Using the comparison theorem, show that ${\displaystyle \underset{-2}{\overset{3}{}}}(x-3)(x+2)dx0$

Since $x-30$ and $x+2v0$ over $-2,3,$ the integrand is $$ negative over $-2,3.$ By the comparison theorem, ${\displaystyle \underset{-2}{\overset{3}{}}}(x-3)(x+2)dx0$