Proving $\int_a^b x^2dx = \frac{b^3 - a^3}{3}$ 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?
 A: Edit : using d=b-a
$\begin{align}
\sum_{i=1}^n \left(a+\dfrac{i(b-a)}{n}\right)^2\left(\dfrac{b-a}{n}\right)&=\left(\dfrac{d}{n}\right)\sum_{i=1}^n \left(a^2+\dfrac{2iad}{n}+\dfrac{i^2d^2}{n^2}\right)\\
&=\left(\dfrac{d}{n}\right)\left(\left(\sum_{i=1}^n a^2\right)+ \left(\sum_{i=1}^n \dfrac{2iad}{n}\right)+\left(\sum_{i=1}^n\dfrac{i^2d^2}{n^2}\right)\right)\\
&=\left(\dfrac{d}{n}\right)\left(na^2+\dfrac{2ad}{n}\sum_{i=1}^n i+\dfrac{d^2}{n^2}\sum_{i=1}^ni^2\right)\\
&=a^2d+\dfrac{2ad^2}{n^2}\dfrac{n(n+1)}{2}+\dfrac{d^3}{n^3}\dfrac{n(n+1)(2n+1)}{6}
\end{align}$
now looking at the limit for $n\to\infty$, all the terms in $\dfrac{1}{n}$ will disappear.
The only terms that survive the limit are $a^2(b-a)+\dfrac{2a(b-a)^2}{2}+\dfrac{2(b-a)^3}{6}=\dfrac{a^3-b^3}{3}$
A: It suffices we prove $$\int_0^1x^2dx=\frac 1 3 $$
Then $$\int_0^ax^2dx=\frac {a^3} 3$$ will follow by substitution and $$\int_a^bx^2dx=\int_0^bx^2dx-\int_0^ax^2dx=\frac{b^3-a^3}3$$
Now, taking an upper Darboux sum for $x^2$ over $[0,1]$ with a regular partition gives $$D(f,\Pi_n)=\frac 1 n\sum_{k=1}^{n}  \frac{k^2}{n^2}=\frac{1}{n^3}\sum_{k=1}^n k^2$$
Now use $\displaystyle\sum_{k=1}^n k^2=\frac{n(2n+1)(n+1)}6$
A: $$\int_a^bx^2dx=\lim_{n\to\infty}\frac{1}{n}\left[\sum_{i=1}^{bn}\left(\frac{i}{n}\right)^2-\sum_{i=1}^{an}\left(\frac{i}{n}\right)^2\right]=\lim_{n\to\infty}\frac{1}{n}\sum_{i=an}^{bn}\left(\frac{i}{n}\right)^2=\lim_{n\to\infty}\frac{(b^3-a^3)*n*(n+1)(2n+1)}{6n^3}=\frac{b^3-a^3}{3}$$
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 $(a,b)$.
A: Convert Riemann definition to a definite integral as below 
$$\int_{x=a}^{b} x^2 dx = \lim_{n\to\infty }\sum_{i=1}^{n} \left [ a+\frac{bi-ai}{n} \right ]^2 \left [ \frac{b-a}{n} \right ] \text{    ........... equation  no 1.} $$
Using 
$$\lim_{n\to\infty}\sum_{k=1}^{n}f\left ( \frac{k}{n} \right)\left [ \frac{1}{n} \right ] =  \int_{x=0}^{1} f(x)dx $$
So equation  1 become as below
$$\int_{x=0}^{1}\left [ a+(b-a)x\right]^2 dx =  \frac{\left [ a+(b-a)x \right ]^3}{3[b-a]}\text{ over }[x=0,1]$$
$\displaystyle\text{So  }\int_{x=a}^{b} x^2 dx =\frac{b^3-a^3}{3}$
You can check this for more information 
http://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/
