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- Mar 10th 2009, 10:17 AMchamprockdefinite integral question
- Mar 10th 2009, 11:03 AMHallsofIvy
The graph of $\displaystyle y= \frac{3}{4}- x- x^2$ is a parabola opening downward. Also $\displaystyle \frac{3}{4}- x- x^2= \frac{-1}{4}(4x^2+ 4x- 3)= \frac{-1}{4}(2x- 3)(2x+ 1)= 0$ when x= 3/2 or -1/2. If a and b are both between -1/2 and 3/2, then the integral from a to b is less than the integral from -1/2 to 3/2. If either or both is outside that interval, you will be

**subtracting**from that.