Explain why would someone factorise $-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$ to get $\frac{1}{6}(3+2)^3$

Question

I am confused by the workflow of my book.

It went from $$-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$$ to $$-\int_{-2}^{3}(x+2)(x-3)\,\mathrm dx$$ and, finally $$\frac{1}{6}(3+2)^3$$

Answer 1

We have that more generally $$\begin{align} -\int_{-a}^{b}(x+a)(x-b)dx&=-\int_{-a}^{b}(x+a)(x+a-(a+b))dx\\&=-\left[\frac{(x+a)^3}{3}-(a+b)\frac{(x+a)^2}{2}\right]_{-a}^{b}\\&=-\frac{(a+b)^3}{3}+\frac{(a+b)(a+b)^2}{2}=\frac{(a+b)^3}{6}. \end{align}$$ This is also a particular the area of a parabolic segment where $w=a+b$, $h=\frac{a+b}{4}$ and the area is $$\frac{4}{3}\cdot\frac{w\cdot h}{2}=\frac{(a+b)^3}{6}.$$