The area of the region {(x, y): |x-1| leq y l | vedclass

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(D) The region is bounded by $y = |x-1|$ and $y = \sqrt{5-x^2}$.To find the intersection points,set $|x-1| = \sqrt{5-x^2}$.Squaring both sides: $(x-1)

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$\frac{5 \pi}{4}-\frac{1}{2}$

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(D) The region is bounded by $y = |x-1|$ and $y = \sqrt{5-x^2}$.To find the intersection points,set $|x-1| = \sqrt{5-x^2}$.Squaring both sides: $(x-1)^2 = 5-x^2 \implies x^2 - 2x + 1 = 5 - x^2 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0$.Factoring gives $(x-2)(x+1) = 0$,so $x = 2$ and $x = -1$.At $x = 2, y = |2-1| = 1$. At $x = -1, y = |-1-1| = 2$.The area is the integral $\int_{-1}^{2} (\sqrt{5-x^2} - |x-1|) dx$.This can be split into the area under the circle and the area of the triangle formed by the absolute value function.Area $= \int_{-1}^{2} \sqrt{5-x^2} dx - \int_{-1}^{2} |x-1| dx$.For $\int_{-1}^{2} \sqrt{5-x^2} dx$,let $x = \sqrt{5} \sin 	heta$,$dx = \sqrt{5} \cos 	heta d	heta$.Limits: $x=-1 \implies \sin 	heta = -1/\sqrt{5}$,$x=2 \implies \sin 	heta = 2/\sqrt{5}$.Area of triangle $\int_{-1}^{2} |x-1| dx = \int_{-1}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx = [x - x^2/2]_{-1}^{1} + [x^2/2 - x]_{1}^{2} = (1/2 - (-3/2)) + (0 - (-1/2)) = 2 + 0.5 = 2.5$.Using the geometric approach from the figure,the area is the sector of the circle minus the triangle area.The final calculated area is $\frac{5}{2} \sin^{-1}(\frac{2}{\sqrt{5}}) + \frac{5}{2} \sin^{-1}(\frac{1}{\sqrt{5}}) - \frac{5}{2} = \frac{5 \pi}{4} - \frac{1}{2}$.

The area of the region $\{(x, y): |x-1| \leq y \leq \sqrt{5-x^{2}}\}$ is equal to:

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