<s>[INST] Evaluate $\int x^2dx$ using darboux sum Let take the partition $P=\{0,\frac{1}{n},\frac{2}{n},...,\frac{n-1}{n},1\}$.
Because $f(x)=x^2$ is strictly increasing $m_i$ we will be on the leftmost side of each interval and $M_i$ on the rightmost side.
$\underline{S}(f,P)=\sum_{i=1}^n m_i*\Delta x_i=\sum_{k=1}^n (\frac{i-1}{n})^2*\frac{1}{n}=\frac{1}{n^3}\sum_{k=1}^n (i-1)^2=\frac{1}{n^3}*(\frac{(n^2-n)}{2})^2=\frac{1}{n^3}*\frac{(n^4-2n^3+n^2)}{4}=\frac{n^2(n^2-2n+1)}{4n^3}=\frac{n^2-2n+1}{4n}=\frac{(n-1)^2}{4n}$
$\overline{S}(f,P)=\sum_{i=1}^n m_i*\Delta x_i=\sum_{k=1}^n (\frac{i}{n})^2*\frac{1}{n}=\sum_{k=1}^n (\frac{i^2}{n^2})*\frac{1}{n}=\frac{1}{n^3}*\sum_{k=1}^ni^2=\frac{1}{n^3}*(\frac{(n*(n+1))}{2})^2=\frac{(n^4+2n^3+n^2)}{4n^3}=\frac{n^2(n^2+2n+1)}{4n^3}=\frac{(n^2+2n+1)}{4n}=\frac{(n+1)^2}{4n}$
Now we take $\lim_{n \to \infty}\underline{S}(f,P)=\lim_{n \to \infty}\frac{(n-1)^2}{4n}=\infty$
$\lim_{n \to \infty}\bar{S}(f,P)=\lim_{n \to \infty}\frac{(n+1)^2}{4n}=\infty$
Which is obviously wrong, where did I get it wrong?
 [/INST] The computation of sum of squares is wrong.
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$
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