The area of the region x , y : y 2 ≤ 4 x , x < 4 , x y x - 1 x - 2 x - 3 x - 4 > 0 , x ≠ 3 is

Given: $y^2\le 4x, x<4$, $\frac{xy\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-4\right)}>0$

Case - I: $y>0$

$\Rightarrow \frac{x\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-4\right)}>0$

$\Rightarrow x\in \left(0, 1\right)\cup \left(2, 3\right) ...\left(i\right)$

Case – II : $y<0$

$\frac{x\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-4\right)}<0$

$\Rightarrow x\in \left(1, 2\right)\cup \left(3, 4\right) ...\left(ii\right)$

So, the required area is given by,

$A={\int }_0^4\sqrt{4x}dx$ {Take the mirror image of strips in the first quadrant to find the area}

$\Rightarrow A=2{\int }_0^4\sqrt{x}dx$

$\Rightarrow A=2·\frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^4$

$\Rightarrow A=2·\frac{2}{3}\left(4^{\frac{3}{2}}\right)$

$\Rightarrow A=2\times \frac{2}{3}\times 8$

$\Rightarrow A=\frac{32}{3}$