# Expected value of Rieman sums of a continous function over all possible patitions of $[0,1]$ into n+1 subintervals.

Let $$(x_1,x_2,...,x_n)$$ be a point chosen at random from the n-dimentional region defined by $$0. Let $$f$$ be a continous function on $$[0,1]$$ with $$f(1)=0$$. Set $$x_0=0$$ and $$x_{n+1}=1$$. Sow that the expected value for the Riemann sum $$\sum_{i=1}^{n}(x_{i+1}-x_i)f(x_{i+1})$$ is $$\int_0^1f(t)P(t)dt$$ where $$P$$ is a polynomial of degree $$n$$, independent of $$f$$, with $$0\leq P(t) \leq 1$$ for $$0\leq t\leq1$$.

My thoughts: The Riemann sum is more or less the integral of $$f(x)$$ from $$0$$ to $$1$$.

• are you sure the summation index is $1$ to $n$? it seems like it "should" be $0$ to $n$. this of course looks like an integral but it equals the integral only if the intervals are even and $n \to \infty$. i don't know how to solve this but suspect $P$ will depend on $n$... – antkam May 4 at 11:13
• According to the question, it is from $1 to$n$. – user 42493 May 4 at 12:17 • That is very strange... why even define$x_0=0$if it never appears? Also, for$n=1$the sum has one term:$(x_2 - x_1) f(x_2) = (1- x_1) f(1) = 0$because it is given that$f(1) = 0\$... – antkam May 4 at 19:59