# 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?

• If you use a uniform partition of size $1/n$ and take Darboux upper and lower sums, you'll end up with a known sum of squares, and ultimately you'll get a limit of $1/3$. You do need to argue that the integral exists, too. – Pedro Tamaroff Dec 2 '13 at 14:44
• Riemann sums are far from the most elegant way to do this problem, but I suspect that most people who teach calculus don't even suspect that. – Michael Hardy Dec 2 '13 at 16:28

$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

$$\int_0^1x^2dx=\lim_{n\to\infty} \frac1n\sum_{r=1}^n \frac{r^2}{n^2}=\lim_{n\to\infty} \frac1{n^3}\sum_{r=1}^nr^2$$

$$\text{Now }\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6$$

So, $\displaystyle\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}\left(\frac16\right)\left(1+\frac1n\right)\left(2+\frac1n\right)=\frac13$

• I wonder if you can explain the third line for me please? How do you got $∑_{r=1} ^n r^2=(n(n+1)(2n+1))/6$ – Diane Vanderwaif Dec 2 '13 at 15:05
• @DianeVanderwaif, math.stackexchange.com/questions/48080/… – lab bhattacharjee Dec 2 '13 at 15:06
• why there is no 6 there on last line from previous line? – Aleksey Dec 2 '13 at 17:22
• @Aleksey, I willingly kept out $\frac16$ from the limit calculation which is legal as it is independent of $n$ – lab bhattacharjee Dec 3 '13 at 3:19

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 $n\to\infty$.

• I have liked this method since I first learned of it. It seems much underappreciated. – Harald Hanche-Olsen Dec 2 '13 at 15:10
• @HaraldHanche-Olsen I agree. The idea is apparantly due to Fermat, so even predating Newton and Leibniz (not to mention Riemann). – mrf Dec 2 '13 at 19:29