The area of the region enclosed by the parabo | vedclass

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(B) To find the intersection points,substitute $y = \frac{x^2}{4}$ into $y^2 = 4x$: $(\frac{x^2}{4})^2 = 4x$ $\frac{x^4}{16} = 4x$ $x^4 = 64x$ $x(x^3

Vedclass · Accepted Answer

$\frac{16}{3}$

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(B) To find the intersection points,substitute $y = \frac{x^2}{4}$ into $y^2 = 4x$: $(\frac{x^2}{4})^2 = 4x$ $\frac{x^4}{16} = 4x$ $x^4 = 64x$ $x(x^3 - 64) = 0$ So,$x = 0$ or $x = 4$. When $x = 0, y = 0$. When $x = 4, y = 4$. The intersection points are $(0,0)$ and $(4,4)$. The area enclosed is given by: $	ext{Area} = \int_{0}^{4} (\sqrt{4x} - \frac{x^2}{4}) dx$ $= \int_{0}^{4} (2\sqrt{x} - \frac{x^2}{4}) dx$ $= [2 \cdot \frac{2}{3} x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3}]_{0}^{4}$ $= [\frac{4}{3} x^{3/2} - \frac{x^3}{12}]_{0}^{4}$ $= (\frac{4}{3} \cdot 8 - \frac{64}{12}) - 0$ $= \frac{32}{3} - \frac{16}{3} = \frac{16}{3} 	ext{ sq. units}$.

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

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