Proving $\int_a^b x^2dx = \frac{b^3 - a^3}{3}$

Question

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?

Answer 1

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 $\displaystyle\sum_{k=1}^n k^2=\frac{n(2n+1)(n+1)}6$

Answer 2

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 $a^2(b-a)+\dfrac{2a(b-a)^2}{2}+\dfrac{2(b-a)^3}{6}=\dfrac{a^3-b^3}{3}$

Answer 3

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/