f-function-contnue-on-0-1-prove-that-lim-n-n-0-1-t-n-f-t-dt-f-1-

Question Number 28679 by abdo imad last updated on 28/Jan/18

f f u n c t i o n c o n t n u e o n [0, 1] . p r o v e t h a t

{l i m}_{n \to + \infty} n \int_{0}^{1} t^{n} f (t) d t = f (1) .

Commented by abdo imad last updated on 29/Jan/18

l e t p u t t^{n} = x \Leftrightarrow t = x^{\frac{1}{n}} a n d I_{n} = n \int_{0}^{1} x f (x^{\frac{1}{n}}) \frac{1}{n} x^{\frac{1}{n} - 1} d x

= \int_{0}^{1} x^{\frac{1}{n}} f (x^{\frac{1}{n}}) d x = \int_{0}^{1} ψ_{n} (x) d x w i t h

ψ_{n} (x) = x^{\frac{1}{n}} f (x^{\frac{1}{n}}) ψ_{n n \to + \infty} \to^{c . s .} f (1) s o

\int_{0}^{1} ψ_{n} (x) {d x}_{n \to + \infty} \to \int_{0}^{1} f (1) d x = f (1) .