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Let $b > 0$. I'd like to calculate $ \int_{0}^b x^2 dx$ using upper and lower Riemann sums. Since the function is continuous on $[0,b]$ I know that it is integrable so I only need to calculate

$$ \mathcal{U}(f) := \inf_{P \in \mathcal{P}} U(f,P)$$ where $\mathcal P $ is the set of all partitions $P = \{x_0 = 0, ...,x_n = b\}$ of $ [0,b]$ and

$$ U(f,P) = \sum_{i = 1}^n M_i(x_i - x_{i-1})$$ with $ M_i = \sup\{f(x) : x \in [x_{i-1},x_i ]\}$.

Calculating this number does not seem to be easy.

If I only consider partitions $P_n$ with $x_k = k \frac{b}{n}$ then I can show that $ \lim_{n \to \infty} U(x^2,P_n) = b^3/3$ using

\begin{align*} U(x^2,P_n) &= \sum_{k=1}^n k^2\frac{b^2}{n^2} \frac{b}{n} \\ &= b^3 \frac{1}{n^3} \frac{n(n+1)(2n+1)}{6} \\ &= b^3 \frac{2n^3 + 3n^2 + n}{6n^3}. \end{align*} This gives the correct answer. It seems obvious that the limit of these partitions approches $\mathcal U(f)$. But why ? Is it sufficient to only consider such partitions ? What are the conditions $f$ should have for this to work ?

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  • $\begingroup$ Any continuous function on $[o,b]$ is integrable, so it is enough to take a sequence of partitions with maximum of the lengths of the subintervals tending to $0$. $\endgroup$ – Kavi Rama Murthy Apr 11 at 7:59
  • $\begingroup$ In this example, considering the lower partitions gives essentially the same sum, and you can use the criterion that if $U(f,P_n)-L(f,P_n)\to0$ then $f$ is integrable with integral $\lim_n U(f,P_n)$. $\endgroup$ – Lord Shark the Unknown Apr 11 at 8:17
  • $\begingroup$ @LordSharktheUnknown because $L(f,P_n) \leq \mathcal L (f) \leq \mathcal U(f) \leq U(f,P_n)$, taking the limit and all inequalities become equalities ? $\endgroup$ – Digitalis Apr 11 at 8:22
  • $\begingroup$ Indeed${}{}{}$! $\endgroup$ – Lord Shark the Unknown Apr 11 at 8:23

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