The area of the region (x,y):x^2+4x+2 y |x+2| is equal to

Question

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

Accepted Answer

${x}^{2}+4x+2\le y\le |x+2|$

The area bounded between

$y={x}^{2}+4x+2=(x+2{)}^{2}-2$

and $y=|x+2|$ is same as area bounded between $y={x}^{2}-2$ and $y=|x|$

For P.O.I $|x{|}^{2}-2=|x|$

$\begin{gathered} \Rightarrow |x|=2\Rightarrow x=\pm 2 \ 	herefore =-{\int }_{2}^{-2}({x}^{2}-2)dx+{\int }_{2}^{-2}|x|dx \ =-2{\int }_{2}^{0}({x}^{2}-2)dx+2{\int }_{2}^{0}x\cdot dx \ =-2{[\frac{{x}^{3}}{3}-2x]}_{2}^{0}+2{[\frac{{x}^{2}}{2}]}_{2}^{0} \ =--2[\frac{8}{3}-4]+2[\frac{4}{2}] \ =-2	imes (\frac{-4}{3})+4 \ =\frac{20}{3} \end{gathered}$

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