Exam 3 Exam 3                              Friday, May 11, 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 2 questions and has a total of 100 marks.

1 (40 Marks) Find the following limit

 lim x® 0+ æè sinxx öø 1/x2 .

Solution. Let f(x) = ([sinx/x])[1/(x2)]. Then we have lnf(x) = [(ln([sinx/x]))/(x2)] = [(lnsinx-lnx)/(x2)].

 lim x® 0+ lnf(x)
 =
 lim x®0+ lnsinx-lnxx2 = lim x® 0+ (lnsinx-lnx)¢2x
 =
 lim x® 0+ xcosx-sinx2x2sinx = lim x® 0+ cosx-xsinx -cosx2(2xsinx+x2cosx)
 =
 lim x® 0+ -sinx 2(2sinx+xcosx) = lim x® 0+ -cosx 2(3cosx-xsinx)
 =
 -1/6.

Hence limx® 0+lnf(x) = -1/6. Since f(x) = elnf(x) and g(x) = lnx is continuous, we have

 lim x® 0+ f(x) = lim x® 0+ elnf(x) = elimx®0+lnf(x) = e-1/6.

2.     (60 Marks)     Evaluate each of the following integrals.

(a)    òxÖ[(5x+3)] dx

(b)    òe2xsin(2x) dx

(c)    ò[1/(x3-5x2+8x-4)] dx

Solution. (a) Let u = 5x+3. Then x = (u-3)/5 and dx = du/5. Hence, we have

 óõ x ____Ö5x+3 dx
 =
 125 óõ (u-3)Öu du
 =
 125 óõ (u3/2-3u1/2 du
 =
 125 æè 25 u5/2-2u3/2 öø +C = 2125 u5/2- 225 u3/2+C
 =
 2125 (5x+3)5/2- 225 (5x+3)3/2+C

(b)

 óõ e2xsin(2x) dx
 =
 12 éë óõ sin(2x) de2x ùû = 12 éë e2xsin(2x)- óõ e2xcos(2x) d(2x) ùû
 =
 12 éë e2xsin(2x)- óõ cos(2x) d(e2x) ùû
 =
 12 éë e2xsin(2x)-e2xcos(2x)+ óõ e2x d cos(2x) ùû
 =
 12 éë e2xsin(2x)-e2xcos(2x)-2 óõ e2xsin(2x) dx ùû .

This implies 2òe2xsin(2x) dx = [1/2][e2xsin(2x)-e2xcos(2x)]+C and

 óõ e2xsin(2x) dx = 14 éë e2xsin(2x)-e2xcos(2x) ùû +C.

(c) Since [1/(x3-5x2+8x-4)] = [1/((x-1)(x-2)2)], we let

 1(x-1)(x-2)2 = Ax-1 + Bx-2 + C(x-2)2 .

Multiplied by (x-1)(x-2)2, we obtain

 A(x-2)2+B(x-1)(x-2)+C(x-1) = 1.

Let x = 1. Then A = 1.

Let x = 2. Then C = 1.

Let x = 0. Then 4A+2B-C = 1. This implies B = -1. Hence, we have

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

Integrating both sides of the above equation, we obtain

 óõ 1(x-1)(x-2)2 dx
 =
 óõ 1x-1 dx- óõ 1x-2 dx+ óõ 1(x-2)2 dx
 =
 ln|x-1|-ln|x-2|- 1x-2 +C.

File translated from TEX by TTH, version 2.79.
On 12 May 2001, 09:32.