The area included between the parabolas y^{2} | vedclass

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Question

(B) The given parabolas are $y^{2} = 5x$ and $x^{2} = 5y$.To find the points of intersection,substitute $y = \frac{x^{2}}{5}$ into the first equation:

Vedclass · Accepted Answer

$\frac{25}{3} 	ext{ sq. units}$

Vedclass · Answer

(B) The given parabolas are $y^{2} = 5x$ and $x^{2} = 5y$.To find the points of intersection,substitute $y = \frac{x^{2}}{5}$ into the first equation:$(\frac{x^{2}}{5})^{2} = 5x \Rightarrow \frac{x^{4}}{25} = 5x \Rightarrow x^{4} = 125x \Rightarrow x(x^{3} - 125) = 0$.Thus,$x = 0$ or $x = 5$. The points of intersection are $(0, 0)$ and $(5, 5)$.The area $A$ bounded by the two curves is given by:$A = \int_{0}^{5} (\sqrt{5x} - \frac{x^{2}}{5}) dx$$A = \sqrt{5} \int_{0}^{5} x^{1/2} dx - \frac{1}{5} \int_{0}^{5} x^{2} dx$$A = \sqrt{5} [\frac{x^{3/2}}{3/2}]_{0}^{5} - \frac{1}{5} [\frac{x^{3}}{3}]_{0}^{5}$$A = \sqrt{5} \cdot \frac{2}{3} \cdot (5)^{3/2} - \frac{1}{15} \cdot (5)^{3}$$A = \frac{2}{3} \cdot 5 \cdot 5 - \frac{125}{15} = \frac{50}{3} - \frac{25}{3} = \frac{25}{3} 	ext{ sq. units}$.Therefore,the correct option is $B$.

The area included between the parabolas $y^{2} = 5x$ and $x^{2} = 5y$ is

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