Finding a better upper bound for an integral of a product of $n$ terms So I'm trying to find and upper bound for the integral
$$
\int\limits_{a}^b \! (x-x_1)^2 \cdots (x-x_n)^2\, \mathrm{d}x,
$$
where $x_i \in [a,b], \enspace \forall i=1,\dots ,n.$ I've tried evaluating a single facotr $(x-x_i)^2$ but all I've gotten was the obvious one:
$$
(x-x_i)^2 \leq (b-a)^2, \enspace \forall i=1,\dots,n,
$$
which as a result gives me:
$$
\int\limits_{a}^b \! (x-x_1)^2 \cdots (x-x_n)^2\, \mathrm{d}x \leq (b-a)^{2n+1}.
$$
That bound is not that good for what I need. Any suggestions on how to make it better? 
 A: From Hölder:
$$
I=\int_{a}^{b}(x-x_1)^2\ldots(x-x_n)^2 dx \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^4 \ldots (x-x_n)^4dx \right)^{1/2} \leq \\
\leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^8 dx \right)^{1/4}\left(\int_{a}^{b}(x-x_3)^{8}\ldots(x-x_n)^8 dx\right)^{1/4} \leq \ldots \\
\leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^{8} \right)^{1/4} \left(\int_{a}^{b}(x-x_3)^{16}dx \right)^{1/8}\ldots \left(\int_{a}^{b} (x-x_n)^{2^{n+1}} dx\right)^{1/2^n} = \\
=\prod_{i=1}^{n}\left(\int_{a}^{b}(x-x_i)^{2^{i+1}}dx \right)^{2^{-i}}
$$
We have to evaluate $\int_{a}^{b}(x-x_i)^{2^{i+1}}dx$. Making the substitution $x-x_i = y$, we have:
$$
\left(\int_{a-x_i}^{b-x_i}y^{2^{i+1}}dy \right)^{1/2^{i}}=\left(\frac{(b-x_i)^{2^{i+1}+1}-(a-x_i)^{2^{i+1}+1}}{2^{i+1}+1} \right)^{1/2^{i}} 
$$
Then, if $0\leq a \leq b$, 
$$
I\leq \left(\frac{(b-x_1)^{5}-(a-x_1)^{5}}{5} \right)^{1/2} \ldots \left(\frac{(b-x_n)^{2^{n+1}+1}-(a-x_n)^{2^{n+1}+1}}{2^{n+1}+1} \right)^{1/2^n} \leq \\
\leq \left(\frac{b^5 - a^5}{5} \right)^{1/2}\ldots \left(\frac{b^{2^{n+1}+1}-a^{2^{n+1}+1}}{2^{n+1}+1} \right)^{1/2^{n}}
$$
Let's check if this is actually decent. Assume $[a,b]=[0,1]$ and take $x_1 = 1/3$ and $x_2 = 2/3$. The integral is:
$$
I=\int_{0}^{1}\left(x-\frac{1}{3}\right)^2 \left(x-\frac{2}{3}\right)^2 dx = \frac{7}{810} \approx 0.008642
$$
The bound you got was $1$ ($(b-a)^{2n+1}=1^3 = 1$). Our gives:
$$
I\leq \left(\frac{1}{5} \right)^{1/2}\left(\frac{1}{9} \right)^{1/4} \approx 0.26
$$
So, not that great, and maybe it got a little bit too complicated, but I think this is fun! Maybe this sparks some other ideas.
