The trapezoid method approximates the curve, between successive points, by a straight line while Simpson's rule uses three points at a time and approximates the curve by the parabola passing through the three points.
Suppose your function is f(x) and you take three points, (x_0, f(x_0), (x_1, f(x_1), (x_2, f(x_2)), equally spaced on the x-axis. For simplicity, take the x-values to be x_0= x_1- h, x_1, and x_2= x_1+ h where h is the distance between succesive x values so the points themselves are (x_1- h, f(x_0)), (x_1, f(x_1)), (x_1+ h,f(x_2)).<br />
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The equation of a parabola through (x_1, f(x_1)) can be written y= a(x- x_1)^2+ b(x- x_1)+ f(x_1). Taking x= x_0= x_1- h we get y= f(x_0)= ah^2+ bh+ f(x_1). Taking x= x_2= x_1+ h we get y= f(x_2)= ah^2- bh+ f(x_1).<br />
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Adding those two equations, f(x_2)+ f(x_0)= 2ah^2+ 2f(x_1) so that <br />
ah^2= \frac{f(x_2)+ f(x_0)}{2}- f(x_1}= \frac{f(x_2)- 2f(x_1)+ f(x_0)}{2}]<br />
and<br />
a= \frac{f(x_0)+ 2f(x_1)+ f(x_2)}{2h^2}<br />
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Subtracting those two equations, f(x_2)- f(x_0)= 2bh so that <br />
b= \frac{f(x_2)- f(x_0)}{2h}<br />
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Now, \int_{x_0}^{x_2}a(x-x_1)^2+ b(x-x_1)+ f(x_1) dx= \int_{x_1-h}^{x_1+h} a(x-x_1)^2+ b(x-x_1)+ f(x_1) dx= (1/3)ah^3+ (1/2)bh^2+ f(x_1)h- (1/3)a(-h)^3- b(-h)^2- f(x_1)(-h)= (2/3)ah^3+ 2f(x_1)hso b really isn&amp;#039;t important. The integral of the parabola through (x_1-h, f(x_0)), (x_1, f(x_1)), (x_1+h,f(x_2)) is &lt;br /&gt;
(2/3)ah^3+ 2f(x_1)h= (2/3)\frac{f(x_0)- 2f(x_1)+ f(x_2)}{2h^2}h^3+ 2f(x_1)h&lt;br /&gt;
= \frac{2(f(x_0)- 2f(x_1)+ f(x_2)}{3}h+ 2f(x_1)h= h\frac{2f(x_0)- 4f(x_1)+ 2f(x_2)}{3}&lt;br /&gt;
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That is the formula for exactly 3 data points. If we have more, we can separte into groups of three and do that for each such interval. Bacause the overlap at the ends, we must have an &lt;b&gt;odd&lt;/b&gt; number of points to do that.&lt;br /&gt;
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For example, if we had only an even number of points, x_0, x_1, x_2, x_3, our first parabola would be over x_0, x_1, x_2 but, even including x_3 as our first endpoint in the next interval we would have only x_2, x_3, not enough for a parabola. But with 5 points, we could have x_0, x_1, x_2, x_2, x_3, x_4. Similarly, with 6 points, an even number, we would have x_0, x_1, x_2, x_2, x_3, x_4, x_4, x_4 and not be able to complete the last parabola but with 7, an odd number, we would have x_0, x_1, x_2, x_2, x_3, x_4, x_4, x_5, x_6.&lt;br /&gt;
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Now see what happens when we &amp;quot;attach&amp;quot; parabolas: on x_0, x_1, x_2 we have &lt;br /&gt;
h\frac{2f(x_0)- 4f(x_1)+ 2f(x_2)}{3}&lt;br /&gt;
while on x_2, x_3, x_4&lt;br /&gt;
h\frac{2f(x_2)- 4f(x_3)+ 2f(x_4)}{3}[/itex]&amp;lt;br /&amp;gt;
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Those add to give&amp;lt;br /&amp;gt;
h\frac{2f(x_0)- 4f(x_1)+ 4f(x_2)- 4f(x_3)+ 2f(x_4)}{3}[/itex]&amp;amp;lt;br /&amp;amp;gt;
so you see why we have &amp;amp;amp;quot;4&amp;amp;amp;quot; multiplying everything except the first and last terms. Taking the integral to be from a to b and using n intervals, h= (b-a)/n so the formula becomes &amp;amp;lt;br /&amp;amp;gt;
(b-a)\frac{2f(x_0)- 4f(x_1)+ 4f(x_2)- 4f(x_3)+ 2f(x_4)}{3n}[/itex]&amp;amp;amp;lt;br /&amp;amp;amp;gt;
Simpson&amp;amp;amp;amp;#039;s rule.
