Asymptotic behaviour of \int_{0}^{1} f(x)x^{n}dx\ \text{as}\ n \rightarrow \infty

obrozenecy6

obrozenecy6

Answered question

2022-01-13

Asymptotic behaviour of 01f(x)xndx as n

Answer & Explanation

Shannon Hodgkinson

Shannon Hodgkinson

Beginner2022-01-14Added 34 answers

Since f is continuous on a closed bounded interval, it has a maximum absolute value M0 on that interval, so you have
|01f(x)xndx|01|f(x)xn|dxM01xndx
=Mn+1.
Medicim6

Medicim6

Beginner2022-01-15Added 33 answers

We know that f(x)max f and so 01f(x)xndx(max f)01xndx=(max f)1n+1.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

To spice it up a bit: We can use Lebesgues

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