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Question Number 129150 by benjo_mathlover last updated on 13/Jan/21

$${\int }_0^{\mathrm{x}}{(\mathrm{x}-\mathrm{u})}^2\mathrm{u}\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}=4\mathrm{x}-6\sin \mathrm{x}+2\mathrm{xcos}\mathrm{x}$$$$\mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by liberty last updated on 13/Jan/21

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\int }_0^{\mathrm{x}}{(\mathrm{x}-\mathrm{u})}^2\mathrm{uf}\left(\mathrm{u}\right)\mathrm{du}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(4\mathrm{x}-6\sin \mathrm{x}+2\mathrm{xcos}\mathrm{x})$$$$\Rightarrow 2{\int }_0^{\mathrm{x}}(\mathrm{x}-\mathrm{u})\mathrm{u}\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}=4-6\cos \mathrm{x}+2\cos \mathrm{x}-2\mathrm{xsin}\mathrm{x}$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\int }_0^{\mathrm{x}}(\mathrm{x}-\mathrm{u})\mathrm{uf}\left(\mathrm{u}\right)\mathrm{du}\right)=6\sin \mathrm{x}-2\sin \mathrm{x}-(2\sin \mathrm{x}+2\mathrm{xcos}\mathrm{x})$$$$\Rightarrow 2{\int }_0^{\mathrm{x}}\mathrm{u}\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}=2\sin \mathrm{x}-2\mathrm{xcos}\mathrm{x}$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\int }_0^{\mathrm{x}}\mathrm{u}\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(2\sin \mathrm{x}-2\mathrm{xcos}\mathrm{x})$$$$\Rightarrow 2\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)=2\cos \mathrm{x}-(2\cos \mathrm{x}-2\mathrm{xsin}\mathrm{x})$$$$\Rightarrow 2\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}\sin \mathrm{x}$$$$\underset{―}{\therefore \mathbf{f}\left(\mathit{x}\right)=\mathbf{sin}\mathit{x}}$$

Commented by benjo_mathlover last updated on 13/Jan/21

$$\mathrm{waw}..\mathrm{superb}...$$

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