How to calculate $\int_a^bx^2 dx$ using summation?

Question

So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.

\begin{align} x_0&=a\\ x_1&=a+\frac{b-a}{n}\\ &\ldots\\ x_{i-1}&=a+(i-1)\frac{b-a}{n}\\ x_i&=a+i\frac{b-a}{n} \end{align}

So I pick left point, which is $x_{i-1}$

I start with \begin{align} \sum\limits_{i=1}^{n} \frac{b-a}{n}f\left(a+\frac{(i-1)(b-a)}{n}\right) &=\frac{b-a}{n}\sum\limits_{i=1}^{n} \left(a+\frac{(i-1)(b-a)}{n}\right)^2\\ &=\frac{b-a}{n}\left(na^2+ \frac{2a(b-a)}{n}\sum\limits_{i=1}^{n}(i-1)+\frac{(b-a)^2}{n^2} \sum\limits_{i=1}^{n}(i-1)^2\right) \end{align}

Here I am stuck because I don't know what$ \sum\limits_{i=1}^{n}(i-1)^2$is (feel like it diverges). Could someone help?

Answer 1

The sum that is of concern is multiplied by $\frac1{n^3}$ and the limit therefore converges. Then, proceeding, we have

$$\sum_{i=1}^n(i-1)^2=\sum_{i=0}^{n-1} i^2$$

We can evaluate the sum on the right-hand side of $(1)$ using a number of approaches. The approach we present here exploits telescoping series. We note that

$$\sum_{n=0}^{n-1}\left((i+1)^3-i^3\right)=n^3 \tag 2$$

If we expand the terms in the sum in $(2)$, we find that

$$3\sum_{n=0}^{n-1}i^2+3\sum_{n=0}^{n-1}i+\sum_{n=0}^{n-1} 1=n^3 \tag 3$$

whereupon solving for the sum $\sum_{n=0}^{n-1}i^2$ and using $\sum_{n=0}^{n-1}i=\frac{n(n-1)}{2}$ reveals

$$\begin{align} \sum_{n=0}^{n-1}i^2&=\frac{n^3-n-3\frac{n(n-1)}{2}}{3}\\\\ &\bbox[5px,border:2px solid #C0A000]{=\frac{n(n-1)(2n-1)}{6}} \end{align}$$

Dividing by $\frac1{n^3}$ and lettgin $n\to \infty$ yields $\frac13$ as expected!

Adding this to the other terms and letting $n \to \infty$ yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{n\,a^2(b-a)}{n}+\frac{2a(b-a)^2\,\frac{n(n-1)}{2}}{n^2}+\frac{(b-a)^3\,\frac{n(n-1)(2n-1)}{6}}{n^3}\right)=\frac{b^3-a^3}{3}}$$

Answer 2

Hint: $$\sum_{i=1}^{n}(i-1)^2 = \sum_{i=1}^{n}(i^2-2i+1) = \sum_{i=1}^{n}i^2 - 2\sum_{i=1}^{n}i + \sum_{i=1}^{n}1$$ You should be able to take it from here.

Answer 3

It is a well-known formula that

$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

So, in your case,

\begin{align*} \sum_{i=1}^n (i-1)^2 &= 0^2 + 1^2 + \cdots + (n-1)^2 \\ &= 1^2 + \cdots + (n-1)^2 \\ \\ &= \frac{(n-1)n(2n-1)}{6} \end{align*}

Answer 4

$$\sum_{i=1}^n (i-1)^2$$ $$=\sum_{i=1}^n (i^2-2i+1)$$ $$=\sum_{i=1}^n i^2-2\sum_{i=1}^ni+\sum_{i=1}^n1$$ $$=\frac{n(n+1)(2n+1)}{6}-2\cdot \frac{n(n+1)}{2}+n$$ $$=\frac{(n^2+n)(2n+1)}{6}-n(n+1)+n$$ $$=\frac{2n^3+3n^2+n}{6}-n^2$$ $$=\frac{2n^3-3n^2+n}{6}$$ $$=\frac{n(n-1)(2n-1)}{6}$$