Area of the region (in sq units) bounded by t | vedclass

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(B) Given curves are $y=\sqrt{x}$ and $x=\sqrt{y}$.Squaring $x=\sqrt{y}$ gives $y=x^2$.We need to find the area bounded by $y=x^2$ and $y=\sqrt{x}$ be

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$\frac{49}{3}$

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(B) Given curves are $y=\sqrt{x}$ and $x=\sqrt{y}$.Squaring $x=\sqrt{y}$ gives $y=x^2$.We need to find the area bounded by $y=x^2$ and $y=\sqrt{x}$ between $x=1$ and $x=4$.In the interval $[1, 4]$,the curve $y=x^2$ lies above the curve $y=\sqrt{x}$.The required area $A$ is given by:$A = \int_1^4 (x^2 - \sqrt{x}) dx$$A = \left[ \frac{x^3}{3} - \frac{2}{3} x^{3/2} ight]_1^4$$A = \left( \frac{4^3}{3} - \frac{2}{3} (4)^{3/2} ight) - \left( \frac{1^3}{3} - \frac{2}{3} (1)^{3/2} ight)$$A = \left( \frac{64}{3} - \frac{2}{3} 	imes 8 ight) - \left( \frac{1}{3} - \frac{2}{3} ight)$$A = \left( \frac{64}{3} - \frac{16}{3} ight) - \left( -\frac{1}{3} ight)$$A = \frac{48}{3} + \frac{1}{3} = \frac{49}{3} 	ext{ sq units}$.

Area of the region (in sq units) bounded by the curves $y=\sqrt{x}$,$x=\sqrt{y}$ and the lines $x=1$,$x=4$ is

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