Ex. 5
As seen above I allowed:
b2= 3.25
b1= 3
H = .5
Once you substitute these numbers into the equation simple arithmetic will allow one to come to an answer
Thus, the area of the first trapezoid would be 1.5625.
Ex. 6
Above I allowed:
b2= 4
b1= 3.25
H = .5
After substituting the variables for values, the equation should look like:
Thus, the Area of second trapezoid would be 1.81.
Since the area under the curve is equal to Atrap1+Atrap2: 1.5625+ 1.81
Then…. 3.37
Based on two trapezoids and on the interval x=0 and x=1.
This approximation is the same as the approximation generated by the computer so one can assume this is correct but if one looks closely at (Ex 3) they will see that this approximation is only a rough estimate and in order to find more accurate results one must add more trapezoids.
(Ex 7) shows the same function with trapezoids increased from 2 to 5. This also decreased the x intervals from .5 to .2.
Ex. 7
Ex. 8 Ex. 9
Ex. 9
Ex. 10 Ex. 11
Ex. 11
Ex. 12
Ex (8-12) illustrates each trapezoid from left to right and in order to find each area I have created a table in order to simplify the process- (Ex 13)
Ex. 13
The order is from left to right starting with the first and ending with the fifth.
Adding all the areas….
0.604+0.620+0.652+0.700+0.764= Atrap= 3.34.
One can assume that this is a more accurate approximation for the area under the curve because with each additional trapezoid one becomes closer to the actual value.
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 f0 and 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 f0 and 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 fa and 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)
)
A8=1.98
2)
A8=17.159
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
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.