Quiz4 Solutions - MAC 2312 Quiz 4 February 7, 2012...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAC 2312 Quiz 4 February 7, 2012 SOLUTIONS 1. Evaluate the indefinite integral integraldisplay 1- x 3 + x dx. Solution: Consider the change of variables w = 3 + x dw = 1 2 x dx = 1 2( w- 3) dx (1) and observe that 1- x = 1- ( w- 3) = 4- w. (2) Therefore, using both (1) and (2) and performing standard simplifications and integrations, the given integral cam be rewritten as follows: integraldisplay 1- x 3 + x dx = 2 integraldisplay (4- w )( w- 3) w dw =- 2 integraldisplay w 2- 7 w + 12 w dw =- 2 integraldisplay parenleftbigg w- 7 + 12 w parenrightbigg dw =- 2 parenleftbigg 1 2 w 2- 7 w + 12 ln | w | parenrightbigg + C. (3) Using the definition of w given in equation (1) gives 1 2 w 2- 7 w + 12 ln | w | = 1 2 (3 + x ) 2- 7(3 + x ) + 12 ln | 3 + x | = x 2- 4 x + 12 ln | 3 + x | + 9 2- 21 , and using this expression in the right-hand side of (3) gives integraldisplay 1- x 3 + x dx =- x + 8 x- 24 ln | 3 + x | + C. 2. Evaluate the indefinite integral integraldisplay 1 x 3 + 6 x dx. Solution: Using the change of variable w = 3 + 6 x dw = 6 2 3 + 6 x dx = 3 w dx (4) we may write x = 1 6 ( w 2- 3) . (5) Therefore, using equations (4) and (5) and performing routine simplifications and a partial fraction decomposition, we may write the given integral as follows...
View Full Document

Page1 / 5

Quiz4 Solutions - MAC 2312 Quiz 4 February 7, 2012...

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