Question Number 219940 by hardmath last updated on 03/May/25
![Let: f : [n−1 , n] → [n , n + 1] be a continuous function Such that: ∫_(n−1) ^( n) (1 + xf^′ (x))dx ≤ nf(n)−(n−1)f(n−1) Then prove: ∫_(n−1) ^( n) (dx/(f(x))) ≤ (2/(n + 1)) , n∈N^∗](https://www.tinkutara.com/question/Q219940.png)
$$\mathrm{Let}: \\ $$$$\mathrm{f}\::\:\left[\mathrm{n}−\mathrm{1}\:,\:\mathrm{n}\right]\:\rightarrow\:\left[\mathrm{n}\:,\:\mathrm{n}\:+\:\mathrm{1}\right] \\ $$$$\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function} \\ $$$$\mathrm{Such}\:\mathrm{that}: \\ $$$$\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}} ^{\:\boldsymbol{\mathrm{n}}} \left(\mathrm{1}\:+\:\mathrm{xf}\:^{'} \left(\mathrm{x}\right)\right)\mathrm{dx}\:\leqslant\:\mathrm{nf}\left(\mathrm{n}\right)−\left(\mathrm{n}−\mathrm{1}\right)\mathrm{f}\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}: \\ $$$$\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}} ^{\:\boldsymbol{\mathrm{n}}} \:\frac{\mathrm{dx}}{\mathrm{f}\left(\mathrm{x}\right)}\:\leqslant\:\frac{\mathrm{2}}{\mathrm{n}\:+\:\mathrm{1}}\:\:\:,\:\:\:\mathrm{n}\in\mathbb{N}^{\ast} \\ $$
Answered by MrGaster last updated on 04/May/25
![∵f:[n−1,n]→[n,n+1]continuously,∴∀x∈[n−1,n]∧f(x)≥n ∴(1/(f(x)))≤(1/n)∧∫_(n−1) ^n (dx/(f(x)))≤∫_(n−1) ^n (1/n)dx=(1/n)∙(n−(n−1))=(1/n) for n∈N^∗ ∧(1/n)≤(2/(n+1))(True ∨ false) ∵(1/n)≤(2/(n+1))⇔n+1≤2n⇔1≤n ∴True So:∫_(n−1) ^( n) (dx/(f(x))) ≤ (2/(n + 1)) [Q.E.D]](https://www.tinkutara.com/question/Q219943.png)
$$\because{f}:\left[{n}−\mathrm{1},{n}\right]\rightarrow\left[{n},{n}+\mathrm{1}\right]\mathrm{continuously},\therefore\forall{x}\in\left[{n}−\mathrm{1},{n}\right]\wedge{f}\left({x}\right)\geq{n} \\ $$$$\therefore\frac{\mathrm{1}}{{f}\left({x}\right)}\leq\frac{\mathrm{1}}{{n}}\wedge\int_{{n}−\mathrm{1}} ^{{n}} \frac{{dx}}{{f}\left({x}\right)}\leq\int_{{n}−\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{n}}{dx}=\frac{\mathrm{1}}{{n}}\centerdot\left({n}−\left({n}−\mathrm{1}\right)\right)=\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{for}\:{n}\in\mathbb{N}^{\ast} \wedge\frac{\mathrm{1}}{{n}}\leq\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\mathrm{True}\:\vee\:\mathrm{false}\right) \\ $$$$\because\frac{\mathrm{1}}{{n}}\leq\frac{\mathrm{2}}{{n}+\mathrm{1}}\Leftrightarrow{n}+\mathrm{1}\leq\mathrm{2}{n}\Leftrightarrow\mathrm{1}\leq{n} \\ $$$$\therefore\mathrm{True} \\ $$$$\mathrm{So}:\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}} ^{\:\boldsymbol{\mathrm{n}}} \:\frac{\mathrm{dx}}{\mathrm{f}\left(\mathrm{x}\right)}\:\leqslant\:\frac{\mathrm{2}}{\mathrm{n}\:+\:\mathrm{1}}\: \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$
Commented by hardmath last updated on 04/May/25
$$ \\ $$My precious magical mathematician, thank you very much, brilliant and perfect solution