Prove for every Riemann mid sum of $f(x)=x$ of $[0,b]$ is the integral $\int_{0}^{b}f(x)dx$?

Question

I found that Riemann mid sum for arbitrary partition is $S(P, f)=\dfrac{1}{2}\sum_{i=1}^n(x_i^2-x_{i-1}^2)=\dfrac{b^2}{2}$, and now I have to show that Riemann sum approaches this value, i.e. chosen points and partition doesn't affect as long norm $|P|$ is small enough.

Answer 1

Well, IF you found that for every partition, the corresponding riemann sum equals $\frac {b^2}2$, then for any $\delta>0$ we already have that

$$ |P|<\delta \implies 0=|S(P,f)-I|<\varepsilon $$

Since the consequent is always true.

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Also, just a remark: If it wasn't the case that for every riemann sum, the 'limit' (i.e, your $I$ here), the integral $\int_0^b f$ would not exist, as, per definition, the symbol $\int_0^b f $ is defined to be $I$ when this is equal for each and every Riemann sum (else, we say $f$ is not integrable over $[0,b]$, and it would make no sense to talk about $f$'s integral).