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# The aim of this particular portfolio investigation is to find the area under a certain curve by dividing it into trapeziums and adding the sum of the areas of these trapeziums to approximate the actual area. This is done instead of using the usual method

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Introduction

The aim of this particular portfolio investigation is to find the area under a certain curve by dividing it into trapeziums and adding the sum of the areas of these trapeziums to approximate the actual area. This is done instead of using the usual method of integration. Let us consider the following function to begin with:- The above graph shows the function  from interval of x=0 and x=1. The area under this curve, as seen above, is divided into 2 trapeziums. If the area under the curve were to be found using the usual method of integration it would be as follows  =  =  =  Another way in which the above value can be approximated is by adding the sum of the area of the two trapeziums. One of the aims of this investigation is to see if there is some kind of relation between the accuracy of the approximation and the number of trapeziums involved to find it.

Area of trapezium =  …………………where a and b = parallel sides of trapezium

h = height of trapezium

Using the above formula the area of each trapezium can be found and then added together to find the approximation of the actual area.

Middle Even further to see the effect that the number of trapezium has on the approximation of the area, the above graph was divided into 20 trapeziums. Height of each trapezium will be,  h=  = 0.05 and the area was calculated as follows:-

Area = area of 1st trapezium + area of 2nd trapezium+area of 3rd trapezium+……………+ area of 20th  trapezium.

=  ( g(x) values of first trap. + g(x) values of second trape.+g(x) values of third trape………….+ g(x)values of twenthyth   trapezium) ……………………………………factorizied

=  ((g(0)+g(0.05)) + (g(0.05)+g(0.1))+ (g(0.1)+g(0.15))+………….+(g(0.95)+g(1)))

=  ((3+3.0025)+(3.0025+3.01)+(3.01+3.0225)+……………..+(3.9025+4))

=  3.33375

It can be concluded from the above calculations that by increasing the number of trapeziums under the area of the curve the approximation will get closer to the actual value. This is because increasing number of trapeziums will lead to smaller areas to be calculated by each trapezium thus minimizing the error formed. A general expression for the area under the curve of  , from x=0 to x= 1, using n trapeziums can be deducted as follows: Given n as the number of trapeziums, the interval (0, 1) is divided into equally sized n intervals. The height, h, will be h= (1-0)/n.

Conclusion

•  when integrated as follows gives the area under the curve from a=1 and b= 3    =  = 8.25

Using eight trapeziums, the same area can be approximated using the general statement derived above:-    =  =  = 8.162

•  when integrated as follows gives the area under the curve from a=1 and b= 3    =  Using eight trapeziums, the same area can be approximated using the general statement derived above:-    =  =  =4.6681

One of the uses of this general statement in other types of functions is it enables to find the area under a graph where integration is not the ideal formula to use. Asymptote graphs are good example. The graph below the function of  where x can never be 1. If the area under the curve of x=1 and x=2 is asked then instead of integration the trapezium rule can be used since x can be approximated to close values such as x= 0.0000001 and so on. This is also true for functions such as f(x) = tan x. The approximation can be as accurate as desired by just increasing the number of trapeziums under a certain curve. One of the limitations of this rule is it can only be applied to continuous functions of interval (a, b).

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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