Inverse of a function related area problem 
My solution is based on the following diagram which I will illustrate in details

Now ${A_1} = \frac{5}{4};{A_3} = 1$
$g\left( {2x} \right) = 2f\left( x \right)$
$g\left( 2 \right) = 2f\left( 1 \right) = 2 \times 1 = 2$
$g\left( 4 \right) = 2f\left( 2 \right)$
$\int\limits_1^8 {xf'\left( x \right)dx}  = \left. {xf\left( x \right)} \right|_1^8 - \int\limits_1^8 {\frac{d}{{dx}}x\left( {\int {f'\left( x \right)dx} } \right)dx} $
$\int\limits_1^8 {xf'\left( x \right)dx}  = 8f\left( 8 \right) - f\left( 1 \right) - \int\limits_1^8 {f\left( x \right)dx} $
Not able to proceed from here
 A: 
Lemma: If $f(x)$ is a continuous and increasing function and $a<b$, then:
$$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\, f(b)-a\, f(a). $$

To prove it, you just need to draw a picture or look at this Wikipedia page.
Now, following my comment, required integral = $ \displaystyle \int_{1}^8 x f'(x) dx= \int_{1}^8 g(x) dx=I\quad $  (say)
And, $\;f(n)=g(n)=n\;$ for $\; n=1,2,4,8.$
Using the above lemma on the given function $f$, we have
\begin{align} 
&\int_{1}^{2} f(x) dx+\int_{f(1)}^{f(2)} g(x) dx=2f(2)-f(1)\\
\implies & \frac5{4} + \int_{1}^{2} g(x) dx = 3\\
\implies & \int_{1}^{2} g(x) dx=\frac7{4} \tag{1}
\end{align}
Now, $\;2f(x)=g(2x)\:$ and $\: \displaystyle\int_{1}^{2} f(x) dx=\frac5{4}$
\begin{align}
\implies & \frac{1}{2}\int_{1}^{2} g(2x) dx=\frac5{4}\\
\implies & \frac{1}{4}\int_{2x=2}^{2x=4} g(2x) d(2x)=\frac5{4}\\
\implies & \frac{1}{4}\int_{2}^{4} g(x) dx=\frac5{4}\\
\implies &\int_{2}^{4} g(x) dx=5 \tag{2}
\end{align}
Again using the lemma on $f$, we have
\begin{align} &\int_{2}^{4} f(x) dx+\int_{f(2)}^{f(4)} g(x) dx=4f(4)-2f(2)\\
\implies & \int_{2}^{4} f(x) dx + \int_{2}^{4} g(x) dx = 16-4\\
\implies & \int_{2}^{4} f(x) dx=12-5=7 \qquad \text{using (2)}
\end{align}
Again, $\;2f(x)=g(2x)\:$ and $\: \displaystyle \int_{2}^{4} f(x) dx=7$
\begin{align}
\implies & \frac{1}{2}\int_{2}^{4} g(2x) dx=7\\
\implies & \frac{1}{4}\int_{2x=4}^{2x=8} g(2x) d(2x)=7\\
\implies & \frac{1}{4}\int_{4}^{8} g(x) dx=7\\
\implies &\int_{4}^{8} g(x) dx=28 \tag{3}
\end{align}
From $(1), (2)$ and $(3)$
\begin{align} I&=\int_{1}^8 g(x) dx\\
&= \int_{1}^2 g(x) dx+\int_{2}^4 g(x) dx+\int_{4}^8 g(x) dx\\
&= \frac7{4} + 5 +28\\
&= \frac{139}4
\end{align}
Since, $\text{HCF}(139,4)=1\;$ required answer is $139+4=143$.
