Core 2 Numerical Integration: Trapezium Rule

Taken from my upcoming Core 2 Revision Guide. This is a brief preview and an explanation on one of the methods of estimating a definite integral. Others include the mid-ordinate rule, Newton-Cotes formula and the Simpsons rule. However none of these need to be dealt with until later units.

Sometimes you may be given an equation that seems impossible to integrate given our current known techniques, and that is because they are. So we need a method of determining the area beneath a curve at given limits. Which is actually not too difficult, if we take shapes that are sort of like the curve, find the area of all of them and then add them up we will get roughly the area beneath the curve at those points.

As you may have guessed by now the area will be estimated using trapeziums. The reason for this is that trapeziums naturally slope and can emulate the shape of a curve (given enough of them). The best way to explain how to do this is explain this is to use an example that we can integrate, which I will do as y = x².

So we have to start by knowing what our limits are (in our case they are at 1 and 4), we also need to know what the height of each of our trapeziums are (the horizontal bit) and what the size of each of the two parallel lines, in our case these are at the points when x = 1, x = 2, x = 3 and x = 4

Once we know each of these we can then find the area of each of the trapeziums, add them together and voila we have an estimate for the area beneath the curve. If you don't remember the area of a trapezium is 0.5(a+b)h where a and b are the parallel sides and h is the height. So the area of each of the trapezium respectively is 0.5*(1+4)*1, 0.5*(4+9)*1 and 0.5*(9+16)*1 which is 2.5, 6.5 and 12.5 so if we add these together (which is 21.5) we get the rough area beneath the graph of y = x².

If we look at how we have found the area of our three trapeziums [0.5*(1+4)*1 + 0.5*(4+9)*1 + 0.5*(9+16)*1] we can see that there are two things that multiply by each other in each trapezium (0.5 and 1) so if we take them out we get: 0.5*1*(1+4+4+9+9+16), this can be further simplified as there are two 4's and two 9's so we get 0.5*1*[(1+16)+2(4+9)] which is were we derive the formula for the rough area beneath the curve at given limits.

Words you need to know is that an ordinate is the y-coordinate at the values of x we have, so for example when x = 1 the ordinate is also 1, when x = 2 the ordinate is 4, etc. And a strip is a trapezium, so if we have 4 strips we have 4 trapezium, also you should note that if there are n strips there will be n+1 ordinates.

So the word equation for an estimate of the definite integration between limits a and b is:

The formula for this (that you will be given in the exam) is:

Just so you know y₀ is the first ordinate, y₁ is the second and y_n is the last.

This is particularly useful when you need to find the integral of something that our current rules for calculus can not give an answer for it, with √(2^x), is an example of where the trapezium rule is useful. In fact using our formula I will solve this between the limits of 3 and 1 with 4 strips.

You will need to know how to make our estimate more accurate, to do this you simply use more strips as the more there are the closer to trapeziums will be to making the actual shape.

If anything I have wrote does not appear to make sense please comment and I will do my best to explain.