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

$$ffunctioncontnueon[0,1].provethat$$$${lim}_{n\to +\mathrm{\infty }}n{\int }_0^1{t}^{n}f\left(t\right)dt=f\left(1\right).$$

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

$$letput{t}^n=x\iff t=x^{\frac{1}{n}}and{I}_n=n{\int }_0^{1}xf\left(x^{\frac{1}{n}}\right)\frac{1}{n}{x}^{\frac{1}{n}-1}dx$$$$={\int }_0^1{x}^{\frac{1}{n}}f\left(x^{\frac{1}{n}}\right)dx={\int }_0^1{\psi }_n\left(x\right)dxwith$$$${\psi }_n\left(x\right)=x^{\frac{1}{n}}f\left(x^{\frac{1}{n}}\right){\psi }_{nn\to +\mathrm{\infty }}{\to }^{c.s.}f\left(1\right)so$$$${\int }_0^1{\psi }_n\left(x\right){dx}_{n\to +\mathrm{\infty }}\to {\int }_0^{1}f\left(1\right)dx=f\left(1\right).$$

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