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Solution

$y = ||x - 1| - 2|$ Critical points where inner terms change sign: $x = -1,, 1,, 3.$ Compute areas of each segment geometrically (each forms triangles): Areas = $1 + 1 + 2 = 4.$

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Solution

Area $= \int_0^{2\pi} |\sin x| \, dx = 2 \int_0^{\pi} \sin x \, dx = 2[ -\cos x ]_0^{\pi} = 2(2) = 4.$

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Solution

For intersection: $y^2 = 4x$ and $y = x \Rightarrow x^2 = 4x \Rightarrow x = 0, 4$ Area $= \int_{0}^{4} (\sqrt{4x} - x),dx = \int_{0}^{4} (2\sqrt{x} - x),dx = \left[\dfrac{4}{3}x^{3/2} - \dfrac{x^2}{2}\right]_{0}^{4} = \dfrac{32}{3} - 8 = \dfrac{8}{3}$

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Solution

The region lies between the circle $x^2 + y^2 = 1$ and the parabola $y^2 = 1 - x$. Converting parabola into standard form: $x = 1 - y^2$. To find points of intersection, substitute $x = 1 - y^2$ in $x^2 + y^2 = 1$: $(1 - y^2)^2 + y^2 = 1 \Rightarrow 1 - 2y^2 + y^4 + y^2 = 1 \Rightarrow y^2(y^2 - 1) = 0.$ $\Rightarrow y = 0, \pm 1.$ Required area (using symmetry about x-axis): $A = 2 \int_{0}^{1} [\sqrt{1 - y^2} - (1 - y^2)] dy.$ Compute separately: $\int_0^1 \sqrt{1 - y^2}dy = \dfrac{\pi}{4}$, $\int_0^1 (1 - y^2)dy = \dfrac{2}{3}.$ Hence $A = 2\left(\dfrac{\pi}{4} - \dfrac{2}{3}\right) = \dfrac{\pi}{2} - \dfrac{4}{3}.$

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Solution

Curves: $y^2 = x$ (right-opening parabola) and $x^2 = y$ (upward parabola). Points of intersection: substitute $y = x^2$ into $y^2 = x$: $(x^2)^2 = x \Rightarrow x^4 - x = 0 \Rightarrow x(x^3 - 1) = 0.$ $\Rightarrow x = 0, 1.$ Between $x = 0$ and $1$: upper curve is $y = \sqrt{x}$, lower is $y = x^2$. Area $A = \int_0^1 (\sqrt{x} - x^2)\,dx = \left[\dfrac{2}{3}x^{3/2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{2}{3} - \dfrac{1}{3} = \dfrac{1}{3}.$