx + x e x.1 dx
0 . 1 1 . 2
= e (.x)(e .x). dx + e .1 x(e x). dx

01
. 1 . 2
(.x)e

.x
x=1 x=2
.x(.1) dx + e .1 xe x
e x dx
.

.

= e

e

x=0 x=10
+ e
1
.x
.1 . e
x=1 x=2
.1 2e 2 . e . e x
.e

=2e . 2.

= e

x=0 x=1
(b)
. 2 .2 f(x) dx = = . .1 .2 f(x) dx + . 1 .1 f(x) dx + . 2 1 f(x) dx . .1 .2 [2 . 2(.x)] dx + . 1 .1 (2 . 2x 2) dx + . 2 1 (2 . 2x) dx

=

(2x + x

2) x=.1
x=.2
+ (2x . 2x

3/3)
x=1
x=.1
+ (2x . x

2) x=2
=2/3

x=1
(c)

2

sin x ln(sin x)dx
2

4

ln(sin x)(. cos x). dx
=

2

4

1

sin x

x=/2
. cos x ln(sin x).
=

x=/4
2

2
11

cos

x
(cos x). dx
= ln +

cos2 x . 1
2

4

22

11
1

(cos x). dx
= ln + 1+

2 x . 1
22

11
cos

1

11

(cos x). dx
ln

.

+ 1+

=

2 cos x . 1 cos x +1

22

11 1
4.
x=/2
ln + cos x + [ln |1 . cos x|. ln |1 + cos x|]
22 2
=

x=/4

1 111
= ln . . ln(3 . 2).
2 222
2. (30 Marks) Evaluate the following inde.nite integrals:
. x3 +1
(a) dx.
(x . 1)4 1
(b) dx.
x(x3 + x2 + x + 1)
x
(c) dx.
b . x
Solution
(a)
. x3 +1
dx (x . 1)4
(x . 1 + 1)3 +1
= dx
(x . 1)4
(x . 1)3 + 3(x . 1)2 + 3(x . 1) + 2
=dx
(x . 1)4
13 3 2
= +