Answer
$\dfrac{125}{6}$
Work Step by Step
We match both functions to find where they intersect
$-x^2+x+6 = 0 \quad \rightarrow \quad x^2-x-6 = 0 \quad \rightarrow \quad (x-3)(x+2) = 0$
Therefore both functions intersect in the points $(-2, 0)$ and $(3, 0)$, thus we take the integral from $-2$ to $3$
${\displaystyle \int_{-2}^{3}} [-x^2+x+6] dx = \bigg[ -\dfrac{x^{2+1}}{2+1} + \dfrac{x^{1+1}}{1+1}+6x \bigg] \Bigg\rvert_{-2}^{3}= \bigg[-\dfrac{x^3}{3} + \dfrac{x^2}{2} + 6x\bigg] \Bigg\rvert_{-2}^{3} =\\ -\dfrac{3^3}{3} + \dfrac{3^2}{2} + 6(3) - \bigg[-\dfrac{(-2)^3}{3} + \dfrac{(-2)^2}{2} + 6(-2)\bigg] = -9 + \dfrac{9}{2}+18-\bigg[\dfrac{8}{3} + 2 -12 \bigg] = \\
\dfrac{27}{2} -\bigg[-\dfrac{22}{3} \bigg] = \dfrac{27}{2} +\dfrac{22}{3} = \dfrac{125}{6}$
