Calculating the Area when inverse function is given

Question

For all the real number, $f\left( x \right)$ is an increasing function that is differentiable while satisfying the following conditions?

(A) $f\left( 1 \right) = 1,\int\limits_1^2 {f\left( x \right)dx} = \frac{5}{4}$

(B) Say g(x) is an inverse function of then, for all real number of x as $x \ge 1$, $g\left( {2x} \right) = 2f\left( x \right)$ If $\int\limits_1^8 {xf'\left( x \right)dx} = \frac{p}{q}$, then what is $p+q=$________?

My solution is as follow It is given that $x \ge 1$ and $f\left( 1 \right) = 1,\int\limits_1^2 {f\left( x \right)dx} = \frac{5}{4}$

${A_2} = 2 \times 2 - \left( {1 \times 1 + \frac{5}{4}} \right) = 4 - \frac{9}{4} = \frac{7}{4}$

$g\left( 1 \right) = g\left( {f\left( 1 \right)} \right) = 1$

$g\left( 2 \right) = 2f\left( 1 \right) = 2$

$g\left( 4 \right) = 2f\left( 2 \right)$

Also $g\left( {2x} \right) = 2f\left( x \right)$ , we need to find the value of $\int\limits_1^8 {xf'\left( x \right)dx} = \frac{p}{q}$

$T = \int\limits_1^8 {xf'\left( x \right)dx} = \left[ {xf\left( x \right)} \right]_1^8 - \int\limits_1^8 {\left( {\left( {\frac{d}{{dx}}x} \right)\int {f'\left( x \right)dx} } \right)dx} $

$T = \int\limits_1^8 {xf'\left( x \right)dx} = \left[ {xf\left( x \right)} \right]_1^8 - \int\limits_1^8 {\left( {\left( {\frac{d}{{dx}}x} \right)\int {f'\left( x \right)dx} } \right)dx} $

$T = \left[ {8f\left( 8 \right) - f\left( 1 \right)} \right] - \int\limits_1^8 {f\left( x \right)dx} \Rightarrow T = 8f\left( 8 \right) - 1 - \int\limits_1^8 {f\left( x \right)dx} $

I have elaborated my steps, how do I proceed

Answer 1

Here is a really big hint.

First, show using induction that $f(2^{n-1})=2^{n-1}$ for all $n\geq 1$.

Next, for each $n\geq 1$, set $$a_n=\int_{2^{n-1}}^{2^n}f(x)\mathrm{d}x$$ Enforce the substitution $x \mapsto \frac{f(x)}{2}$ on the integral for $a_{n+1}$ then follow up with integration by parts to get a nice recurrence relation: $$\begin{eqnarray*}a_{n+1}&=&\int_{2^n}^{2^{n+1}}f(x)\mathrm{d}x \\ &=& \int _{2^{n-1}}^{2^n}2x\cdot \frac{2\mathrm{d}x}{f'\left(f^{-1}(2x)\right)} \\ &=& \int_{2^{n-1}}^{2^n}2x\frac{\mathrm{d}}{\mathrm{d}x}\left(f^{-1}(2x)\right)\mathrm{d}x \\ &=& \int_{2^{n-1}}^{2^n}2x\frac{\mathrm{d}}{\mathrm{d}x}\left(2f(x)\right)\mathrm{d}x \\ &=& \int_{2^{n-1}}^{2^n}4xf'(x)\mathrm{d}x \\ &=& 4xf(x)\Bigg|_{x=2^{n-1}}^{x=2^n} - 4\int_{2^{n-1}}^{2^n}f(x)\mathrm{d}x \\ &=& 4^{n+1}-4^{n}-4a_n\end{eqnarray*}$$ Express $\int_1^8$ as $\int_1^2+\int_2^4+\int_4^8$ then use the fact that $a_1=5/4$ to get your answer.