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I would be finding the approximation of the integral on the graph between the shaded portion of the graph above which is from 0 to 1 using numerical integration. This problem was chosen because it cannot be integrated by any analytical method, therefore approximation method would be used. Due to this I suggest that this problem will be appropriate for numerical solution. Strategy To solve this problem, I am going to use knowledge of numerical integration studied in the Numerical Methods textbook. Numerical integration is a method used to approximate an area under the graph.

According to syllabus on NM module, the approximate methods of definite integrals may be determined by numerical integration using; 1. Mid – point rule 2. Trapezium rule 3. Simpson’s rule Since there are lots mathematical functions which can not be integrated in real life, an alternative approach to these problems are to sub-divide the area under the graph into strips or shapes such as rectangles which approximately covers the area. I intend to use the trapezium and Mid-point rule as these are on of the basis in which the approximation can be found.

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Note: X is in radians. 1. 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. 2. 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 hi??. Absolute error = Khi??

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= Khi?? Then, Absolute error in M= ki?? = 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 hi??. It is a second order method.

Also trapezium rule is similar to the mid- point rule. Error is also proportional to hi??. 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 hi??. 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.

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