Integral of a Kind!

First, let's work with the first and the third integrals alone; use the substitution $x \to \frac{1}{x}$ in the third one:

$\displaystyle \int_{\frac{1}{e}}^{\tan x } \frac{x}{1+x^2} dx + \int_{\frac{1}{e}}^{\cot x } \frac{dx}{x(1+x^2)}$

$\displaystyle \overset{\text{Substitution}}{=} \int_{\frac{1}{e}}^{\tan x } \frac{x}{1+x^2} dx + \int_{\tan x}^{e} \frac{x}{1+x^2} dx = \int_{\frac{1}{e}}^{e} \frac{x}{1+x^2} dx = 1$

The last step is easily done by basic integration techniques..

Now, our given equation is thus equivalent to:

$\displaystyle \int _{ 0 }^{ 1 }{ f(x) \left( 4{ x }^{ 2 } - f(x) \right) dx } = \frac{4}{5}$

Let $a= f(x) , b = 4x^2 - f(x)$ . The AM-GM* inequality states:

$\displaystyle \frac{a+b}{2} \geq \sqrt{ab}$

Substitute $a$ and $b$ :

$\displaystyle 2x^2 \geq \sqrt{f(x) (4x^2 - f(x) ) }$

Square bothe sides:

$\displaystyle f(x) (4x^2 - f(x) ) \leq 4x^4$

Using a property of definite integrals:

$\displaystyle \int_0^1 f(x) (4x^2 - f(x) ) dx \leq \int_0^1 4x^4 dx$

$\displaystyle <=> \int_0^1 f(x) (4x^2 - f(x) ) dx \leq \frac{4}{5}$

See that we are searching for an $f(x)$ satisfying the equality case of the above inequality, which is given by $a=b$ :

$\displaystyle f(x) = 4x^2 - f(x)$

Therefore : $\boxed{f(x) = 2x^2}$

So , $f(5) = 50$ is the final answer.

$.............................................$

• In order to use AM-GM inequality, we should prove that both $a$ and $b$ are non-negative.

Suppose that $f(x) >4x^2 \; \text{or} \; f(x) <0 \; \text{for} \; x\in [0,1]$ . A quick observation on $a$ and $b$ tells us that one of them is going to be less than zero, and the other is greater than $4x^2$ , which implies that $ab<0$ .

This contradicts our equation; because the integrand can't be negative for all $x$ . So our supposition is false. Hence $f(x) \in [0,4x^2]$ , which implies that $a$ and $b$ are non-negative.