Evaluate $\int ^1 _0 x^2 \, dx$ without using the Fundamental Theorem of Calculus Evaluate $\int ^1 _0 x^2 \, dx$ without using the Fundamental Theorem of Calculus.
I can find the Riemann sum of some partition and the intermediate points, what should I do next?
 A: $$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\int_0^1x^2dx=\lim_{n\to\infty} \frac1n\sum_{r=1}^n \frac{r^2}{n^2}=\lim_{n\to\infty} \frac1{n^3}\sum_{r=1}^nr^2$$
$$\text{Now }\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6$$
So, $\displaystyle\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}\left(\frac16\right)\left(1+\frac1n\right)\left(2+\frac1n\right)=\frac13$
A: A cute way to do it while avoiding sums of squares is to use a geometric progression. (The same method can easily be adapted to $\int_0^1 x^n\,dx$ for other choices of $n$ without having to cook up formulas for $\sum k^n$.)
By general theory, $x \mapsto x^2$ is Riemann integrable, so we can compute the integral as a limit of any Riemann sum, as long as the "size of the partition" tends to $0$.
Fix an integer $n$ and choose points $x_j = \dfrac{r^j}{n}$ ($0 \le j \le n$) where $r > 0$ is chosen so that $r^n = n$.
Then a corresponding Riemann sum is
\begin{align}
\sum_{j=0}^{n-1} f(x_j)(x_{j+1}-x_{j}) &=
\sum_{j=0}^{n-1} \frac{r^{2j}}{n^2}\left( \dfrac{r^{j+1}}{n} - \dfrac{r^{j}}{n} \right) \\
&= \frac{1}{n^3}\sum_{j=0}^{n-1} (r^{3j+1}-r^{3j}) \\
&= \frac{1}{n^3} \cdot \frac{r^{3n+1}-r^{3n}-r+1}{r^3-1} \\
&= \frac{1}{n^3} \cdot \frac{(n^3-1)(n^{1/n}-1)}{n^{3/n}-1}.
\end{align}
I'll leave it as an exercise to check that this tends to $\dfrac13$ as $n\to\infty$.
