Let f(x) be a continuous function in R such that $f\left(x\right)+f\left(y\right)=f(x+y).$ If $\underset{0}{\overset{3}{\int }}f\left(x\right)dx=kthen\underset{-3}{\overset{3}{\int }}f\left(x\right)dx=$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}(\mathrm{x}+\mathrm{y}) \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=px \\ \underset{0}{\overset{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\overset{3}{\int }}\left|px\right|=k \\ \Rightarrow {\left(\mathrm{P} \frac{{\mathrm{x}}^2}{2}\right)}_0^3=\mathrm{k} \\ \Rightarrow \frac{\mathrm{P}}{2}\left(9\right)=\mathrm{k} \Rightarrow \mathrm{P}=\frac{2\mathrm{k}}{9} \\ \underset{-3}{\overset{3}{\int }} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} =\underset{-3}{\overset{3}{\int }} \frac{2\mathrm{kx}}{9}\mathrm{dx} \\ =\frac{2\mathrm{k}}{9}\left(0\right)=0 \end{gathered}$$