# Math Help - Integral. #5

1. ## Integral. #5

Problem :

if f is continuous on [0,a] and $f(x) \cdot f(a-x) = 1$ , Evaluate :

$\int_0^a \dfrac{dx}{1+f(x)}$.

Final answer should be $\dfrac{a}{2}$

2. Let $I = \int_0^a \dfrac{dx}{1+f(x)}$

Make the change of variable $x = a - u$ so your integral becomes

$I = - \int_a^0 \dfrac{du}{1+f(a-u)}$

Now your your $f(x) \cdot f(a-x) = 1$ relation

so

$I = \int_0^a \dfrac{du}{1 + 1/f(u)} = \int_0^a \dfrac{f(u)}{1 + f(u)}du$

Now replace the $u$ with $x$ and add the two integrals

$2I = \int_0^a \dfrac{dx}{1+f(x)} + \int_0^a \dfrac{f(x)dx}{1+f(x)} = \int_0^a dx$

and solve for I.

3. See here for a similar result.