Question:

<div class='solution'> <p>$x^2-x=x(x-1)$: negative on $(0,1)$, non-negative elsewhere on $[-1,2]$.</p> <p>$$I=\int_{-1}^0(x^2-x)dx+\int_0^1(x-x^2)dx+\int_1^2(x^2-x)dx$$</p> <p>$$=\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1+\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_1^2$$</p> <p>$$=\left(0-(-\frac{1}{3}-\frac{1}{2})\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{8}{3}-2-(\frac{1}{3}-\frac{1}{2})\right)$$</p> <p>$$=\frac{5}{6}+\frac{1}{6}+\left(\frac{7}{3}-2+\frac{1}{6}\right)=\frac{5}{6}+\frac{1}{6}+\frac{14-12+1}{6}=\frac{5+1+3}{6}=\frac{9}{6}=\frac{3}{2}$$</p> <p>Checking: $\frac{11}{6}$ vs $\frac{3}{2}$. Recompute piece 3: $\frac{8}{3}-2-\frac{1}{3}+\frac{1}{2}=\frac{7}{3}-\frac{3}{2}=\frac{14-9}{6}=\frac{5}{6}$. Total: $\frac{5}{6}+\frac{1}{6}+\frac{5}{6}=\frac{11}{6}$. ✓</p> </div>