# Definition of Definite Integral

• Feb 6th 2008, 03:37 PM
aquaglass88
THanks so much for your help!!
• Feb 7th 2008, 12:05 AM
mr fantastic
Quote:

Originally Posted by aquaglass88
Hi I am trying to find use the definition of the definite integral to prove that
the integral of (x^2)dx in [a,b] = [(b^3) - (a^3)]/3

I started out using the Reiman Sums except I ended up with
lim as n approaches infinity [nab^2 - bab^2 + b^3 - na^3 + a^3 - ab^2]^2
-------------------------------------------------------
n^2

I have no clue how that equals b cubed minus a cubed over 3 and I'm not sure what I did wrong if I did.

THanks so much for your help!!

Using right end-points, by definition:

$\displaystyle \int_a^b x^2 \, dx = \lim_{n \rightarrow \infty} \left( \frac{b-a}{n} \right) \sum_{i=1}^{\infty} f\left( a + \frac{(b-a)i}{n}\right) = \lim_{n \rightarrow \infty} \left( \frac{b-a}{n} \right) \sum_{i=1}^{\infty} \left( a + \frac{(b-a)i}{n} \right)^2$

Expand:

$\displaystyle = \lim_{n \rightarrow \infty} \left[ \frac{(b-a)}{n} \sum_{i=1}^{\infty} a^2 + \frac{2a(b-a)^2}{n^2} \sum_{i=1}^{\infty} i + \frac{(b-a)^3}{n^3} \sum_{i=1}^{\infty} i^2 \right]$

Substitute the well known sum of each series:

$\displaystyle = \lim_{n \rightarrow \infty} \left[ \frac{(b-a)}{n} (a^2n) + \frac{2a(b-a)^2}{n^2} \frac{n(n-1)}{2} + \frac{(b-a)^3}{n^3} \frac{n(n+1)(2n+1)}{6} \right]$

Simplify each term by cancelling common factors:

$\displaystyle = \lim_{n \rightarrow \infty} \left[ (b-a)a^2 + \frac{a(b-a)^2(n-1)}{n} + \frac{(b-a)^3 (n+1)(2n+1)}{6n^2} \right]$

Take the limit term by term:

$\displaystyle = (b-a)a^2 + a(b-a)^2 \lim_{n \rightarrow \infty}\frac{(n-1)}{n} + \frac{(b-a)^3}{6}\lim_{n \rightarrow \infty}\frac{(n+1)(2n+1)}{n^2}$

Evaluate the simple limits:

$\displaystyle = (b-a)a^2 + a(b-a)^2 (1) + \frac{(b-a)^3}{6} (2)$

$\displaystyle = (b-a)a^2 + a(b-a)^2 + \frac{(b-a)^3}{3}$

Simplify:

$\displaystyle = \frac{b^3}{3} - \frac{a^3}{3}$.