Question Number 204645 by cortano12 last updated on 24/Feb/24
$$\:\:\mathrm{Let}\:{f}\::\:\left[\:\bar {\mathrm{1}}\infty\right)\:\rightarrow\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{differentiable}\: \\ $$$$\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\: \\ $$$$\:\mathrm{3}\underset{\mathrm{1}} {\overset{\mathrm{x}} {\int}}\:{f}\left({t}\right)\:{dt}\:=\:{x}\:{f}\left({x}\right)−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:,\mathrm{x}\in\left[\mathrm{1},\infty\right)\: \\ $$$$\:\mathrm{find}\:\mathrm{tbe}\:\mathrm{value}\:\mathrm{of}\:{f}\left({e}\right)\: \\ $$
Commented by universe last updated on 24/Feb/24
$$\frac{\mathrm{4}{e}^{\mathrm{2}} }{\mathrm{3}}? \\ $$
Answered by mr W last updated on 24/Feb/24
$${y}={f}\left({x}\right) \\ $$$$\mathrm{3}{y}={y}+{xy}'−{x}^{\mathrm{2}} \\ $$$${y}'−\frac{\mathrm{2}}{{x}}{y}={x} \\ $$$$−\mathrm{2}\int\frac{{dx}}{{x}}=−\mathrm{2ln}\:{x}=\mathrm{ln}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{y}=\int\left({x}×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}+{C}=\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow{y}={x}^{\mathrm{2}} \left(\mathrm{ln}\:{x}+{C}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1}^{\mathrm{2}} \left(\mathrm{0}+{C}\right)\:\Rightarrow{C}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{y}={f}\left({x}\right)={x}^{\mathrm{2}} \left(\mathrm{ln}\:{x}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow{f}\left({e}\right)={e}^{\mathrm{2}} \left(\mathrm{ln}\:{e}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{4}{e}^{\mathrm{2}} }{\mathrm{3}} \\ $$
Commented by cortano12 last updated on 25/Feb/24
$$\mathrm{yes} \\ $$