This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CARLETON UNIVERSITY FINAL EXAMINATION December 2007 SOLUTIONS DURATION: 3 HOURS Department Name and Course Number: Mathematics and Statistics, MATH 2007 Course Instructor(s): Dr. L. Haque, Dr. S. Melkonian AUTHORIZED MEMORANDA NONPROGRAMMABLE CALCULATORS ONLY This examination may be released to the Library. This examination paper may not be taken from the examination room. [Marks] 1. Evaluate lim x → e x x 1 x 2 . [4] Solution: By L’Hospital’s rule, lim x → e x x 1 x 2 = lim x → e x 1 2 x = lim x → e x 2 = 1 2 . 2. Evaluate the following integrals. [25] Solution: (a) 4 x √ 1 x 2 dx = 2 √ 1 x 2 ( 2 x ) dx = 2 u 1 / 2 du = 4 3 u 3 / 2 + C = 4 3 (1 x 2 ) 3 / 2 + C ( u = 1 x 2 , du dx = 2 x ). (b) 4 x cos(2 x ) dx = 2 x sin(2 x ) 2 sin(2 x ) dx = 2 x sin(2 x ) + cos(2 x ) + C . (c) cos 3 ( x ) sin( x ) dx = cos 3 ( x )[ sin( x )] dx = u 3 du = 1 4 u 4 + C = 1 4 cos 4 ( x ) + C ( u = cos( x ) , du dx = sin( x ))....
View
Full
Document
This note was uploaded on 01/18/2012 for the course MATH 2007 taught by Professor S.melkonian during the Fall '11 term at Carleton CA.
 Fall '11
 S.Melkonian
 Calculus, Statistics

Click to edit the document details