Question Number 208652 by efronzo1 last updated on 20/Jun/24
Answered by Berbere last updated on 20/Jun/24
$${a},{b}\:{solution}\:{of}\:−\mathrm{3}{x}^{\mathrm{3}} +\mathrm{2}{x}={c} \\ $$$${S}_{\mathrm{1}} =\int_{\mathrm{0}} ^{{a}} {c}−\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right)=\int_{{a}} ^{{b}} \left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} −{c}\right){dx} \\ $$$$={ca}−{a}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}{a}^{\mathrm{4}} ={b}^{\mathrm{2}} −{a}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}\left({b}^{\mathrm{4}} −{a}^{\mathrm{4}} \right)−{c}\left({b}−{a}\right) \\ $$$${b}\left({b}−\frac{\mathrm{3}}{\mathrm{4}}{b}^{\mathrm{3}} −{c}\right)=\mathrm{0} \\ $$$${c}={b}−\frac{\mathrm{3}}{\mathrm{4}}{b}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{b}−\mathrm{3}{b}^{\mathrm{3}} ={b}−\frac{\mathrm{3}}{\mathrm{4}}{b}^{\mathrm{3}} \\ $$$$\Rightarrow{b}−\frac{\mathrm{9}}{\mathrm{4}}{b}^{\mathrm{3}} =\mathrm{0}\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{9}}\Rightarrow{b}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${c}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{9}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$−\mathrm{3}{x}^{\mathrm{3}} +\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{3}}{x}+\frac{\mathrm{4}}{\mathrm{27}}= \\ $$$$\left({x}−\frac{\mathrm{2}}{\mathrm{3}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{x}−\frac{\mathrm{2}}{\mathrm{9}}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{12}}{\mathrm{9}};\frac{−\frac{\mathrm{2}}{\mathrm{3}}+\sqrt{\frac{\mathrm{12}}{\mathrm{9}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}={a} \\ $$$${a}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}};{b}=\frac{\mathrm{2}}{\mathrm{3}};{c}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$ \\ $$