Answer
$\frac{13}{6}$
Work Step by Step
Setup the integration to find the area bounded by the two functions
$\int_0^1 [(-x+2) -(x^2-1)]dx$
$\int_0^1(-x^2-x+3)dx$
$[-\frac{1}{3}x^3 - \frac{1}{2}x^2 + 3x]_0^1$
$(-\frac{1}{3}-\frac{1}{2} +3)- (0)$
$\frac{13}{6}$
