The area of the region {(x, y): y^2 leq 4x, x | vedclass

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(D) Given the region defined by $y^2 \leq 4x$,$x  0$.For $y > 0$,we need $\frac{x(x-1)(x-2)}{(x-3)(x-4)} >

Vedclass · Accepted Answer

$\frac{32}{3}$

Vedclass · Answer

(D) Given the region defined by $y^2 \leq 4x$,$x  0$.For $y > 0$,we need $\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$. Using the wavy curve method,the intervals for $x$ are $(0, 1) \cup (2, 3)$.For $y The area is given by the sum of integrals of $2\sqrt{x}$ over these intervals:$	ext{Area} = \int_0^1 2\sqrt{x} dx + \int_2^3 2\sqrt{x} dx + \int_1^2 2\sqrt{x} dx + \int_3^4 2\sqrt{x} dx$Combining these,we get:$	ext{Area} = \int_0^4 2\sqrt{x} dx = 2 	imes \frac{2}{3} [x^{3/2}]_0^4 = \frac{4}{3} 	imes (4^{3/2} - 0) = \frac{4}{3} 	imes 8 = \frac{32}{3}$.

The area of the region $\{(x, y): y^2 \leq 4x, x < 4, \frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0, x \neq 3\}$ is

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