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

Numerical integration coursework

Extracts from this document...


Numerical integration coursework



For this coursework, I am going to use my knowledge of numerical methods to produce an approximation to an area which does not have an analytic solution. I will be finding the approximation, to an appropriate degree of accuracy, of the integral shown above. On the graph below is the area that I will be approximating underneath the curve of y=image01.png

 from x=0 to x=2. Note that throughout my method I worked in radians.


This problem is appropriate for numerical solution as I chose my graph to be a polynomial curve involving a square root so that there would be no analytical solution. Due to the fact that I cannot yet integrate functions like this approximating methods will have to be used. According to the numerical methods module the three approximation methods to be used are:

Mid-point rule- this method was adopted because it is used to approximate the area underneath the graph by dividing it up into individual rectangles.

Trapezium rule- this method was adopted because it is used to approximate the area underneath the graph, this is done so by dividing it up into individual trapeziums.

Simpson’s rule- I have realised that out of the

...read more.


Sn= 2(Mn) +Tn


Extrapolating my answers to the Simpson’s rule to infinity will be the most accurate answer that I can give.  

Formula view of my method for approximations

Here I will briefly show how I used algorithms on Microsoft Excel to reach my approximations:image24.png

Mid-point rule

Trapezium rule


Simpson’s rule


Error analysis

As all three rules are actually only approximations there is always error involved, they are all gradually converging towards the actual solution. In reference to my graph the integral section is concave which means that the trapezium rule gives an overestimate and the mid-point rule gives an underestimate. As the number of strips increases the trapezium rule will tend to the solution from above and the mid-point rule will tend from below.

Error in the mid-point rule

In the mid-point rule the error is proportional to the width of the rectangle squared, or in mathematical terms, absolute error Mn= kh2, wherek is the constant. If the mid-point rule with n strips has a strip width of h, then the mid-point rule with 2n strips has a strip width ofh/2.

Hence:  image02.png

Therefore when you double the amount of strips (n) the error decreases by about a factor of 4, the error multiplier is therefore 0.25.

Error in the trapezium rule

The error connected to the trapezium rule is the same as the error in the mid-point rule.

...read more.



Using this formula we can now extrapolate our answers of Mⁿ to infinity.



Trapezium rule

Similarly you can use the formula for image13.png

 in the same way, just replacing M with T due to the error multiplier being the same.




Simpson’s rule

If you follow through the same steps as we did for the mid-point you similarly reach a formula for extrapolating Simpson’s rule to infinity:





So in conclusion I can confidently quote my solution to the integral image00.png

  to 8 significant figures:


This is due to the three approximations that I reached (17 sig figs):







There was however some limitation. The software, Microsoft Excel, only calculated to 15 decimal places inevitably producing minimal amounts of error with each calculation done. To further improve my results I could have used better mathematical software which worked to more decimal places, but for a problem like this Excel was more than sufficient.

The fact that the trapezium rule and the mid-point rule give the same solution up to 8 significant figures guarantees that my solution is valid to this many significant figures. This is because, as I have indicated in the paragraphs dealing with error analysis, in the integral section of my graph the trapezium rule tends towards the solution from above and the mid-point rule from below meaning the solution is between the two. Simpson’s rule, being a weighted average, is more accurate and therefore likewise proves the solution.

Overall then my solution is proved by 3 different rules.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics 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 AS and A Level Core & Pure Mathematics essays

  1. The open box problem

    X 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 V 60.5 100 121.5 128 122.5 108 87.5 64 40.5 20 5.5 0 Here is another table focusing in between 1.5 and 2.5 as the maximum volume clearly lies between those two values.

  2. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    The trapezium rule makes use of trapeziums instead of rectangles. Like the mid-point rule it is easier to describe by using a diagram: Figure 1.1 shows 4 trapezia, each one of length one (h), approximating the area under the curve, between 1 and 5.

  1. MEI numerical Methods

    D.P because of its display screen, however excel is accurate to 15 D.P. My calculation was done to 9 D.P. This is because I felt it 9D.P would be close to the real root and it would still be neat and easy to understand.

  2. Math Portfolio Type II - Applications of Sinusoidal Functions

    8. Using the data that you obtained for the number of daylight hours per day in both Toronto and Miami, complete the following table Toronto Miami Mean number of daylight hours 12.21h 12.15h Mean number of daylight hours for Toronto = 12.21h Mean number of daylight hours for Miami = 12.15h

  1. Experimentally calculating the wavelength of an He-Ne laser by means of diffraction gratings

    The following is the calculation of the relative uncertainty due to the use of the meter-stick in each calculation: 600 lines/mm 1st Fringe x = 2.080 � 0.002m, uncertainty = 0.002/2.080 x 100 � 0.096% L = 4.410 � 0.002m, uncertainty = 0.002/4.410 x 100 � 0.045% 0.096% + 0.045%

  2. Functions Coursework - A2 Maths

    It also implies that x<1.879395, confirming that x is not greater than or equal to 1.87940 to five decimal places. If x?1.879395, then x would be at least 1.87940 to five decimal places. Thus the root has been proven to be 1.87939 to five decimal places as a result of

  1. Methods of Advanced Mathematics (C3) Coursework.

    If the intervals were smaller then it would have worked. Newton Raphson Method In this method an estimate of the route is taken a line equal to x is taken up until it hits the curve and then a tangent is drawn down to the x-axis.

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    Microsoft excel can be set up in order to reduce human error and increase efficient of calculation. A spread sheet is set up for n a f(a) B f(b) 1 0 =B2^3-5*B2^2+4*B2+2 2 =D2^3-5*D2^2+4*D2+2 2 =IF(G2>0,F2,B2) =B3^3-5*B3^2+4*B3+2 =IF(G2>0,D2,F2) =D3^3-5*D3^2+4*D3+2 3 =IF(G3>0,F3,B3)

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