Answer
$F^{'}(x)=x^{2}-2x$
Work Step by Step
$F(x)=\int_{-2}^{x}(t^{2}-2t)$
$F^{'}(x)=\frac{d}{dx}\int_{-2}^{x}(t^{2}-2t)$
$=x^{2}-2x$
This is valid since the function $f(t)=t^{2}-2t$ is a continuous function over the implied domain, $\mathbb{R}$
