Integral exercise asked by Katrine Linda solved by Substitution

Exercise:

Show that, $\int_{0}^{a}{\frac{f\left( x \right)}{f\left( x \right)+f\left( a-x \right)}=}\frac{a}{2}$

Solution: Let $I=\int_{0}^{a}{\frac{f\left( x \right)}{f\left( x \right)+f\left( a-x \right)}}$  (*)

Now by using the property $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( \left( a+b \right)-x \right)dx}$

So $I=\int_{0}^{a}{\frac{f\left( a-x \right)}{f\left( a-x \right)+f\left( x \right)}dx}$      (**)

Adding (*) and (**) we get :

$2I=\int_{0}^{a}{\frac{f\left( x \right)+f\left( a-x \right)}{f\left( a-x \right)+f\left( x \right)}dx=\int_{0}^{a}{dx}}=a$ thus $I=\frac{a}{2}$