TITLE: Proving $\int_a^b x^2dx = \frac{b^3 - a^3}{3}$
QUESTION [4 upvotes]: Can anyone help me prove $\int_a^b x^2dx = \frac{b^3 - a^3}{3}$, the long way? I know exactly what to do, but the algebra involved is just too much for me and I keep making a mistake somewhere and getting a different result every time...
I need to prove it using the Riemann defintion of an integral, for a start it would be:
$$\int_a^b x^2dx = \displaystyle \lim_{n \to\infty} \sum_{i = 1}^n\left[{a+\frac{bi - ai}{n}}\right]^2\left[\frac{b - a}{n}\right]$$
right? And I need to do so many steps to prove it..is there an easier way or will I just have to go through all the steps?

REPLY [1 votes]: $$\int_a^bx^2dx=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{bn}\left(\frac{i}{n}\right)^2-\sum_{i=1}^{an}\left(\frac{i}{n}\right)^2\right]=\lim_{n\to\infty}\frac{1}{n}\sum_{i=an}^{bn}\left(\frac{i}{n}\right)^2=\lim_{n\to\infty}\frac{(b^3-a^3)*n*(n+1)(2n+1)}{6n^3}=\frac{b^3-a^3}{3}$$
The integral on the interval $(a,b)$ is the same as the area under the curve which is the sum of all of the y-values between $a$ and $b$ multiplied by an infinitely small width ($\frac{1}{n}$)
$\frac{1}{n}\left[f(a)+f(a+\frac{1}{n})+f(a+\frac{2}{n})+\cdots+f(b-\frac{2}{n})+f(b-\frac{1}{n})+f(b)\right]$
and so you can write it out as $\frac{1}{n}$ times the quantity of the sum of the y-values on the interval $(0,b)$ minus the sum of the y-values on the interval $(0,a)$ and combine those two summations into one. Then, convert the summation into a limit, do the limit, and you're left with $\frac{b^3-a^3}{3}$ as the equation for the definite integral on the interval $(a,b)$.
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