1. ## Trapezoidal Rule

English is my second language:

I want to increase the accuracy of the trapezoidal rule for approximating the area beneath a curve by increasing the number of trapeziums the area is segregated into.

My problem is I have very limited knowledge of how to use the sigma function. This is what I have so far:

Area of trapezium = lim(Δx->0) 1/2*Δx*(f(x)+f(x+Δx)) where f(x) is some function of x.

My justification is that by making the Δx very small, the area must be segregated into more trapeziums but I know that this is only half of the solution that I desire - how do I fit this into a sigma function and what constraints do I use?

2. Yes, the area of a single trapezium is given by $\frac{f(x)+ f(x+ \Delta x)}{2}\Delta x$. If you add one more trapezium, from $x+ \Delta x$ to $x+ 2\Delta x$ you get $\left(\frac{f(x)+ f(x+ \Delta x)}{2}\Delta x\right)+ \left(\frac{f(x+ \Delta x)+ f(x+ 2\Delta x)}{2}\Delta x\right)$ which reduces to $\left(f(x)+ 2f(x+ \Delta x)+ f(x+ 2\Delta x)\right)\frac{\Delta x}{2}$.

If you use a third trapezium you get
$\left(f(x)+ 2f(x+ \Delta x)+ f(x+ 2\Delta x)\right)\frac{\Delta x}{2}+ \left(\frac{f(x+ 2\Delta x)+ f(x+ 3\Delta x)}{2}\Delta x\right)$ which reduces to $\left(f(x)+ 2f(x+ \Delta x)+ 2f(x+ 2\Delta x)+ f(x+ 3\Delta x)\right)\frac{\Delta x}{2}$.

Do you see the point? Everytime you add one more trapezium, from n to n+1, you add a new " $f(x+ n\Delta x)+ f(x+ (n+1)\Delta x)$" making the coefficient of $f(x+ n\Delta x)$ 2.

That is, you have $\left(f(x)+ 2\f(x+ \Delta x)+ 2f(x+ 2\Delta x)+ 2f(x+ 3\Delta x)+ \cdot\cdot\cdot+ 2f(x+ (n-1)\Delta x)+ f(x+ n\Delta x)$, all but the first and last terms multiplied by 2. And that is always multiplied by $\frac{\Delta x}{2}$.

You can write that using the $\Sigma$ notation as
$\left(f(x)+ 2\sum_{i=1}^{n-1} f(x+ i\Delta x)+ f(x+ n\Delta x)\right)\frac{\Delta x}{2}$.

If you are integrating f(x) from x= a to x= b and want to use n intervals, then $\Delta x= \frac{b- a}{n}$

3. Thank you very much, that has cleared up a lot of my confusion!

I have one more quick question: If i want to integrate between b and a using this model, can I introduce the limit n->infinity to improve the accuracy? If i can, does this necessitate any other alterations to the model?

4. Well, the limit as n goes to infinity would not exactly "improve the accuracy"- it would make it exactly correct, giving the Riemann integral. Taking larger and larger n will improve the accuracy (assuming your calculator or computer can handle the increased number of decimal places).

5. Excellent, thank you for all your help.