Find the area of the region that is enclosed between the curves y = x2 and
y = x + 6.

Integrate:
[tex]\displaystyle\int_{-2}^3((x+6)-x^2)\,\mathrm dx = \int_{-2}^3(6+x-x^2)\,\mathrm dx[/tex]
[tex]=\left(6x+\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{-2}^3[/tex]
[tex]=\left(18+\dfrac92-9\right) - \left(-12+2+\dfrac83\right)[/tex]
[tex]=\boxed{\dfrac{125}6}[/tex]