Answer
$g(4)=3$
$g'(4)=0$
Work Step by Step
$g(x)=\int_{0}^{x}f(t)dt$
$g(4)=\int_{0}^{4}f(t)dt$
Since $1$ is between $0$ and $4$ it follows:
$g(4)=\int_{0}^{1}f(t)dt+\int_{1}^{4}f(t)dt$
From the graph of $f$, the integral $\int_{0}^{1}f(t)dt$ represents the area of the region bounded between $f$, $t$ axis on $[0,1]$ which is the right triangle with heigh $2$ units and base $1$ unit.
$\int_{0}^{1}f(t)dt=\frac{1}{2} \cdot 2 \cdot 1=1$
Since the the function $f$ is negative on $[0,1]$ it follows that the integral of $f$ on $[0,1]$ is negative so:
$\int_{0}^{1}f(t)dt=-1$
From the graph of $f$, the integral $\int_{1}^{4}f(t)dt$ represents the area of the region bounded between $f$, $t$ axis on $[1,4]$ which is the trapezoid with heigh $2$ units and bases $3$ unit and $1$ unit.
$\int_{1}^{4}f(t)dt=\frac{1}{2} \cdot 2 \cdot (3+1)=4$
$g(4)=\int_{0}^{1}f(t)dt+\int_{1}^{4}f(t)dt \to g(4)=-1+4=3 $ ------------------------------------------------------------------ $g(x)=\int_{0}^{x}f(t)dt$
Using the fundamental theorem of calculus it follows:
$g'(x)=f(t)$
Substitute $x=4$ into the equation:
$g'(4)=f(4)$
From the graph of $f$, the value of $f(4)$ is $0$. So:
$g'(4)=0$
