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Question Number 191199 by 073 last updated on 20/Apr/23

Commented by 073 last updated on 20/Apr/23

$$\mathrm{sheed}\mathrm{area}=??$$

Answered by cortano12 last updated on 20/Apr/23

$$\left(1\right){\mathrm{x}}^2=\frac{1}{\mathrm{x}}\Rightarrow \mathrm{x}=1$$$$\left(2\right)\mathrm{Area}=\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^2\mathrm{dx}+\underset{1}{\stackrel{2}{\int }}\frac{1}{\mathrm{x}}\mathrm{dx}+\underset{1/2}{\stackrel{1}{\int }}(2-\frac{1}{\mathrm{x}})\mathrm{dx}+\underset{1}{\stackrel{\sqrt{2}}{\int }}(2-{\mathrm{x}}^2)\mathrm{dx}$$$$\mathrm{Area}=\frac{1}{3}+\ln 2+(2.1-0)-(2.\frac{1}{2}+\ln 2)+2(\sqrt{2}-1)-\frac{1}{3}(2\sqrt{2}-1)$$$$\mathrm{Area}=\frac{1}{3}+2-1+2\sqrt{2}-2-\frac{2\sqrt{2}}{3}+\frac{1}{3}$$$$\mathrm{Area}=\frac{4\sqrt{2}-1}{3}$$

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