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Question Number 154359 by ZiYangLee last updated on 17/Sep/21

$$\mathrm{If}f\left(0\right)=1,f\left(2\right)=3,f{}^{\prime }\left(2\right)=5,$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\int }_0^{1}x{f}^{″}\left(2x\right)dx.$$

Answered by aleks041103 last updated on 17/Sep/21

$$\int {xf}^{″}\left(2x\right)dx=\int xd\left(\frac{1}{2}{f}^{\prime }\left(2x\right)\right)=$$$$=\frac{1}{2}{xf}^{\prime }\left(2x\right)-\frac{1}{2}\int f^{\prime }\left(2x\right)dx=$$$$=\frac{1}{2}{xf}^{\prime }\left(2x\right)-\frac{1}{4}f\left(2x\right)+C$$$$\Rightarrow {\int }_0^1{xf}^{″}\left(2x\right)dx=(\frac{1}{2}{xf}^{\prime }\left(2x\right)-\frac{1}{4}f\left(2x\right)){\mid }_0^1=$$$$=\frac{1}{2}{f}^{\prime }\left(2\right)-\frac{1}{4}(f\left(2\right)-f\left(0\right))=$$$$=\frac{5}{2}-\frac{2}{4}=2$$

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