ma241exam1-c-key

ma241exam1-c-key - Spring 2007 Therkelsen MA 241 Exam 1(C...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Spring 2007 Therkelsen MA 241 - Exam 1 (C) KEY 1. Evaluate 1 Z 0 x 4 x 2 dx Solution: Let u =4 x 2 , then du = 2 xdx . If we also transform the limits of integration, 4 (1) 2 = 3 and 4 (0) 2 ,wehave 1 Z 0 x 4 x 2 dx = 1 2 3 Z 4 1 u du = 1 2 (2 u ) 3 4 = u 3 4 = 3 ( 4) =2 3 2. Integrate Z x sin 4 xdx Solution: Let u = x and dv = sin 4 . Then du = dx and v = 1 4 cos 4 x . Integration by parts (and appropriate substitutions) yields the result Z x sin 4 = 1 4 x cos 4 x + Z 1 4 cos 4 = 1 4 x cos 4 x + 1 16 sin 4 x + C 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
3. Integrate Z x 2 2 x 1 ( x 1)( x 2 +1) dx Solution: We fnd the coefficients in the Following partial Fractions decomposition x 2 2 x 1 ( x 1)( x 2 = A x 1 + Bx + C x 2 +1 Multiplying both sides by ( x 1)( x 2 + 1) yields x 2 2 x 1= A ( x 2 +1)+( + C )( x 1) = Ax 2 + A + 2 + Cx C =( A + B ) x 2 +( B + C ) x A C ) We thereFore have the Following system oF linear equations A + B =1 B + C = 2 A C = 1
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/25/2011 for the course MA 241 taught by Professor Mccollum during the Spring '08 term at N.C. State.

Page1 / 5

ma241exam1-c-key - Spring 2007 Therkelsen MA 241 Exam 1(C...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online