The area of the region \ (x, y): y ^ 2 ≤ 4x, x 0, x ≠ 3\ is

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

Case - I: $y > 0$

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

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

Case – II : $y < 0$

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

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

So, the required area is given by,

$A = {∫}_{0}^{4}\sqrt{4x}dx$ {Take the mirror image of strips in the first quadrant to find the area}

$⇒ A = 2{∫}_{0}^{4}\sqrt{x}dx$

$⇒ A = 2 \cdot \dfrac{2}{3}{\left[x\dfrac{3}{2}\right]}_{0}^{4}$

$⇒ A = 2 \cdot \dfrac{2}{3}\left(4\dfrac{3}{2}\right)$

$⇒ A = 2 \times \dfrac{2}{3} \times 8$

$⇒ A = \dfrac{32}{3}$