The area of region enclosed by the parabolas& | vedclass

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Question

For intersection points, substitute

$y^{2}=4 x$ or $y=2 \sqrt{x}$

${x^{2}=4 y} $

or   ${x^{2}=4 	imes 2 \sqrt{x}}$

or $\quad x^

Vedclass · Accepted Answer

$\frac{{16}}{3}$

Vedclass · Answer

For intersection points, substitute

$y^{2}=4 x$ or $y=2 \sqrt{x}$

${x^{2}=4 y} $

or   ${x^{2}=4 	imes 2 \sqrt{x}}$

or $\quad x^{4}=64 x$

or $\quad x\left(x^{3}-64ight)=0$

$	ext { or } \quad x=0$

$	ext { and } \quad x=\pm 4$

When $x=0, y^{2}=4 	imes 0$ or $y=0$

When $x=4, y^{2}=4 	imes 4$ or $y=4$

Points of intersection are $(0,0)$ and $(4,4)$

Given, $y^{2}=4 x$ or $y=2 \sqrt{x}=f(x)$

$	ext { and } \quad y=\frac{1}{4} x^{2}=g(x)$

where $f(x) \geq g(x)$ in $(0,4)$

$	herefore$ Required Area $=\int_{0}^{4}[f(x)-g(x)] d x$

${\left.=\int_{0}^{4}\left[2 \sqrt{x}-\frac{1}{4} x^{2}ight)ight] d x} $

${=\left[\frac{4}{3} x^{3 / 2}-\frac{1}{12} x^{3}ight]_{0}^{4}} $

${=\frac{4}{3} 	imes 4^{3 / 2}-\frac{1}{12} 	imes 4^{3}} $

${=\frac{16}{3} 	ext { sq. units }}$

The area of region enclosed by the parabolas ${y^2} = 4x$ and ${x^2} = 4y$ is

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