Question Number 200748 by Rupesh123 last updated on 22/Nov/23
Answered by MM42 last updated on 22/Nov/23
$$\left.{s}=\mathrm{4}\int_{\mathrm{0}} ^{{c}} \left({c}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){dx}=\mathrm{4}\left({c}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{{c}} \\ $$$$=\mathrm{4}\left({c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}{c}^{\mathrm{3}} \right)=\frac{\mathrm{8}}{\mathrm{3}}{c}^{\mathrm{3}} =\frac{\mathrm{64}}{\mathrm{3}}\Rightarrow{c}=\mathrm{2}\:\checkmark \\ $$
Answered by cortano12 last updated on 23/Nov/23
$$\:\:\Leftrightarrow\:\mathrm{2x}^{\mathrm{2}} −\mathrm{2c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\Leftrightarrow\:\frac{\mathrm{64}}{\mathrm{3}}=\:\frac{\mathrm{16c}^{\mathrm{2}} .\mathrm{4c}}{\mathrm{6}.\mathrm{4}} \\ $$$$\:\Rightarrow\mathrm{64}\:=\mathrm{8c}^{\mathrm{3}} \:;\:\mathrm{c}=\mathrm{2} \\ $$