• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12

Numerical integration can be described as set of algorithms for calculating the numerical value of a definite integral. Definite integrals arise in many different areas and calculus is a tool

Extracts from this document...




Numerical integration can be described as set of algorithms for calculating the numerical value of a definite integral. Definite integrals arise in many different areas and calculus is a tool for evaluating them; with numerous applications in science and engineering as well as mathematical analysis. However, calculus cannot always be applied; there are functions which do not have antiderivaties. One such example is

image32.png;this is an important function since it will be used in this coursework.

Using my knowledge of numerical integration I shall produce an approximation to the function image33.png ,seeing as it cannot be integrated. The area will be evaluated between the values of 0 and 1 (radians). The graph below gives a visual representation of the area which I shall be calculating.


It is often complicated to find the analytic solution to many differential equations. However, to our benefit there are many methods for finding the approximate solutions to differential equations. These methods are referred to as polynomials: the mid-point rule, trapezium rule and Simpson’s rule. Before explaining the methods in detail, we should note that all of these methods presented do not produce exact solutions, only approximate ones.  

Lissaman R. (2004), suggests that the midpoint rule uses rectangles to approximate the area underneath a curve. Below is a diagram which makes use of the mid-point rule:image41.png

In figure 1.0 five rectangles, each with the same width, are used to approximate the area under the graph of a function f(x) between x = 0 and x = 1.

The widths of the rectangles are donated h. The height of the first rectangle in the example is at the mid-point of the interval 0 to 0.2; which is 0.1 (represented by dashed line in figure 1.0).

...read more.




On the other hand, the trapezium rule is an underestimate and each gap is a quarter to the one on its left. Likewise, the trapezium rule is also a second order polynomial and only underestimates when the curve is convex.image11.pngimage10.pngimage09.pngimage12.pngimage06.pngimage06.png


Notice, that the mid-point rule is a mirror image of the trapezium rule and vice versa. This is because the mid-point overestimates whilst the trapezium rule underestimates. Therefore, by using both methods we can confine the exact value between the two different polynomials. This means by obtaining values of Tn and Mn where n is a large number, we can find an accurate approximation to the integral image32.png as we know the exact value is between that of Tn and Mn.


Most of this coursework will be carried out using Microsoft Excel; an electronic spreadsheet program used for organizing and manipulating data. The program is capable of working accurately up to 16 decimals places; this is also the amount I have chosen to work with in my calculations. Due to its manipulative ability Excel makes the handling of data very easy and saves an enormous amount of time. This is due to the fact that you don’t need to calculate everything; once the formula is entered and two consecutive calculations are complete, the cells can be dragged down and the answers required appear.

Excel is able to do this as it follows the formula and is judicious. The reason I chosen to use a program instead of a calculator is due to its ability to use 16 decimal places whereas a calculator can only give answers accurate to 9 decimal places.

...read more.


n+Tn)/3. The average is twice as close to Mn as it is to Tn. This can be justified by the differences in the errors associated with the two polynomials. This is explained below:

Estimate the area of I=image38.pngimage39.pngusing the trapezium rule and mid-point rule, only using 1 strip.

Trapezium rule:         Mid-point Rule:

 h= (4-0)/1= 4           h= (4-0)/1= 4











T1 = ½ * 4 * (f(0)+f(4)) = 32        M1= 4* (f(2)) = 16

The area of image39.pngcan be worked out from integration and is exactly 21 1/3. Therefore the absolute error in T1 is (32-21 1/3) = 10 2/3, whereas that in M1 is (21 1/3- 16) = 5 1/3. From this we can see that the absolute error in T1 is twice as much as that in M1.


You can see from inspecting the errors in the trapezium sum and the midpoint sum that the midpoint sum is about twice as accurate as the trapezoidal sum and opposite in sign. This explains the weighting of the formula (2Mn+Tn)/3. By using the Simpson’s rule we will have cancellations of the errors and should thus get a much more accurate approximation.

The advantage of using the Simpson’s rule is that it’s a fourth- order polynomial. As mentioned before Simpson’s fits a parabola between successive triples of pairs. The absolute error is proportional to h4 so it is able to achieve a more accurate approximation to the curve,image42.png.

Absolute error ∝ h4 image37.png Absolute error = kh4

As a result the absolute error is Sn= kh4 whereas the absolute error in S2n= k(h/2)4. This means the errors reduces by a scale factor of 1/16 between each successive Sn values. In theory, the results I have obtained for Sn should be more accurate and reach the exact value in fewer calculations. This is seen to be the case by looking at the spreadsheet attached to this coursework.


...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. MEI numerical Methods

    This creates the secant method, this requires two approximations, x(0) and x(1). Secant method: Evaluating this formula means, the first approximation x0 multiplied by the function of the second approximation x1, minus the function of the first approximation x0 multiplied by the second approximation x1.

  2. Math Portfolio Type II - Applications of Sinusoidal Functions

    January 1 1 10.55h February 1 32 10.98h March 1 60 11.62h April 1 91 12.44h May 1 121 13.13h June 1 152 13.65h July 1 182 13.74h August 1 213 13.35h September 1 244 12.67h October 1 274 11.92h November 1 305 11.16h December 1 335 10.64h The times

  1. The open box problem

    X 0.5 0.6 0.7 0.8 0.8 1.0 1.1 1.2 1.3 1.4 1.5 V 18 20.064 21.672 22.848 23.616 24 24.024 23.712 23.088 22.176 21 I will now use a graph as well to show the volume and value of x and see if my hypothesis works.

  2. Investigation of circumference ratio - finding the value of pi.

    Known: Segment AC=1 Segment BC=1 Angle ACD=7.5 Area of triangle ACD: Area of : 6-sided: 12-sided: 24-sided: 48-sided: Then we could make a formula for calculate the area of an n-sided equilateral polygon, which is inscribed in a circle of a radius r.

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

    In this case this method will fail to find the target root we want. If we want to find the target root then we have to set up upper and lower bound very close to the target root in order to find the target root.

  2. Functions. Mappings transform one set of numbers into another set of numbers. We could ...

    have to factorise first ==> If we have fractions in the numerator or denominator, multiply to remove the fraction Multiplying and Dividing ==> Multiply the numerators and multiply the denominators. ==> Cancel where necessary ==> Where one fraction is divided by another, invert/flip the second fraction and multiply.

  1. C3 Coursework: Numerical Methods

    This is illustrated by the graph below. This graph shows the equation of the line y=x�+10x�+4.8x+0.576 The roots appear to be between -1 and 0. By taking increments of 0.1 between -1 and 0 it will be possible to use decimal search to attempt to look for a change in sign.

  2. Numerical integration coursework

    You then multiply this by the height, which you can get by doing the function of the mid-point of each rectangle, and finally add up the area found of them. Hence you get this formula: Mn= h (f(mâ)+f(mâ)+…+f(mn)) I used this formula and many algorithms when working on Excel to

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