The area (in square units) bounded by the cur | vedclass

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(B) The given curves are:$y^2 = 4x$ ... $(i)$$x^2 = 4y$ ... (ii)From (ii),we have $y = \frac{x^2}{4}$. Substituting this into $(i)$:$(\frac{x^2}{4})^2

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$\frac{16}{3}$

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(B) The given curves are:$y^2 = 4x$ ... $(i)$$x^2 = 4y$ ... (ii)From (ii),we have $y = \frac{x^2}{4}$. Substituting this into $(i)$:$(\frac{x^2}{4})^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.Thus,$x = 0$ or $x = 4$. The intersecting points are $(0,0)$ and $(4,4)$.The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:$	ext{Area} = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx$$= \int_0^4 (2x^{1/2} - \frac{x^2}{4}) dx$$= [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12}]_0^4$$= [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_0^4$$= [\frac{4}{3}(4)^{3/2} - \frac{4^3}{12}] - [0 - 0]$$= [\frac{4}{3} \cdot 8 - \frac{64}{12}]$$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3} 	ext{ sq. units.}$

The area (in square units) bounded by the curves $y^2=4x$ and $x^2=4y$ in the plane is

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