{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz10-10.40-soln.

# quiz10-10.40-soln. - b Evaluate the definite integral& =...

This preview shows page 1. Sign up to view the full content.

MATH 101-B Fall 2004 Quiz 10 (10:40 – 12:30 Group) Dec. 17, 2004 Time : 15 minutes Name: ___________________ Student No: ___________ Follow the directions. No work = No credit!! Problem (6 + 4 pts) a. If = a xdx 0 3 cos , sin(-a) = ? Express your reasoning in words. Solution: Since cos is an even function: & - = 0 0 cos cos a a xdx xdx sin is an antiderivative of cos, so from FTC: 3 cos ) sin( ) 0 sin( 0 = = - - & - a xdx a Therefore sin(-a) = -3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: b. Evaluate the definite integral & = + 1 ? ) 5 ( dx x x x Solution: We first find an antiderivative for the function in the integral, 3 2 5 ln 4 5 ln 1 2 / 3 1 5 ln 5 2 / 3 5 ln 5 5 2 / 3 5 ln 5 5 1 2 / 3 1 2 / 3 + =--+ = ± ± ² ³ ´ ´ µ ¶ + = ± ± ² ³ ´ ´ µ ¶ + + + = ± ± ² ³ ´ ´ µ ¶ + & & x dx x x C x dx x x x x x x...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online