Ah, ok. Yes, c=4 since f is a continuous function on [-5,c], then the average value of f from -5 to c = [tex]\frac{1}{c-(-5)}\[ \int_{-5}^{c} f(x)\,dx.\] [/tex]. Since the average value = 0, then it's just the integral from -5 to 4(since it's a graph, yeah the areas cancel out)
i also looked at the mean value theorem. it deals with f'(x)=0 ?
No, check out the MVT for integrals also known as the average value of a function.