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Question Number 131517 by benjo_mathlover last updated on 05/Feb/21

$$\mathrm{Given}{\int }_0^3\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^3(2\mathrm{x}-1)\mathrm{dx}+{\int }_0^3\left({\int }_0^3\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\right)\mathrm{dx}$$$$\mathrm{find}\underset{-1}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.$$

Answered by EDWIN88 last updated on 05/Feb/21

$$\mathrm{let}\underset{0}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\ell \Rightarrow \ell ={[{\mathrm{x}}^2-\mathrm{x}]}_0^3+{\int }_0^3\ell \mathrm{dx}$$$$\ell =6+3\ell ;\ell =-3\mathrm{or}\underset{0}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=-3$$$$\mathrm{then}-3=\underset{0}{\stackrel{3}{\int }}(2\mathrm{x}-1)\mathrm{dx}+\underset{0}{\stackrel{3}{\int }}-3\mathrm{dx}$$$$\Rightarrow -3=\underset{0}{\stackrel{3}{\int }}(2\mathrm{x}-4)\mathrm{dx}\Rightarrow \underset{0}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{3}{\int }}(2\mathrm{x}-4)\mathrm{dx}$$$$\mathrm{by}\mathrm{theorem}\underset{\mathrm{a}}{\stackrel{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{a}}{\stackrel{\mathrm{b}}{\int }}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{dx}$$$$\mathrm{we}\mathrm{have}\underset{0}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{3}{\int }}\mathrm{f}(3-\mathrm{x})\mathrm{dx};\mathrm{so}$$$$\mathrm{f}(3-\mathrm{x})=2\mathrm{x}-4\mathrm{or}\mathrm{f}\left(\mathrm{x}\right)=2(3-\mathrm{x})-4=2-2\mathrm{x}$$$$\mathrm{Now}\underset{-1}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{-1}{\stackrel{1}{\int }}(2-2\mathrm{x})\mathrm{dx}={[2\mathrm{x}-{\mathrm{x}}^2]}_{-1}^1$$$$=\left(1\right)-(-3)=4.$$$$\mathrm{check}\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=2-2\mathrm{x}\Rightarrow \underset{0}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{3}{\int }}(2-2\mathrm{x})\mathrm{dx}$$$$={[2\mathrm{x}-{\mathrm{x}}^2]}_0^3=6-9=-3\left(\mathrm{true}\right)$$

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