## Thomas' Calculus 13th Edition

Integrate with respect to x, and then plug in the limits: $=\frac{1}{3}x^3-x^2+3x$ with limits from -1 to 1 $=\frac{1}{3}(1)^3-1+3-\frac{1}{3}(-1)^3+(-1)^2-3(-1)=1/3+2+1/3+1+3=20/3$