Question Number 214925 by efronzo1 last updated on 24/Dec/24
Commented by GDVilla last updated on 24/Dec/24
$$\mathrm{Why}\:\mathrm{this}\:\mathrm{question}\:\mathrm{in}\:\mathrm{arabic}\:\mathrm{translate}\:\mathrm{pls} \\ $$
Answered by TonyCWX08 last updated on 24/Dec/24
$${Q}\mathrm{1}. \\ $$$${x}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{{x}} \\ $$$${x}^{\mathrm{6}} ={x} \\ $$$${x}^{\mathrm{6}} −{x}=\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{5}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:{or}\:{x}=\mathrm{1} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\sqrt[{\mathrm{3}}]{{x}}−{x}^{\mathrm{2}} \right){dx} \\ $$$$=\left(\frac{\mathrm{3}{x}\sqrt[{\mathrm{3}}]{{x}}}{\mathrm{4}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\right)}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$ \\ $$$${Q}\mathrm{2}. \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{4}}=\mathrm{2} \\ $$$$\mathrm{2}{x}−\mathrm{4}=\mathrm{4} \\ $$$$\mathrm{2}{x}=\mathrm{8} \\ $$$${x}=\mathrm{4} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\left(\mathrm{2}−\sqrt{\mathrm{2}{x}−\mathrm{4}}\right){dx} \\ $$$$=\left(\mathrm{2}{x}−\frac{\left(\mathrm{2}{x}−\mathrm{4}\right)\sqrt{\mathrm{2}{x}−\mathrm{4}}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\mathrm{4}} {\right)}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 24/Dec/24
$$\left(\mathrm{1}\right) \\ $$$${A}_{{green}} =\mathrm{1}×\mathrm{1}−\frac{\mathrm{1}×\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}×\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\left(\mathrm{2}\right) \\ $$$${A}_{{green}} =\mathrm{2}×\mathrm{4}−\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$