The area enclosed between the parabolas {y^2 | vedclass

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(D) The given equations are ${y^2 = 4x}$ and ${x^2 = 4y}$.To find the points of intersection,substitute ${y = \frac{x^2}{4}}$ into ${y^2 = 4x}$:$(\fra

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$\frac{16}{3} 	ext{ sq. unit}$

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(D) The given equations are ${y^2 = 4x}$ and ${x^2 = 4y}$.To find the points of intersection,substitute ${y = \frac{x^2}{4}}$ into ${y^2 = 4x}$:$(\frac{x^2}{4})^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.Thus,the intersection points are ${x = 0}$ and ${x = 4}$.The area $A$ enclosed between the curves is given by the integral:$A = \int_{0}^{4} (\sqrt{4x} - \frac{x^2}{4}) \, dx$$A = \int_{0}^{4} (2\sqrt{x} - \frac{x^2}{4}) \, dx$$A = [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12}]_{0}^{4}$$A = [\frac{4}{3} x^{3/2} - \frac{x^3}{12}]_{0}^{4}$$A = (\frac{4}{3} \cdot 8 - \frac{64}{12}) - (0 - 0)$$A = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} 	ext{ sq. unit}$.

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

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