Question

Solve this:
Q. If f and g are continuous on [0, a] satisfying f (a - x) = f (x) and g (x) + g (a - x) = a, then show that ${\int }_0^{a}f\left(x\right)g\left(x\right)dx=\frac{a}{2}{\int }_0^{a}f\left(x\right)dx$

Answer 1

$$\begin{gathered} Given: \\ f\left(a-x\right)=f\left(x\right)\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right) \\ g\left(x\right)+g\left(a-x\right)=a\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(2\right) \\ I={\int }_0^{a}f\left(x\right)g\left(x\right)dx\_\_\_\_\_\_\_\_\_\_\_\_\_\left(3\right) \\ Weknow{\int }_a^{b}h\left(x\right)dx={\int }_a^{b}h\left(a+b-x\right)dx \\ I={\int }_0^{a}f\left(a+0-x\right)g\left(a+0-x\right)dx \\ I={\int }_0^{a}f\left(a-x\right)g\left(a-x\right)dx \\ Fromequation1 \\ I={\int }_0^{a}f\left(x\right)g\left(a-x\right)dx\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(4\right) \\ Addingequation3and4 \\ 2I={\int }_0^{a}f\left(x\right)g\left(x\right)dx+{\int }_0^{a}f\left(x\right)g\left(a-x\right)dx \\ ={\int }_0^a\left(f\left(x\right)g\left(x\right)+f\left(x\right)g\left(a-x\right)\right)dx \\ ={\int }_0^{a}f\left(x\right)\left(g\left(x\right)+g\left(a-x\right)\right)dx \\ Fromequation2 \\ 2I={\int }_0^{a}f\left(x\right)adx \\ 2I=a{\int }_0^{a}f\left(x\right)dx \\ I=\frac{a}{2}{\int }_0^{a}f\left(x\right)dx \end{gathered}$$