# Approximate integration

• September 27th 2011, 12:56 PM
minmin1
Approximate integration
Can anyone help me with what seems like an easy problem?

Show that:

$(1/3)T_n + (2/3)M_n = S_2_n$
• September 27th 2011, 04:55 PM
HallsofIvy
Re: Approximate integration
Well, it would help if you told us what $T_m$, $M_n$, and $S_$ mean!
• September 27th 2011, 09:31 PM
tangibleLime
Re: Approximate integration
I think he's using Simpson's rule. If I'm correct, it looks like he's trying to show the relation between the midpoint rule and the trapezoidal rule with Simpson's rule.

Midpoint rule: $M = (b-a)f(\frac{a+b}{s})$

Trapezoidal rule: $T = \frac{1}{2}(b-a)(f(a)+f(b))$

Weighted average:

$\frac{2M+T}{3}$

This weighted average is Simpson's rule.

Simpson's rule:

$\int_{a}^{b} f(x) dx \approx \frac{b-a}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right]$

Looking at his equation,

$\frac{1}{3}T_{n} + \frac{2}{3}M_{n} = \frac{2M+T}{3}$ = weighted average = Simpson's Rule

Not totally sure what these subscripts are, but there's a start.