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# Numerical integration coursework

Extracts from this document...

Introduction

Numerical integration coursework

Problem

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=

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

Middle

Sn= 2(Mn) +Tn

3

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:

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:

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.

Conclusion

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

Trapezium rule

Similarly you can use the formula for

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:

Interpretation

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

to 8 significant figures:

10.193205

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

10.1932054791824

10.1932054077805

10.1932054458615

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.

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

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