If-f-a-b-R-0-lt-a-b-f-continuous-Then-prove-that-b-4047-a-4047-4047-a-b-f-2-x-2024-dx-1-1012-a-2024-b-2024-f-x-dx-

Question Number 219846 by hardmath last updated on 02/May/25

$$\mathrm{If}\:\:\:\mathrm{f}:\left[\mathrm{a},\mathrm{b}\right]\rightarrow\mathbb{R} \\ $$$$\:\:\:\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b} \\ $$$$\mathrm{f}\:-\:\mathrm{continuous} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{b}^{\mathrm{4047}} \:−\:\mathrm{a}^{\mathrm{4047}} }{\mathrm{4047}}\:+\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\:^{\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{2024}} \right)\:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{1012}}\:\int_{\boldsymbol{\mathrm{a}}^{\mathrm{2024}} } ^{\:\boldsymbol{\mathrm{b}}^{\mathrm{2024}} } \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$

Answered by MrGaster last updated on 03/May/25

$$\int_{{a}^{\mathrm{2024}} } ^{{b}^{\mathrm{2024}} } {f}\left({y}\right){dy}=\int_{{a}} ^{{b}} {f}\left({x}^{\mathrm{2024}} \right)\centerdot\mathrm{2024}{x}^{\mathrm{2023}} {dx} \\ $$$$\left(\int_{{a}} ^{{b}} {f}\left({x}^{\mathrm{2024}} \right)\centerdot\mathrm{2024}{x}^{\mathrm{2023}} {dx}\right)^{\mathrm{2}} \leq\left(\int_{{a}} ^{{b}} {f}^{\mathrm{2}} \left({x}^{\mathrm{2024}} \right){dx}\right)\left(\int_{{a}} ^{{b}} \left(\mathrm{2024}{x}^{\mathrm{2023}} \right)^{\mathrm{2}} {dx}\right) \\ $$$$\int_{{a}} ^{{b}} \left(\mathrm{2024}{x}^{\mathrm{2023}} \right){dx}=\mathrm{2024}^{\mathrm{2}} \centerdot\frac{{b}^{\mathrm{4047}} −{a}^{\mathrm{4047}} }{\mathrm{4047}} \\ $$$$\left(\int_{{a}^{\mathrm{2024}} } ^{{b}^{\mathrm{2024}} } {f}\left({y}\right){dy}\right)\leq\left(\int_{{a}} ^{{b}} {f}^{\mathrm{2}} \left({x}^{\mathrm{2024}} \right){dx}\right)\centerdot\mathrm{2024}^{\mathrm{2}} \centerdot\frac{{b}^{\mathrm{4047}} −{a}^{\mathrm{4047}} }{\mathrm{4047}} \\ $$$$\int_{{a}} ^{{b}} {f}^{\mathrm{2}} \left({x}^{\mathrm{2024}} \right)\geq\frac{\left(\int_{{a}^{\mathrm{2024}} } ^{{b}^{\mathrm{2024}} } {f}\left({y}\right){dy}\right)^{\mathrm{2}} }{\mathrm{2024}^{\mathrm{2}} \centerdot\frac{{b}^{\mathrm{4047}} −{a}^{\mathrm{4947}} }{\mathrm{4047}}} \\ $$$$\frac{{b}^{\mathrm{4047}} −{a}^{\mathrm{4047}} }{\mathrm{4047}}+\frac{\left(\int_{{a}^{\mathrm{2024}} } ^{{b}^{\mathrm{2024}} } {f}\left({y}\right){dy}\right)^{\mathrm{2}} }{\mathrm{2024}^{\mathrm{2}} \centerdot\frac{{b}^{\mathrm{4047}} −{a}^{\mathrm{4047}} }{\mathrm{4047}}}\geq\frac{\mathrm{2}}{\mathrm{2024}}\int_{{a}^{\mathrm{2024}} } ^{{b}^{\mathrm{2024}} } {f}\left({y}\right){dy} \\ $$$$\because\frac{\mathrm{2}}{\mathrm{2024}}=\frac{\mathrm{1}}{\mathrm{1012}} \\ $$$$\frac{\mathrm{b}^{\mathrm{4047}} \:−\:\mathrm{a}^{\mathrm{4047}} }{\mathrm{4047}}\:+\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\:^{\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{2024}} \right)\:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{1012}}\:\int_{\boldsymbol{\mathrm{a}}^{\mathrm{2024}} } ^{\:\boldsymbol{\mathrm{b}}^{\mathrm{2024}} } \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$

Commented by hardmath last updated on 03/May/25

$$ \\ $$Perfect solutions as always, thank you very much, my dear professor

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