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Question Number 96837 by mathmax by abdo last updated on 05/Jun/20

$$\mathrm{determine}\mathrm{f}\mathrm{continue}\mathrm{on}[\mathrm{a},\mathrm{b}]\mathrm{wich}\mathrm{verify}{\left({\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\right)}^2={\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{f}}^2\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{let}\mathrm{b}=\mathrm{t}\left(\mathrm{afixed}\right)\Rightarrow \frac{\mathrm{d}}{\mathrm{dt}}{\left({\int }_{\mathrm{a}}^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\right)}^2=\frac{\mathrm{d}}{\mathrm{dx}}{\int }_{\mathrm{a}}^{\mathrm{t}}{\mathrm{f}}^2\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow$$$$2{\int }_{\mathrm{a}}^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{f}}^2\left(\mathrm{t}\right)\Rightarrow 2{\int }_{\mathrm{a}}^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}\left(\mathrm{t}\right)\Rightarrow$$$${2\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)={\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)=0\Rightarrow \mathrm{f}=\mathrm{c}$$$${\left({\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{cdx}\right)}^2={\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{c}}^2\mathrm{dx}\Rightarrow (\mathrm{c}{(\mathrm{b}-\mathrm{a})}^2={\mathrm{c}}^2(\mathrm{b}-\mathrm{a})\Rightarrow {\mathrm{c}}^2{(\mathrm{b}-\mathrm{a})}^2={\mathrm{c}}^2(\mathrm{b}-\mathrm{a})$$$$\mathrm{so}\mathrm{if}\mathrm{a}\ne \mathrm{b}\Rightarrow \mathrm{c}=0\mathrm{so}\mathrm{f}=0$$

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