Answer
$g''(4) = -2$
Work Step by Step
$g(x) = \int_{0}^{x}f(t)~dt$
Then we can say that $g'=f$ and $g'' = f'$
We can find $g''(4)$ which is $f'(4)$. This value is the slope of the graph of $f$ at the point $t = 4$:
$g''(4) = f'(4) = \frac{-2-2}{5-3} = \frac{-4}{2} = -2$