The area of the region \(x, y): y^2 <= 4 x, x0, x ≠ 3\ is

Correct option is (4) (32 / 3) \(x, y) ; y^2 0 => (x(x-1)(x-2) / (x-3)(x-4))>0 x in (0,1) union (2,3) Case (ii) $ y

The area of the region x, y y^2 4 x, x<4, \frac{x y(x-1... | null (Mathematics)

Correct option is (4)

\frac{32}{3}

\begin{aligned} & \left\{(x, y) ; y^2 \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0, x \neq 3\right\} \\ & y^2 \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \end{aligned}

Case (i)

y>0 \Rightarrow \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0

x \in(0,1) \cup(2,3)

Case (ii)

\mathrm{y}<0 \Rightarrow \frac{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2)}{(\mathrm{x}-3)(\mathrm{x}-4)}<0

x \in(1,2) \cup(3,4)

\text { Area }=\int_0^1 2 \sqrt{x} d x+\left|\int_1^2 2 \sqrt{x} d x\right|+\int_2^3 2 \sqrt{x} d x+\left|\int_3^4 2 \sqrt{x} d x\right|

\begin{aligned} & =\frac{4}{3}\left(\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left|\left[\mathrm{x}^{\frac{3}{2}}\right]_1^2\right|+\left[\mathrm{x}^{\frac{3}{2}}\right]^3+\left|\left(\mathrm{x}^{\frac{3}{2}}\right)^4\right|\right) \\ & \mathrm{A}=\frac{4}{3}(1+2 \sqrt{2}-1+3 \sqrt{3}-2 \sqrt{2}+4 \sqrt{4}-3 \sqrt{3})=\frac{4}{3} \times 8=\frac{32}{3} \end{aligned}