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Revision:OCR Core 2 - The trapezium ruleTSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 2 - The trapezium rule
10. The trapezium ruleThe needs for approximationThere are some functions that cannot be integrated exactly (and some which one will not be able to integrate due to the level of knowledge that one has upon the topic of integration). Therefore it is useful to be able to give an approximation, and say whether it is above or below what one might expect the real value to be. (In later study one might be able to predict the accuracy of these estimates). The trapezium rule: simple formConsider that there is a function of x: If one was to plot the graph The idea of one trapezium is simple. Take the two bounds as the two vertical edges of the trapezium, and then take a line between them as the sloping line at the top. The x-axis is the width, and base line. One can see that this is a rather crude approximation for larger areas, but consider the idea of multiple trapezia, this is much more accurate. The simple form of the trapezium rule refers to the single trapezium, and can be shown:
(The area of a trapezium is the sum of the parallel sides multiplied by the width, multiplied by a half). This gives the most basic approximation, and can be improved (in most cases). The trapezium rule: general formOne can consider that as one allows the number of trapezia to increase, and therefore their widths to decrease, one will have a better approximation of the value of the integral. One thing that one should note is that good setting-out of work is crucial, as sometimes one might be expected to deal with 8 trapezia, and therefore the working can become incomprehensible if it is not properly organised. Consider now that between the bounds, a, and b, one has an indefinite number of trapezia, denoted:
Where:
Now one can begin the formal representation: The one trapezium example is:
Hence, one can use sigma notation to show the total area:
Hence:
Example 1. Use the trapezium rule with 4 intervals to calculate an estimate for the definite integral:
and give the percentage accuracy. This is a simple matter of careful calculation. The coordinates on the x-axis will be: (0, 0); (1, 0); (2, 0); (3, 0); (4, 0). Now the coordinates on the curve are: (0, 0); (1, 1); (2, 4); (3, 9); (4, 16). Hence the areas:
Hence:
Now:
Hence the accuracy is:
Accuracy of the trapezium ruleOne cannot (at this stage) give, in all cases, a good approximation of the accuracy, however, one will be expected to be able to tell whether the estimate is an overestimate, or an underestimate. It is quite simple to detect these things. First one should draw a sketch graph, and then connect lines in the same manner as one would was one drawing out the trapezia. If the tops of the trapezia are above the curve, there is an overestimate, and if the trapezia are below the curve, it is an underestimate. The determining factor is whether the curve is having a negative second derivative or not. Consider that a negative second derivative will mean that the curve will be having less gradient at any given interval in the positive x-direction, and therefore the trapezia will be underneath the curve. If the curve bends upwards, the trapezia will extrude, and thus give an overestimate. The easiest way to predict these things is to draw an accurate sketch graph, or study that graph one is provided with.
Also SeeRead these other OCR Core 2 notes:
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![\frac{1}{2} (x_{1} - x_{0}) \left[ f(x_{1}) + f(x_{0}) \right]](http://www.thestudentroom.co.uk/latexrender/pictures/ef/efaf9a2b8a297b241c4ee78c0d505f81.png "\frac{1}{2} (x_{1} - x_{0}) \left[ f(x_{1}) + f(x_{0}) \right]")
![\displaystyle \sum _{n = 1}^{n} \left{ \frac{1}{2} (x_{n} - x_{n - 1}) \left[ f(x_{n}) + f(x_{n - 1}) \right] \right}](http://www.thestudentroom.co.uk/latexrender/pictures/e4/e48ec301265293c1f8f899a659ce7772.png "\displaystyle \sum _{n = 1}^{n} \left{ \frac{1}{2} (x_{n} - x_{n - 1}) \left[ f(x_{n}) + f(x_{n - 1}) \right] \right}")
![\displaystyle \int _{a}^{b} f(x) \ dx \approx \sum _{n = 1}^{n} \left{ \frac{1}{2} (x_{n} - x_{n - 1}) \left[ f(x_{n}) + f(x_{n - 1}) \right] \right}](http://www.thestudentroom.co.uk/latexrender/pictures/cf/cf8328cb9a4652e0bbd1f5e24ca27c0e.png "\displaystyle \int _{a}^{b} f(x) \ dx \approx \sum _{n = 1}^{n} \left{ \frac{1}{2} (x_{n} - x_{n - 1}) \left[ f(x_{n}) + f(x_{n - 1}) \right] \right}")
![\displaystyle \int _{0}^{4} \left( x^{2} \right) \ dx = \left[ \frac{1}{3} x^{3} \right] _{0}^{4} = 21 \frac{1}{3} - 0 = 21 \frac{1}{3}](http://www.thestudentroom.co.uk/latexrender/pictures/d6/d6b0d7e333467a6ebb9cb127c726f19e.png "\displaystyle \int _{0}^{4} \left( x^{2} \right) \ dx = \left[ \frac{1}{3} x^{3} \right] _{0}^{4} = 21 \frac{1}{3} - 0 = 21 \frac{1}{3}")
