If the area of the region (x,y):|x^2-2| y x is A, then 6A+16√2 is equal to __________.

Question

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

Answer: 27

Accepted Answer

${(x,y):|{x}^{2}-2|\le y\le x}$

On solving, $|{x}^{2}-2|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$

$\begin{gathered} A=\int _{1}^{\sqrt{2}}(x-(2-{x}^{2}))dx+\int _{\sqrt{2}}^{2}(x-({x}^{2}-2))dx \ A={[\frac{{x}^{2}}{2}-2x+\frac{{x}^{3}}{3}]}_{\sqrt{2}}^{1}+{[\frac{{x}^{2}}{2}-\frac{{x}^{3}}{3}+2x]}_{2}^{\sqrt{2}} \end{gathered}$

$\begin{gathered} A=(1-2\sqrt{2}+\frac{2\sqrt{2}}{3})-(\frac{1}{2}-2+\frac{1}{3})+ (2-\frac{8}{3}+4) \ -(1-\frac{2\sqrt{2}}{3}+2\sqrt{2}) \end{gathered}$

$\begin{gathered} A=-4\sqrt{2}+\frac{4\sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \ A=\frac{-8\sqrt{2}}{3}+\frac{9}{2}\Rightarrow A=\frac{-16\sqrt{2}+27}{6} \ 6A=-16\sqrt{2}+27 \ 6A+16\sqrt{2}=27 \end{gathered}$

Solutions · 1