Answer
$g'(x)=2+\sin x$
See image for graph of $g(x)$
Work Step by Step
a) $\frac{d}{dx}\int_0^x(2+\sin t)dt=2+\sin x$
b) $\int_0^x(2+\sin t)dt=F(x)-F(0)=2x-\cos x-2\times0+\cos0=2x-cosx+1$
$\frac{d}{dx}(2x-cosx+1)=2+\sin x$
See image for graph of $g(x)$
