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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

image03.png

(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

image04.png

(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

image15.png

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

...read more.

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 image11.png. N= number of trapezoids

Hence….

image12.png

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:

image13.png

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:image14.png

One can factor out the image16.png and 2 in order to have a more concise way of finding the solution. One is then left with:

image17.png

To make the above formula easier to use one can simplify to:

image18.png

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

image19.png

image20.png

image21.png

1)

image19.png

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

-

...read more.

Conclusion

/tr>

2.75

12.469

24.938

3

13.5

-

image28.png

image23.png

image24.png

image29.png

A8=17.159

image21.png

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

-

image30.png

image23.png

image24.png

image32.png

A8=4.6875

image19.png

image33.png

image20.png

image34.png

image21.png

image35.png

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


image19.png

image37.png

image38.png

0.0007


image20.png

image39.png

image40.png

0.1335


image21.png

image41.png

image42.png

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:

image43.pngimage01.png

image44.png

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.

image46.pngimage02.png

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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