Show that $H_i=H_{n-i}$ and $\sum H_i=1$ We define $$H_i=\frac{1}{n}\frac{(-1)^{n-1}}{i!(n-1)!}\int_{0}^{n}\prod_{j=0,j\neq i}^{n}(x-j)dx$$
This is called the Newton-Cotes coefficient.
Here is the exercise:
First, convince yourself that $H_i=H_{n-i}$.
Then show that $\sum_{i=0}^{n}H_i=1$
What I did:
I tried proving that $H_i=H_{n-i}$ but no success.
if its true, then 
$$\frac{(-1)^{n-1}}{i!(n-1)!}\int_{0}^{n}\frac{x(x-1)(x-2)...(x-n)}{x-i}dx=\frac{(-1)^{n-1}}{(n-i)!(n-1)!}\int_{0}^{n}\frac{x(x-1)(x-2)...(x-n)}{x-n+i}$$
and it is unclear to me why that is true.
But even if it is, why does that imply that the sum is equal to $1$?
 A: Well, with the commented amendment to the question, here's a little trick for the first part:
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$Best to see graphically, easiest to prove by substitution.
Still working on the second part.
A: Here is a way to show $\sum H_i = 1$.
Let us start by defining $$P(x) = \prod_{j=0}^n (x-j)$$
then
$$H_i = \frac{1}{n}\frac{(-1)^{n-i}}{i!(n-i)!}\int_0^ndx \frac{P(x)}{x-i}$$
so
$$\sum_{i=0}^n H_i = \frac{1}{n\cdot n!}\int_0^n dx P(x) \sum_{i=0}^n \frac{(-1)^{n-i}}{x-i}{n\choose i} $$
To evaluate the last sum lets introduce the generating function
$$g(y) = \sum_{i=0}^n \frac{(-1)^{n-i}}{x-i}{n\choose i}y^{x-i} $$
where $g(1)$ is the value we are after. Taking the derivative we find
$$g'(y) = (-1)^ny^{x-1}\sum_{i=0}^n {n\choose i}\left(-\frac{1}{y}\right)^i = y^{x-1-n}\left(1-y\right)^n$$
so by integrating up we finally get (the integral is the definition of the $\beta$ function)
$$g(1) = \int_0^1 dy y^{x-1-n}\left(1-y\right)^n = \frac{\Gamma(x-n)n!}{\Gamma(x+1)} = \frac{n!}{P(x)}$$
where we have used $\Gamma(x+1) = x\Gamma(x) = x(x-1)\Gamma(x-1) = \ldots = P(x)\Gamma(x-n)$ to arrive at the final expression. Inserting this back into the sum above we arrive at
$$\sum_{i=0}^n H_i = \frac{1}{n\cdot n!}\int_0^n n! dx = 1$$
