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Question Number 204645 by cortano12 last updated on 24/Feb/24

$$\mathrm{Let}f:\left[\overline{1}\mathrm{\infty }\right)\to \mathrm{R}\mathrm{be}\mathrm{a}\mathrm{differentiable}$$$$\mathrm{function}\mathrm{such}\mathrm{that}f\left(1\right)=\frac{1}{3}\mathrm{and}$$$$3\underset{1}{\stackrel{\mathrm{x}}{\int }}f\left(t\right)dt=xf\left(x\right)-\frac{x^3}{3},\mathrm{x}\in [1,\mathrm{\infty })$$$$\mathrm{find}\mathrm{tbe}\mathrm{value}\mathrm{of}f\left(e\right)$$

Commented by universe last updated on 24/Feb/24

$$\frac{4e^2}{3}?$$

Answered by mr W last updated on 24/Feb/24

$$y=f\left(x\right)$$$$3y=y+{xy}^{\prime }-x^2$$$$y^{\prime }-\frac{2}{x}y=x$$$$-2\int \frac{dx}{x}=-2\ln x=\ln \frac{1}{x^2}$$$$\frac{1}{x^2}y=\int \left(x\times \frac{1}{x^2}\right)dx+C=\ln x+C$$$$\Rightarrow y=x^2(\ln x+C)$$$$\frac{1}{3}=1^2(0+C)\Rightarrow C=\frac{1}{3}$$$$\Rightarrow y=f\left(x\right)=x^2(\ln x+\frac{1}{3})$$$$\Rightarrow f\left(e\right)=e^2(\ln e+\frac{1}{3})=\frac{4e^2}{3}$$

Commented by cortano12 last updated on 25/Feb/24

$$\mathrm{yes}$$

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