Exam 2 Exam 2                              Friday, April 20, 2001



NAME:

Student Number:------            No.----           Marks ----






Instructions:

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 5 questions and has a total of 100 marks.

3. Show all of your work. Your work must justify the answer that you give.

1. (35 Marks)    Evaluate each of the following indefinite integrals.


(a)    (15 Marks)   
1



4x2+4x-1
 dx                     (b)    (20 Marks)   
1
x3-x2+x-1
 dx.

Solution. (a)



1



4x2+4x-1
 dx
=

1



(2x+1)2-22
 dx = (1/2)
1



(2x+1)2-22
 d(2x+1)
=
(1/2) ln|(2x+1)+

 

(2x+1)2-22
 
|+C.

(b) Step 1. Factorizing the polynomial f(x) = x3-x2+x-1, we have


1
x3-x2+x-1
= 1
(x-1)(x2+1)
.

Step 2. Get partial fraction decomposition.

Let [1/((x-1)(x2+1))] = [A/(x-1)]+[(Bx+C)/(x2+1)].

Multipling the above equation by (x-1)(x2+1), we obtain


A(x2+1)+(Bx+C)(x-1) = 1.

Taking x = 1 in the above equation, we obtain A = 1/2.

Taking x = 0 in the above equation, we obtain C = -1/2.

Taking x = 2 in the above equation, we obtain 5A+2B+C = 1. This implies B = -1/2. Hence, we have


1
(x-1)(x2+1)
= 1/2
x-1
+ -x/2-1/2
x2+1
.

Step 3. Evaluate the indefinite integral.



1
(x-1)(x2+1)
 dx
=

1/2
x-1
 dx+
-x/2-1/2
x2+1
 dx
=
1
2
ln|x-1|- 1
2

x+1
x2+1
 dx
=
1
2
ln|x-1|- 1
4

1
x2+1
 d(x2+1)- 1
2

1
x2+1
 dx
=
1
2
ln|x-1|- 1
4
ln(x2+1)- 1
2
arctanx+C.

2. (20 Marks)    Determine whether the following integrals converges or diverges. If it converges, find its value.





3 
1
(x-2)q
 dx    (1 q < ).

Solution.

(1). When q = 1, [1/(x-2)] dx = ln| x-2|+C. Hence, we have



b

3 
1
x-2
 dx = ln(x-2)|3b = ln(b-2)-ln(3-2) = ln(b-2) .

and limb 3b[1/(x-2)] dx = limbln(b-2) = . This implies 3 [1/(x-2)] dx = limb3b [1/(x-2)] dx = and 3 [1/(x-2)] dx diverges.

2. When q > 1, we have [1/((x-2)q)] dx = [1/(1-q)](x-2)1-q+C. Let F(x) = [1/(1-q)](x-2)1-q. Then



b

3 
1
(x-2)q
 dx = F(x)|3b = 1
1-q
(b-2)1-q- 1
(1-q)
(3-2)1-q = 1
1-q
(b-2)1-q- 1
(1-q)

So, we have



lim
b  

b

3 
1
(x-2)q
 dx =
lim
b  
1
1-q
(b-2)1-q = 0- 1
1-q
= 1
q-1
.

Hence, 3[1/((x-2)q)] dx = limb 3b[1/((x-2)q)] dx = [1/(q-1)]. The integral 3[1/((x-2)q)] dx converges.

3.     (15 Marks)     Evaluate the following limit.

Solution.


n
  ___
n+1
 
-n
=
n
  ___
n+1
 
-n

  ___
n+1
 
+n


  ___
n+1
 
+n
= n

  ___
n+1
 
+n
=
1
1+   _____
1+1/n
 
1/2, n .

Hence, limn n([(n+1)]-n) = 1/2.

4.     (20 Marks) Evaluate k = 0([1/(k2+3k+2)]+[(2k)/(3k)])

Solution. (1) Evaluate k = 0[1/(k2+3k+2)].

Noting that


1
k2+3k+2
= 1
(k+1)(k+2)
= 1
k+1
- 1
k+2
,

we obtain the partial sum


Sn = (1- 1
2
)+( 1
2
- 1
3
)+...+( 1
n+1
- 1
n+2
) = 1- 1
n+2
.

and limn Sn = 1. Hence, k = 0[1/(k2+3k+2)] = 1.

(2) Evaluate k = 0[(2k)/(3k)].

The series k = 0[(2k)/(3k)] is a geometric series with a0 = 1 and r = [2/3]. It follows that




k = 0 
2k
3k
= a0
1-r
= 1
1-2/3
= 3.

(3) k = 0([1/(k2+3k+2)]+[(2k)/(3k)]) = k = 0[1/(k2+3k+2)]+k = 0 [(2k)/(3k)] = 1+3 = 4.

5.     (10 Marks) Determine whether the following series converges or diverges.




n = 1 
nsin(1/n).

Solution. Noting that nsin(1/n) = [(sin(1/n))/(1/n)]. we have



lim
n  
nsin(1/n) =
lim
n  
sin(1/n)
1/n
= 1 0.

It follows from the nth term test that the series n = 1nsin(1/n) diverges.


File translated from TEX by TTH, version 2.79.
On 20 Apr 2001, 14:23.