If the area of the region {(x, y): |x^2-2| le | vedclass

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(C) The region is defined by $|x^2-2| \leq y \leq x$. First,find the intersection points of $y = x^2-2$ and $y = x$: $x^2-x-2 = 0 \implies (x-2)(x+1)

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) The region is defined by $|x^2-2| \leq y \leq x$. First,find the intersection points of $y = x^2-2$ and $y = x$: $x^2-x-2 = 0 \implies (x-2)(x+1) = 0$,so $x=2$ or $x=-1$. Also,$y = |x^2-2|$ intersects $y=x$ when $x^2-2 = x$ (for $x^2 \geq 2$,i.e.,$x \geq \sqrt{2}$) or $2-x^2 = x$ (for $x^2 For $x^2 0$). For $x^2 \geq 2$,$x^2-x-2=0 \implies x=2$. The area $A$ is given by: $A = \int_{1}^{\sqrt{2}} (x - (2-x^2)) dx + \int_{\sqrt{2}}^{2} (x - (x^2-2)) dx$ $A = \int_{1}^{\sqrt{2}} (x^2+x-2) dx + \int_{\sqrt{2}}^{2} (-x^2+x+2) dx$ $A = [\frac{x^3}{3} + \frac{x^2}{2} - 2x]_{1}^{\sqrt{2}} + [-\frac{x^3}{3} + \frac{x^2}{2} + 2x]_{\sqrt{2}}^{2}$ $A = ((\frac{2\sqrt{2}}{3} + 1 - 2\sqrt{2}) - (\frac{1}{3} + \frac{1}{2} - 2)) + ((-\frac{8}{3} + 2 + 4) - (-\frac{2\sqrt{2}}{3} + 1 + 2\sqrt{2}))$ $A = (\frac{2\sqrt{2}}{3} - 2\sqrt{2} - 1 + \frac{5}{6}) + (\frac{10}{3} + \frac{2\sqrt{2}}{3} - 2\sqrt{2} - 1)$ $A = (-\frac{4\sqrt{2}}{3} - \frac{1}{6}) + (\frac{7}{3} - \frac{4\sqrt{2}}{3}) = \frac{13}{6} - \frac{8\sqrt{2}}{3}$ Then $6A = 13 - 16\sqrt{2}$. Therefore,$6A + 16\sqrt{2} = 13 + 14 = 27$.

If the area of the region $\{(x, y): |x^2-2| \leq y \leq x\}$ is $A$,then $6A + 16\sqrt{2}$ is equal to $...........$.

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