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.

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

(a)    (15 Marks)    ó
õ
1
 Ö 4x2+4x-1
dx                     (b)    (20 Marks)    ó
õ
1
x3-x2+x-1
dx.

Solution. (a)

ó
õ
1
 Ö 4x2+4x-1
dx
 =
ó
õ
1
 Ö (2x+1)2-22
dx = (1/2) ó
õ
1
 Ö (2x+1)2-22
d(2x+1)
 =
 (1/2) ln|(2x+1)+ Ö (2x+1)2-22 |+C.

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

 1x3-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) = 1/2x-1 + -x/2-1/2x2+1 .

Step 3. Evaluate the indefinite integral.

 óõ 1(x-1)(x2+1) dx
 =
 óõ 1/2x-1 dx+ óõ -x/2-1/2x2+1 dx
 =
 12 ln|x-1|- 12 óõ x+1x2+1 dx
 =
 12 ln|x-1|- 14 óõ 1x2+1 d(x2+1)- 12 óõ 1x2+1 dx
 =
 12 ln|x-1|- 14 ln(x2+1)- 12 arctanx+C.

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

 óõ ¥ 3 1(x-2)q dx    (1 £ q < ¥).

Solution.

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

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

and limb® ¥ò3b[1/(x-2)] dx = limb®¥ln(b-2) = ¥. This implies ò3¥ [1/(x-2)] dx = limb®¥ò3b [1/(x-2)] dx = ¥ and ò3¥ [1/(x-2)] dx diverges.

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

 óõ b 3 1(x-2)q dx = F(x)|3b = 11-q (b-2)1-q- 1(1-q) (3-2)1-q = 11-q (b-2)1-q- 1(1-q)

So, we have

 lim b® ¥ óõ b 3 1(x-2)q dx = lim b® ¥ 11-q (b-2)1-q = 0- 11-q = 1q-1 .

Hence, ò3¥[1/((x-2)q)] dx = limb® ¥ò3b[1/((x-2)q)] dx = [1/(q-1)]. The integral ò3¥[1/((x-2)q)] dx converges.

3.     (15 Marks)     Evaluate the following limit.

Solution.

 Ön æè ___Ön+1 -Ön öø
 =
 Ön æè ___Ön+1 -Ön öø æè ___Ön+1 +Ön öø

 æè ___Ön+1 +Ön öø
= Ön
 æè ___Ön+1 +Ön öø
 =
1
 1+ _____Ö1+1/n
® 1/2, n® ¥.

Hence, limn® ¥Ön(Ö[(n+1)]-Ön) = 1/2.

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

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

Noting that

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

we obtain the partial sum

 Sn = (1- 12 )+( 12 - 13 )+...+( 1n+1 - 1n+2 ) = 1- 1n+2 .

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

(2) Evaluate åk = 0¥[(2k)/(3k)].

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

 ¥å k = 0 2k3k = a01-r = 11-2/3 = 3.

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

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

 ¥å 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 ¹ 0.

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

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