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Question-169242




Question Number 169242 by Skabetix last updated on 26/Apr/22
Answered by Skabetix last updated on 26/Apr/22
please help
pleasehelp
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
Iffiscontinousin[0,1],Msuchthatf(x)∣⩽M.So,01xnf(x)dx∣⩽01xnf(x)dxM01xndx=Mn+1Ifwetakethelimitwehavelimn+01xnf(x)dx=0
Commented by Skabetix last updated on 26/Apr/22
thank you sir
thankyousir

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