Evaluate $\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$, for integrable $f$ Let $f : [0, 1] \to \mathbb{R}$ be Riemann integrable. Evaluate
$$\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$$
I have tried applying by parts by assuming
$$\\ \int_{0}^1 f(x) \ dx = F(1)-F(0)$$
but it doesn't seem to get me anywhere. Do I have to use partitions and if so how do I go about it?
 A: Since $f$ is integrable, say $\int_0^1 f(x)\,dx = a$, it is bounded, say $m\le f(x)\le M$ for $x\in [0,1]$. In particular, $g(x) = f(x)-m \ge 0$. Then we have
$$
\int _0^1 x^{n+1} f(x)\,dx = \int _0^1 m x^{n+1}\,dx + \int_0^1 (f(x)-m)x^{n+1}\,dx
$$The first integral clearly vanishes in the limit. For the second integral, we can use the First MVT for integrals to deduce
$$
\int _0^1 (f(x)-m)x^{n+1}\,dx = c^{n+1} \int _0^1 f(x)-m\,dx = (a-m)c^{n+1},
$$for $c\in(0,1)$; this also vanishes in the limit.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x}:\
{\Large ?}}$.

As $\ds{n \to \infty}$, the main contribution to the integral comes from $\ds{x \lesssim 1}$. So, I'll make the change $\ds{x \mapsto 1 - x}$ such that the main contribution comes from $\ds{x \gtrsim 0}$. Namely, 
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{1}\pars{1 - x}^{n + 1}
\,\on{f}\pars{1 - x}\,\dd x
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{1}\expo{\pars{n + 1}\ln\pars{1 - x}}
\,\,\,\on{f}\pars{1 - x}\,\dd x
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{\infty}\expo{-\pars{n + 1}x}
\,\,\,\on{f}\pars{1 - 0}\,\dd x
\end{align}
where I used the
Laplace's Method.
\begin{align}
&\mbox{Then,}
\\ &\
\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x} =
\lim_{n \to \infty}\,\,{\on{f}\pars{1} \over n + 1} =
\bbx{\large 0} \\ &
\end{align}
