The parabola y^2=4x+1 divides the disc x^2+y^ | vedclass

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(B) Given the parabola $y^2=4x+1$ and the circle $x^2+y^2=1$.The intersection points are found by substituting $y^2=4x+1$ into $x^2+y^2=1$:$x^2 + (4x+

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$\frac{2}{3}$

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(B) Given the parabola $y^2=4x+1$ and the circle $x^2+y^2=1$.The intersection points are found by substituting $y^2=4x+1$ into $x^2+y^2=1$:$x^2 + (4x+1) = 1 \Rightarrow x^2 + 4x = 0 \Rightarrow x(x+4) = 0$.Since the parabola vertex is at $x = -1/4$,the intersection occurs at $x=0$ (where $y = \pm 1$).The area $A_1$ (shaded region) is given by:$A_1 = 2 \left( \int_{-1/4}^{0} \sqrt{4x+1} \, dx + \int_{0}^{1} \sqrt{1-x^2} \, dx \right)$Calculating the integrals:$2 \int_{-1/4}^{0} (4x+1)^{1/2} \, dx = 2 \left[ \frac{2}{3} \cdot \frac{1}{4} (4x+1)^{3/2} \right]_{-1/4}^{0} = \frac{1}{3} [1 - 0] = \frac{1}{3}$.$2 \int_{0}^{1} \sqrt{1-x^2} \, dx = 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2} \sin^{-1}(x) \right]_{0}^{1} = 2 \left[ 0 + \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi}{2}$.So,$A_1 = \frac{1}{3} + \frac{\pi}{2}$.The total area of the circle is $\pi$. Thus,$A_2 = \pi - A_1 = \pi - (\frac{1}{3} + \frac{\pi}{2}) = \frac{\pi}{2} - \frac{1}{3}$.Finally,$|A_1 - A_2| = |(\frac{1}{3} + \frac{\pi}{2}) - (\frac{\pi}{2} - \frac{1}{3})| = |\frac{2}{3}| = \frac{2}{3}$.

The parabola $y^2=4x+1$ divides the disc $x^2+y^2 \leq 1$ into two regions with areas $A_1$ and $A_2$. Then,$|A_1-A_2|$ equals

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