• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...

Introduction

Shady areas

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:-

image00.png

The above graph shows the functionimage01.pngimage01.png 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

image24.pngimage24.png

                                                                = image32.pngimage32.png

                                                      =  image40.pngimage40.png

                                                                = image02.pngimage02.png

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 =image09.pngimage09.png …………………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.

...read more.

Middle

image26.png

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= image27.pngimage27.png = 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.  

       =  image22.pngimage22.png( 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

     = image28.pngimage28.png((g(0)+g(0.05)) + (g(0.05)+g(0.1))+ (g(0.1)+g(0.15))+………….+(g(0.95)+g(1)))

      = image28.pngimage28.png((3+3.0025)+(3.0025+3.01)+(3.01+3.0225)+……………..+(3.9025+4))

= image29.pngimage29.png3.33375

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 image01.pngimage01.png, from x=0 to x= 1, using n trapeziums can be deducted as follows:

image30.png

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.

...read more.

Conclusion

  • image49.pngimage49.png  when integrated as follows gives the area under the curve from a=1 and b= 3

image50.pngimage50.png

image51.pngimage51.png

           = image52.pngimage52.png

           = 8.25

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

image03.pngimage03.png

image04.pngimage04.png

       = image05.pngimage05.png

            =  image06.pngimage06.png

              = 8.162

  • image07.pngimage07.pngwhen integrated as follows gives the area under the curve from a=1 and b= 3

image08.pngimage08.png

image10.pngimage10.png

           =image11.pngimage11.png

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

image12.pngimage12.png

image04.pngimage04.png

       = image05.pngimage05.png

            =  image06.pngimage06.png

               =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 image13.pngimage13.png where x can never be 1. image14.png

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).

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math Portfolio: trigonometry investigation (circle trig)

    Highlighted means 0 and number with E will be just defined as undefined values. This later was verified by TI-83 plus calculator. Once again, upon the analysis of the above table of values, the values of sin? are found to have a specific range; no smaller than -1 and no bigger than 1 (-1?sin??1).

  2. Infinite Summation - In this portfolio, I will determine the general sequence tn with ...

    by counting the infinite sum for variable a and x. When x = 1 and a = 2: Sn = eln2 = 2 = 21 When x = 2 and a = 2: Sn = e2ln2 = 4 = 22 When x = 3 and a = 2: Sn =

  1. Maths Portfolio Shady Areas

    The table below will show the differences in the areas that each number of trapeziums produces. Number of trapeziums Total area Total difference T1 3.5 0 T2 3.37 0.125 T4 3.34375 0.03125 T5 3.34 0.00375 When we take the total difference and square it, then add the two of the same squared differences.

  2. Population trends. The aim of this investigation is to find out more about different ...

    This model is , this is a linear equation, many points (represented by circles) are not close to the line and therefore the model is not likely to represent what happens next or what happened before because population doesn't normally increase constantly, this is the case for population in China.

  1. Moss's Egg. Task -1- Find the area of the shaded region inside the two ...

    ?BAE falls into both larger circles of the diagram. The length AB is the radius of these two circles, which we know to be 6cm in length. Therefore, the radius for determining the area of this sector is 6 cm.

  2. Investigating the Ratios of Areas and Volumes around a Curve

    The following table demonstrates the results obtained using this program for several values of n, a, and b. Results which do not correlate with the observed pattern (and cannot be put down to the calculator's internal rounding errors) are shown in bold.

  1. Function Transformation Investigation

    between the original graph and the transformed one. The stretch happens along the y axis. The second graph represents a sine graph and its transformation done by multiplying the output. The stretch is a vertical one. The two circled points help us notice that it is a stretch of 1/2, because the output was multiplied by 2.

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    added together. The area is approaching around the value of 3. However, more calculations with additional trapezoids are needed to determine a more precise value. Eight Trapezoids Trapezoid h a b Area 1 0.125 3 3.015625 0.375976562 2 0.125 3.015625 3.0625 0.379882812 3 0.125 3.0625 3.140625 0.387695312 4 0.125 3.140625

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work