Note: X is in radians.
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Mid – point rule: The mid point rule was adopted, because it is used to approximate the area underneath the graph using rectangles. Below is the formula which is involved in the calculation.
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Trapezium rule; This is also similar to the mid – point rule. It was adopted, because it would help me to approximate the region under the graph using strips of trapezium and calculating their area. Below is the formula which is involved in the calculation.
Formula application
This was done on a spreadsheet on Microsoft excel. Various algorithms were used to get approximations. The copy of the spreadsheet where the formula was also displayed was also shown. (See page 6 and 7)
Use of technology
Microsoft excel was used to devise my spreadsheets. Excel can only handle decimals accurately up to 15 decimal places. My calculation was made to 15 decimals places. Excel was used because it made my work easier, this because you don’t need to calculate everything, once the formula is given and some calculations are done, you just need to drag the cell down and the answers emerges. This because excel is astute. Also it was used because it calculated to more decimal places than a calculator. I also used the autograph version 3 to draw the graph s as this would make my graph neat and easier to understand.
Error analysis
I suggest that there would be any error multiplier of about 0.25. Also there would be an error of about 3% due to the limitation which might have occurred.
According to my Numerical methods text book, I realised that in the mid – point rule, error is proportional to h².
Absolute error = Kh²
Therefore, if two successive mid – point estimates to an integral are taken, the first using n and the second 2n strips. Where h is the strip width which corresponds to n strips. Therefore when using n strips, the strip width would be h, then the midpoint rule of 2n strips would have a strip width of , then
Absolute error in M= Kh²
Then,
Absolute error in M= k² =
This means that halving h, or equivalently doubling n will reduce the error by a factor of = 0.25. Since the absolute error is proportional to h². It is a second order method.
Also trapezium rule is similar to the mid- point rule. Error is also proportional to h². Therefore this also means halving h, or equivalently doubling n will reduce the error by a factor of = 0.25. Since the absolute error is proportional to h². It is a second order method.
Viewing error in terms of differences and ratio differences
When the values for the number of strips double, the ratio difference between successive estimates is the same. This assumption was made according to the theory above. The table below gives an example to justify the allegation.
This shows that doubling the number of stripes is four times as close to the exact value.
T T T T
For mid-point rule
Interpretation
The approximation to the area under the graph using the mid – point rule was found to be 0.779919455488404 (15s.f), while the approximation for the trapezium rule was found to be 0.779923113159401 (15s.f). The solutions of the two result were the same after (5s.f) and the solution was found to be 0.77992.
Overall I think my result was about 95% valid. This occurred due to some limitation which might have occurred. The software program I used which was Microsoft Excel and it only calculates accurately up to 15 decimal places. Starting from “M16”, I started extrapolating to get better estimates using ratio differences. Here I assumed that the ratio difference would continue to be 0.25, according to the theory, so I suggest error might have occurred because some values of the ratio of differences may not be 0.25 precisely.
The problem specified. can also be written as ³. Microsoft Excel might find it complicated to solve square roots. So I suggest error might have occurred, therefore affecting the validity of my result.
For example;
)² -
If I had more time I could have increased the accuracy of my result by finding the real answer and not using extrapolated values to find “M16”, “T32” ,“M32”and “T64”, as this could have improved the validity of my result and therefore making my answer to the solution more accurate by producing the solution to a higher degree of significant figures. Also to improve the validity of my result Simpson’s rule could have been used. This is because in Simpson’s rule error is proportional to , which means it is a fourth order method and also when you halve the width of the strip or you equivalently doubled the number of strips, this would reduced the absolute error with a scale factor of 0.0625.