Question

given that, g(x) = sin xExplain reason if the question cannot be solved.

Answer 1

Given$g(x)=sinx;\text{ }\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx$Solution$\int sinx=-cosx+C$$\begin{gathered} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sinxdx=-cos(\frac{\pi}{2})--cos(-\frac{\pi}{2}) \\ =-cos(\frac{\pi}{2})+cos(-\frac{\pi}{2}) \\ \\ =0 \end{gathered}$The final answer$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx=0$