Q:

Estimating $\int_0^1 x^{2n}(1-2x)^{2n}e^xdx$

To find a bound for $\int_0^1 x^{2n}(1-2x)^{2n}e^xdx$ I did the following:
\begin{align}
&x(1-2x)\leq\text{max}[x(1-2x)]\\
&x(1-2x)\leq\  1/8 \:\:\: \text  { the maximum occurs at $x=1/4$ }\\  
&x^{2n}(1-2x)^{2n}\leq  \left( 1/8 \right)^{2n}\\
& x^{2n}(1-2x)^{2n} e^x \leq  \left( 1/8 \right)^{2n} e^x\\
& \int_0^1 x^{2n}(1-2x)^{2n} e^x dx\leq \left( 1/8 \right)^{2n} \int_0^1 e^xdx\\
& \int_0^1 x^{2n}(1-2x)^{2n} e^x dx\leq \left( 1/8 \right)^{2n} (e-1)
\end{align}
I believe something is terrible wrong with that.

A:

When $x\in [0,1]$, $(1-2x)^{2n}\le 1$ and $e^x\le e$, by first mean value theorem for definite integrals,
\begin{align*}
I_n&\le\max_{x\in[0,1]}\{(1-2x)^{2n}e^x\}\int_0^1x^{2n}~\mathrm dx\\
&=\frac{e}{2n+1}\to0.
\end{align*}

