Answer
$-2$
Work Step by Step
$g(x)=\int_{0}^{x}f(t)~dt$
Using the fundamental theorem of calculus it follows:
$g'(x)=f(x)$
Differentiate the above equation with respect to $x$:
$g''(x)=f'(x)$
Substitute $x=4$ into the equation:
$g''(4)=f'(4)$
Notice that $f'(4)$ represents the slope of the tangent line to the graph of $f$ at $x=4$.
Notice that $4$ is between $3$ and $5$ so took the part of $f$ that is between $3$ and $5$.
From the graph of $f$, $f$ represent the line that passes through the points $(3,2)$ and $(5,-2)$ so the tangent line to $f$ at $4$ is the slope of the linear function $f$ for $x$ between $3$ and $5$ .
So the slope of the line $f$ for $x$ between $3$ and $5$ is:
$f'(x)=\frac{-2-2}{5-3}=\frac{-4}{2}=-2$
Substitute $x=4$ into the equation:
$f'(4)=-2$
So:
$g''(4)=f'(4) \to g''(4)=-2$
