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Question Number 104383 by abony1303 last updated on 21/Jul/20

$$\mathrm{When}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{a}\mathrm{differentiable}\mathrm{function}$$$$\mathrm{satisfying}x\cdot f\left(x\right)=x^2+{\int }_0^x(x-t)\cdot f{}^{\prime }\left(t\right)dt$$$$\mathrm{Find}\Rightarrow f\left(1\right)$$

Commented by abony1303 last updated on 21/Jul/20

$$\mathrm{Pls}\mathrm{help}$$

Answered by mathmax by abdo last updated on 21/Jul/20

$$\mathrm{xf}\left(\mathrm{x}\right)={\mathrm{x}}^2+{\int }_0^{\mathrm{x}}(\mathrm{x}-\mathrm{t}){\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{x}=1\Rightarrow \mathrm{f}\left(1\right)=1+{\int }_0^1(1-\mathrm{t}){\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{dt}\mathrm{by}\mathrm{parts}$$$${\int }_0^1(1-\mathrm{t}){\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{dt}={\left[(1-\mathrm{t})\mathrm{f}\left(\mathrm{t}\right)\right]}_0^1-{\int }_0^1(-1)\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$$$$=\mathrm{f}\left(0\right)+{\int }_0^1\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\mathrm{but}\mathrm{f}\left(0\right)=0\Rightarrow {\int }_0^1(1-\mathrm{t}){\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{dt}={\int }_0^1\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\Rightarrow$$$$\mathrm{f}\left(1\right)=1+{\int }_0^1\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$$$$$$

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