The parabola y^2=4x divides the area of the c | vedclass

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(A) The intersection points of $y^2=4x$ and $x^2+y^2=5$ are found by substituting $y^2=4x$ into the circle equation:$x^2+4x-5=0$$(x+5)(x-1)=0$Since $x

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$\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$

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(A) The intersection points of $y^2=4x$ and $x^2+y^2=5$ are found by substituting $y^2=4x$ into the circle equation:$x^2+4x-5=0$$(x+5)(x-1)=0$Since $x \ge 0$ for the parabola,we have $x=1$,which gives $y=\pm 2$.The area of the smaller part is the area bounded by the parabola and the circle,which is symmetric about the $x$-axis.Area $= 2 	imes \left[ \int_0^1 \sqrt{4x} \, dx + \int_1^{\sqrt{5}} \sqrt{5-x^2} \, dx ight]$$= 2 	imes \left[ \frac{4}{3} x^{3/2} \Big|_0^1 + \left( \frac{x}{2} \sqrt{5-x^2} + \frac{5}{2} \sin^{-1} \frac{x}{\sqrt{5}} ight) \Big|_1^{\sqrt{5}} ight]$$= 2 	imes \left[ \frac{4}{3} + \left( 0 + \frac{5}{2} \sin^{-1}(1) ight) - \left( \frac{1}{2} \sqrt{4} + \frac{5}{2} \sin^{-1} \frac{1}{\sqrt{5}} ight) ight]$$= 2 	imes \left[ \frac{4}{3} + \frac{5\pi}{4} - 1 - \frac{5}{2} \sin^{-1} \frac{1}{\sqrt{5}} ight]$$= 2 	imes \left[ \frac{1}{3} + \frac{5}{2} \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{\sqrt{5}} ight) ight]$$= \frac{2}{3} + 5 \cos^{-1} \frac{1}{\sqrt{5}}$Since $\cos^{-1} \frac{1}{\sqrt{5}} = \sin^{-1} \frac{2}{\sqrt{5}}$,the area is $\frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} ight)$.

The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ into two parts. The area of the smaller part is equal to:

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