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How to Value Your Rich Elderly Aunt

Philip Cooper, 26 November 2002, 10 am – 3.30 pm

Notes for Task 3

The Trapezium Rule

To find the area under the curve using the Trapezium Rule, we have to find the areas of the individual trapezia and then add them altogether.
Area of a trapezium = $\frac{(a + b)h}{2}$ where a and b are the lengths of the parallel sides, and h is the height of the trapezium.

$\begin{displaymath} \par area under curve \par = \frac{1}{2} [f(x_1) + f(x_2)](... .... \par + \frac{1}{2} [f(x_{n-1}) + f(x_n)](x_n - x_{n-1}) \par\end{displaymath}$

Now if the heights of all the trapezia are the same, then

$\begin{displaymath} \par x_2 - x_1 = x_3-x_2 = ... = x_{i+1} - x_i = ... = x_n - x_{n-1} \end{displaymath}$

and we can write the area as

$\begin{displaymath} \par area under curve \par = \sum_{i=1}^{n-1} [\frac{1}{2} [f(x_i) + f(x_{i+1})](x_{i+1} - x_i)] \par\end{displaymath}$

Special Case of the Trapezium Rule

We can simplify this formula for the special case where the end-points of the interval across which we want to estimate the area are integers, and the spacings between the trapezia are of length 1.

So,

$\begin{displaymath} \par area under curve = \frac{1}{2} [f(x_1) + f(x_2)](x_2 - ... ... \par + ... + \frac{1}{2} [f(x_{n-1}) + f(x_n)](x_n - x_{n-1}) \end{displaymath}$

$\begin{displaymath} \par = \frac{1}{2} ([f(x_1) + f(x_2)] + [f(x_2) + f(x_3)] + ... \par + [f(x_{n-1}) + f(x_n)]) \end{displaymath}$

since $\par x_{i+1} - x_i = 1$

$\begin{displaymath} \par = \frac{1}{2} (f(x_a) + 2[f(x_2)] + f(x_3) + ... \par + f(x_{n-1})] + f(x_b)) \end{displaymath}$

giving the following neat formula

$\begin{displaymath} \par\frac{1}{2} f(x_a) + \sum_{x=a+1}^{x=b-1} f(x) + \frac{1}{2} f(x_b)\end{displaymath}$

since 1#1

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