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Let $(x_1,x_2,...,x_n)$ be a point chosen at random from the n-dimentional region defined by $0<x_1<x_2<...<x_n<1$. 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 $1$.

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  • $\begingroup$ 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 $n$... $\endgroup$ – antkam May 4 at 11:13
  • $\begingroup$ According to the question, it is from $1 to $n$. $\endgroup$ – user 42493 May 4 at 12:17
  • $\begingroup$ 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 $f(1) = 0$... $\endgroup$ – antkam May 4 at 19:59

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