What is the area of the region bounded by the curve $y = x^{2} + x - 6$ and the $x$-axis?

Explanation

1. 01

   $x^{2} + x - 6 = (x + 3)(x - 2)$: the curve meets the $x$-axis at $x = -3$ and $x = 2$ and lies below it between them.

2. 02

   $\int_{-3}^{2} (x^{2} + x - 6)\,dx = -\frac{125}{6}$, so the area is $\frac{125}{6}$ square units.