Integral approximation -

Question

Whole day I can not figure out how can be proved the equality: $$\int_0^1 x^2 dx = \frac{1}{n} \sum_{i=1}^n \left(\frac{2i-1}{2n}\right)^2 + \frac{1}{12n^2}$$

Can someone help me, what should I use for the proof?

Answer 1

We can use the three identities $$ \sum_{i=1}^n1=n $$ $$ \sum_{i=1}^ni=\frac{n^2+n}2 $$ $$ \sum_{i=1}^ni^2=\frac{2n^3+3n^2+n}6 $$ to evaluate $$ \begin{align} \frac1n\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2 &=\frac1{4n^3}\sum_{i=1}^n(4i^2-4i+1)\\ &=\frac1{4n^3}\left(\frac{4n^3+6n^2+2n}3-2n^2-2n+n\right)\\ &=\frac1{4n^3}\frac{4n^3-n}3\\ &=\frac13-\frac1{12n^2} \end{align} $$ Therefore, $$ \frac1n\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2+\frac1{12n^2}=\frac13 $$

Answer 2

We know that $$\int_0^1 x^2 \, dx = \frac{1}{3}$$Thus, proof by induction would be most efficient if we just make this replacement in the formula. So, prove $$\frac{1}{n}\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2 + \frac{1}{12n^2} = \frac{1}{3}$$for all $n\in\mathbb{N}$.

Proof by induction:

First, we must show the base case works for $n=1$. So, $$\frac{1}{1}\sum_{i=1}^1\left(\frac{2i-1}{2\times1}\right)^2 + \frac{1}{12\times(1)^2} = \frac{1}{4} + \frac{1}{12} = \frac{1}{3}$$Awesome! Now assume the formula works for $n =k$. So, $$\frac{1}{k}\sum_{i=1}^k\left(\frac{2i-1}{2k}\right)^2 + \frac{1}{12k^2} = \frac{1}{3}$$We must now show the formula works for $n = k+1$. So,

$$\begin{align}\frac{1}{k+1}\sum_{i=1}^{k+1}\left[\frac{2i-1}{2(k+1)}\right]^2 + \frac{1}{12(k+1)^2} & =\frac{1}{k+1}\sum_{i=1}^{k+1}(2i-1)^2[2(k+1)]^{-2} + \frac{1}{12(k+1)^2}\\ & = \frac{(2(k+1))^{-2}}{(k+1)}\sum_{i=1}^{k+1}(2i-1)^2+\frac{1}{12(k+1)^2}\\ & = \frac{1}{4(k+1)^3}\left[\frac{1}{3}(k+1)(2k+1)(2k+3)\right] + \frac{1}{12(k+1)^2}\\ & = \left[\frac{1}{3} - \frac{1}{12(k+1)^2}\right] + \frac{1}{12(k+1)^2}\\ & = \frac{1}{3} + 0 \\ & =\frac{1}{3}\end{align}$$

Therefore, $n=k+1$ also works. Thus, by mathematical induction, $$\frac{1}{n}\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2 + \frac{1}{12n^2} = \int_0^1x^2\,dx = \frac{1}{3}$$for all $n\in\mathbb{N}$.

Take careful note that this proof depends on the fact that $\displaystyle \sum_{i=1}^{k+1}(2i-1)^2 = \frac{1}{3}(k+1)(2k+1)(2k+3)$, which is the sum of the first $k+1$ squared odd numbers, which there's tons of proofs of that on the internet. You could've also proved it by foiling out the $(2i-2)^2$ and summing each one of those individually.