g(x)=int_{-pi}^{x}sin(t)dt\n\ng(pi)=

Explanation:

Step1: Apply fundamental theorem of calculus

By the fundamental theorem of calculus, if $g(x)=\int_{a}^{x}f(t)dt$, then $g'(x) = f(x)$. Here $a = -\pi$ and $f(t)=\sin(t)$, so $g'(x)=\sin(x)$.

Step2: Evaluate $g'(\pi)$

Substitute $x = \pi$ into $g'(x)$. We get $g'(\pi)=\sin(\pi)$. Since $\sin(\pi)=0$, then $g'(\pi)=0$.

Answer:

$0$