# Show that $\int\limits_0^af(x)g(x)dx=2\int\limits_0^af(x)dx,$ if $$f$$ and $$g$$ are defined as $$f(x) = f(a - x)$$ and $$g(x) + g(a - x)=4$$

Toolbox:
• (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
• (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int \limits_0^a f(x)g(x)dx-----(1)$

Now applyimg property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$

$I=\int \limits_0^a f(a-x)g(a-x)dx$

$I=\int \limits_0^a f(x)g(a-x)dx------(2)$

Adding equ (1) and equ (2)

$2I=\int \limits_0^a f(x)g(x)+f(x)g(a-x)dx$

$I=\int \limits_0^a f(x)\{g(x)+g(a-x)\}dx$

But $g(x)+g(a-x)=4$

Therefore$2I=\int f(x) \times 4dx$

$2I=4 \int f(x)dx$

Therefore $I=2\int f(x)dx$