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

Question

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.

Answer 1

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*}