Question Number 169242 by Skabetix last updated on 26/Apr/22
Answered by Skabetix last updated on 26/Apr/22
$${please}\:{help} \\ $$
Answered by kowalsky78 last updated on 26/Apr/22
![If f is continous in [0,1], ∃M∈ℜ such that ∣f(x)∣≤M. So, ∣∫_0 ^1 x^n f(x)dx∣≤∫_0 ^1 x^n ∣f(x)∣dx≤ M∫_0 ^1 x^n dx=(M/(n+1)) If we take the limit we have lim_(n→+∞) ∫_0 ^1 x^n f(x)dx=0](https://www.tinkutara.com/question/Q169246.png)
$${If}\:{f}\:{is}\:{continous}\:{in}\:\left[\mathrm{0},\mathrm{1}\right],\:\exists{M}\in\Re\:{such}\:{that}\:\mid{f}\left({x}\right)\mid\leqslant{M}.\:{So}, \\ $$$$\mid\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {f}\left({x}\right){dx}\mid\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \mid{f}\left({x}\right)\mid{dx}\leqslant\:{M}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx}=\frac{{M}}{{n}+\mathrm{1}} \\ $$$${If}\:{we}\:{take}\:{the}\:{limit}\:{we}\:{have} \\ $$$${li}\underset{{n}\rightarrow+\infty} {{m}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {f}\left({x}\right){dx}=\mathrm{0} \\ $$
Commented by Skabetix last updated on 26/Apr/22
$${thank}\:{you}\:{sir} \\ $$