 Level: International Baccalaureate
 Subject: Maths
 Word count: 1171
Math Portfolio Shay Areas
Extracts from this document...
Introduction
Dickerson 1
In order for one to understand, the process and subsequent work one must understand the purpose of this assignment, which is, to find a rule to approximate the area under the curve. Considering the following: g(x) = x2+3 (Ex 1)
Ex. 1
(Ex 2) allows one to see what area we are trying to find. The area is represented is represented by the green diagonals. This example is still from intervals x=0 to x=1.
Ex. 2
(Ex 3) shows us the estimation of the area by adding two trapezoids underneath the curve. One can observe that the area is not exact but still a good approximation. Using technology, I have found that the sums of the areas of both trapezoids are equal to 3.375. The next step will show one how to calculate this without the use of technology.
Ex. 3
To calculate the area one must know the formula for area of a trapezoid. The area of a trap is equal to the height (H) multiplied by the first base(b1)added to the second base(b2) divided by 2
Middle
3.37
T4
3.34375
T5
3.34
As a result of creating a table, one can see a pattern more clearly. One learns know that the height will always depend on the interval and number of trapezoids being drawn. Therefore, the height equals . N= number of trapezoids
Hence….
Then I also noticed that b1 and b2 for the first trapezoid would always be f0and f∆x. Then that each successive trapezoid will contain the b2 of the previous triangle for its b1, and then the b2 will be equal to f2∆x; adding one to the constant in front of the ∆x with each triangle. Based on the discovery the formula for the first trapezoid would be:
Also, b1 and b2 are equal to f0and f∆x and the next trapezoids b2 will be that of the previous trapezoids b1. b2 is also equal to f∆x.. This changes the formula:
This only applies to the area of one trapezoid; One needs the sum of the total trapezoids under the curve in order to find a accurate answer.
In order to simplify this:
One can factor out the and 2 in order to have a more concise way of finding the solution. One is then left with:
To make the above formula easier to use one can simplify to:
Because one will use the base multiple times excluding the first and last b1... Which are faand fb respectively. Just as done in u substitution the 2 can divided under either one and the answer will stay the same. Either general statement is correct but the first requires more information and be time consuming with a many trapeziums. While the second is easier to use but it could be somewhat less accurate.
I have considered the following using my general statement on the interval form x=1 to x=3
1)
X  y  2g 
1  0.629960525   
1.25  0.731004435  1.462009 
1.5  0.825481812  1.650964 
1.75  0.914826428  1.829653 
2  1  2 
2.25  1.081687178  2.163374 
2.5  1.160397208  2.320794 
2.75  1.236521861  2.473044 
3  1.310370697   
Conclusion
2.75
12.469
24.938
3
13.5

A8=17.159
x  y  2y 
1  3   
1.25  3.875  7.75 
1.5  3.75  7.5 
1.75  3  6 
2  2  4 
2.25  1.125  2.25 
2.5  0.75  1.5 
2.75  1.25  2.5 
3  3   
A8=4.6875
I used my calculator to compare the trapeziums rule to the actual area. I pressed Y on my calculator and placed the function in my equation then graphed it. Next I pressed second function trace, entered the lower limit as one and the upper as 3, the calculator then calculated the area under the curve
Area  
Y=  Trapezium  Integration  Difference 
0.0007  
0.1335  
0.0205 
The trapezium rule was very accurate in determining the area and one could find a closer approximation by adding more trapeziums under the curve.
Scope and Limitations:
Sin curves are the only exception I found to the rule not working. The fluctuation between axes is too difficult for the rule to calculate. The graphs go from positive to negative and they areas negate each other.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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