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Question Number 191303 by Mingma last updated on 22/Apr/23

Answered by mahdipoor last updated on 22/Apr/23

$$\int f\left(x\right)=\int {xf}^{{}^{\prime }}\left(x\right)-\sqrt{2x-x^2}=[xf\left(x\right)-\int f\left(x\right)]-\int \sqrt{2x-x^2}$$$$\Rightarrow {\int }_0^{1}f\left(x\right)=\frac{1}{2}{\left[xf\left(x\right)\right]}_0^1-{\int }_0^1\sqrt{2x-x^2}=$$$$0-{\int }_0^1\sqrt{2x-x^2}$$$$$$

Answered by cortano12 last updated on 22/Apr/23

$$\Rightarrow \mathrm{x}\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x}\right)+\sqrt{2\mathrm{x}-{\mathrm{x}}^2}$$$$\Rightarrow \mathrm{x}\mathrm{df}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\sqrt{-({\mathrm{x}}^2-2\mathrm{x}+1)+1}\mathrm{dx}$$$$\Rightarrow \underset{0}{\stackrel{1}{\int }}\mathrm{x}\mathrm{df}\left(\mathrm{x}\right)=\underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{0}{\stackrel{1}{\int }}\sqrt{1-{(\mathrm{x}-1)}^2}\mathrm{dx}$$$$\Rightarrow {\left[\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\right]}_0^1-\underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{0}{\stackrel{1}{\int }}\sqrt{1-{(\mathrm{x}-1)}^2}\mathrm{dx}$$$$\Rightarrow -2\underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\pi }{4}$$$$\Rightarrow \therefore \underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{\pi }{8}$$

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