The area of the region {(x, y): x^2+4x+2 leq | vedclass

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Question

(C) The given region is defined by $x^2+4x+2 \leq y \leq |x+2|$.Let $u = x+2$,then $x = u-2$. The region becomes $(u-2)^2 + 4(u-2) + 2 \leq y \leq |u|

Vedclass · Accepted Answer

$20/3$

Vedclass · Answer

(C) The given region is defined by $x^2+4x+2 \leq y \leq |x+2|$.Let $u = x+2$,then $x = u-2$. The region becomes $(u-2)^2 + 4(u-2) + 2 \leq y \leq |u|$,which simplifies to $u^2-4u+4+4u-8+2 \leq y \leq |u|$,or $u^2-2 \leq y \leq |u|$.The points of intersection are $u^2-2 = |u|$.For $u \geq 0$,$u^2-u-2 = 0 \Rightarrow (u-2)(u+1) = 0 \Rightarrow u = 2$.For $u Thus,the intersection points are $u = \pm 2$.The required area is $\int_{-2}^{2} (|u| - (u^2-2)) \, du$.Since the integrand is an even function,the area is $2 \int_{0}^{2} (u - u^2 + 2) \, du$.$= 2 \left[ \frac{u^2}{2} - \frac{u^3}{3} + 2u \right]_{0}^{2}$.$= 2 \left( \frac{4}{2} - \frac{8}{3} + 4 \right) = 2 \left( 2 - \frac{8}{3} + 4 \right) = 2 \left( 6 - \frac{8}{3} \right) = 2 \left( \frac{18-8}{3} \right) = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$.

The area of the region $\{(x, y): x^2+4x+2 \leq y \leq |x+2|\}$ is equal to

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