• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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

/tr>

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.

## 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

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

ï¿½Dï¿½ï¿½\$ï¿½Ef ï¿½ï¿½3ï¿½ï¿½a8Kyï¿½ï¿½ï¿½ï¿½ï¿½r... ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ËÇ³ï¿½ï¿½ï¿½_PCï¿½"hï¿½Ø"ï¿½dï¿½4ï¿½ï¿½ï¿½ï¿½Uï¿½ Mï¿½Xï¿½ï¿½ï¿½*oTï¿½ï¿½i484%ï¿½Ô´ï¿½tltï¿½ï¿½ï¿½ r Kï¿½ï¿½"Lï¿½(tm)>7ï¿½7_ï¿½D[ï¿½[sÙï¿½'*Õ°3ï¿½7wï¿½qï¿½wrrvqquusswwï¿½ï¿½ï¿½ï¿½ï¿½ï¿½3/4ï¿½ï¿½ï¿½ï¿½Ûjï¿½cï¿½"ï¿½ï¿½/ï¿½ï¿½" ï¿½ï¿½ï¿½&ï¿½qï¿½ï¿½V^ï¿½ï¿½ï¿½:ï¿½Cï¿½Õcï¿½ï¿½ï¿½-Pï¿½(tm)d',ï¿½1/4(tm)ï¿½4ï¿½eï¿½4ï¿½tï¿½ Zï¿½n]ï¿½*ï¿½ Ú§ï¿½Ø3ï¿½ï¿½'ï¿½ï¿½@ï¿½Aï¿½<-1/4ï¿½|ï¿½aï¿½#ï¿½GÕ]93/4Yï¿½[,yBï¿½D(r)T3/4Lï¿½\(c)Bï¿½Rï¿½ï¿½Jï¿½ï¿½)"ï¿½gï¿½ï¿½Nï¿½0ï¿½ï¿½ï¿½8ï¿½Rï¿½ï¿½p1/4ï¿½ï¿½¹ï¿½-ï¿½Îï¿½Zï¿½.}ï¿½ï¿½5ï¿½ï¿½hï¿½x{XGYg[ï¿½ï¿½ï¿½sï¿½[=,7ï¿½{ï¿½ï¿½ ï¿½ï¿½ ï¿½+o?ï¿½ï¿½fhï¿½-ï¿½3/4ï¿½ï¿½ï¿½'Øaï¿½Nï¿½Tï¿½xÆï¿½ï¿½gï¿½<ï¿½ï¿½ï¿½ï¿½kï¿½ï¿½ï¿½ï¿½ï¿½}S=/Z^-ï¿½ï¿½`ï¿½}zï¿½ï¿½ï¿½ï¿½ï¿½(tm)ï¿½wï¿½ï¿½ï¿½ï¿½ï¿½Kï¿½_ï¿½`ï¿½Qq'ï¿½ï¿½ï¿½ï¿½ï¿½Ïµï¿½....,Û¬(}ï¿½*3/4ï¿½ï¿½"ï¿½zï¿½ï¿½?ï¿½7]ï¿½jï¿½ï¿½wï¿½ï¿½''%Úï¿½ï¿½Cï¿½P\$ï¿½ sï¿½ï¿½ï¿½ * ï¿½}Ïï¿½ï¿½ï¿½dp_ï¿½"*bÉÆï¿½@ï¿½Jï¿½CÏO(c)geï¿½ï¿½"ï¿½ï¿½ï¿½Vï¿½ï¿½"ï¿½eï¿½}ï¿½g"ï¿½ï¿½o'p\pTï¿½\$b,ï¿½ V'ï¿½Hï¿½"1/2ï¿½ï¿½ï¿½'ï¿½ï¿½ï¿½ï¿½1/4ï¿½ï¿½ï¿½"ï¿½1/22ï¿½ï¿½Ueï¿½VCï¿½ ï¿½Zï¿½ï¿½(c):!ï¿½"ï¿½lï¿½ T eï¿½"ï¿½9Lï¿½Mï¿½ï¿½fKï¿½"-ï¿½V=ï¿½-lï¿½wï¿½ï¿½ï¿½ï¿½ï¿½Sluï¿½\$ï¿½ï¿½\0.ï¿½(r)/ÜÜ¯ï¿½>ï¿½'ï¿½"ï¿½ï¿½ï¿½ï¿½[ï¿½ï¿½Hï¿½ï¿½ï¿½Ò·Ï¯ï¿½`@h uï¿½l0Cï¿½-ï¿½'ï¿½maeï¿½ï¿½(r)'*Q"ï¿½...ï¿½1Eï¿½~qï¿½ï¿½"sï¿½ï¿½Iï¿½ï¿½Lï¿½3)WR3ï¿½Ú¤qï¿½}Hï¿½ï¿½8''eï¿½m"Xï¿½ï¿½~ï¿½\ï¿½' ...ï¿½xï¿½ï¿½S "ï¿½ ï¿½l|;ï¿½|lï¿½z(r)ï¿½ï¿½"Xï¿½j(c)Q(tm)mï¿½ï¿½Ê"iU Nï¿½>yï¿½Ùê±ï¿½sï¿½u-ï¿½ï¿½ï¿½[ï¿½7n5ï¿½7_ j9r(c)ï¿½ï¿½ï¿½"ï¿½ï¿½ï¿½>j'whwwï¿½_ï¿½ï¿½1/2Ý£|3ï¿½ï¿½ï¿½ï¿½ï¿½-ï¿½1/2ï¿½ï¿½ï¿½wï¿½ï¿½ï¿½.ï¿½'+>pï¿½~ï¿½4ï¿½t ;(r)ï¿½ï¿½ï¿½Iï¿½Dï¿½Óï¿½ωSï¿½/_&ï¿½:ï¿½ï¿½ï¿½ï¿½ï¿½[ï¿½(tm)ï¿½wWgWï¿½ï¿½j>1/4^ï¿½ï¿½ï¿½_Kï¿½ _ï¿½W)kï¿½uï¿½ï¿½Ç×·ï¿½ï¿½ï¿½Xï¿½0Cjï¿½Rpï¿½Bï¿½ï¿½qhÖÏ¢È¨ï¿½hï¿½#ï¿½ï¿½ï¿½ï¿½(ï¿½"ï¿½ï¿½"/qï¿½ï¿½|ï¿½ï¿½ï¿½(r)ï¿½ï¿½(C!ï¿½ï¿½ï¿½Sï¿½ï¿½+ï¿½uv ï¿½4ï¿½ ï¿½{/O)ï¿½E3/4~ï¿½ï¿½ï¿½ï¿½ï¿½"V"-ï¿½ï¿½iRï¿½_ï¿½ï¿½ï¿½>Iw)YiHzLï¿½V6Qï¿½Bï¿½O~MaDï¿½N)]ï¿½IEJVï¿½TkVï¿½ï¿½pï¿½"ï¿½ï¿½ï¿½ï¿½(r)ï¿½IÔµï¿½ï¿½ï¿½[ï¿½68gï¿½nï¿½l,cï¿½1yazï¿½ì¹*...ï¿½%ï¿½rÚªï¿½:ï¿½ï¿½ï¿½ï¿½gvMï¿½(tm) ï¿½Âßï¿½ï¿½+\B]5ï¿½HnSï¿½vï¿½xï¿½yï¿½x3/4ï¿½sï¿½+ï¿½ï¿½'ï¿½Jï¿½ï¿½ï¿½ï¿½gï¿½ï¿½ï¿½?Pï¿½ï¿½ - ) ï¿½ [ ï¿½È4ï¿½"D Gï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½eï¿½ï¿½ï¿½o\$ï¿½'ï¿½&ï¿½\$-\$×¦xï¿½rï¿½>ï¿½[ï¿½f- ï¿½ï¿½dï¿½gï¿½eqgmdOï¿½Ê¹1/4ï¿½27ï¿½@ï¿½Aï¿½<ï¿½C"ï¿½ï¿½ï¿½ï¿½/ï¿½ï¿½*h<Z|,ï¿½xbaLQT1'"D-F-Eï¿½GTWzï¿½tï¿½ï¿½8euï¿½ï¿½Lï¿½Ùªï¿½ï¿½ï¿½ï¿½Tï¿½[ï¿½ï¿½7ï¿½5ï¿½3/4ï¿½ï¿½tï¿½yï¿½ï¿½ï¿½Rï¿½ï¿½ï¿½W%ï¿½%ï¿½ï¿½ï¿½"::Ëºï¿½ï¿½ï¿½o"ï¿½ï¿½ï¿½eï¿½ ï¿½5ï¿½9w{dHï¿½nï¿½1/2ï¿½ï¿½ï¿½GfFï¿½Mï¿½ï¿½>^ï¿½88ï¿½ï¿½yJï¿½ï¿½ï¿½"ï¿½iï¿½ ï¿½1/2_[(^\v\{ï¿½ï¿½ï¿½ï¿½-ï¿½4ï¿½ ï¿½Lp.ï¿½Aï¿½ï¿½#ï¿½C [2ï¿½ï¿½f/ Ð(yï¿½ï¿½?xï¿½>;ï¿½ï¿½Z0ï¿½x,ï¿½?Ì¡Pï¿½Ôï¿½z_aVXï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Åï¿½Gï¿½ï¿½ï¿½:Poï¿½\$ï¿½ ï¿½ï¿½>ï¿½ï¿½Dï¿½aï¿½1Æxï¿½ï¿½ï¿½ï¿½`ï¿½ï¿½ï¿½Øï¿½yï¿½ï¿½ w ï¿½ /ï¿½ï¿½_ï¿½oï¿½gßï¿½ï¿½f(tm)Aï¿½Qï¿½9K&'ï¿½3ï¿½vï¿½}t tï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½)0%ï¿½Øï¿½Dï¿½Tï¿½,ï¿½|ï¿½Åe'5' ï¿½Vï¿½nï¿½ï¿½ï¿½ï¿½4ï¿½(r)

2. ## Shady Areas. In this investigation you will attempt to find a rule to approximate ...

[0,1] Sum 2 3.375 3.375 5 3.34 3.34 10 3.33 3.336 20 3.33 3.33375 First I calculated the area manually by hand. Then I tested these results with the use of technology, which in this case was the Riemann Sum Application on a TI-84+, which I used to create the table above.

1. ## Math portfolio: Modeling a functional building The task is to design a roof ...

So the width of the base of the rectangular building = 150m And the length of the rectangular building = 72m Here height of the structure varies from 50%of 150m to 75% of 150mwhich is from 75m to 112.5m

2. ## Math Portfolio: trigonometry investigation (circle trig)

160 0.34202014 -0.93969 -0.3639702 -0.36397023 180 1.2251E-16 -1 -1.225E-16 -1.2251E-16 200 -0.3420201 -0.93969 0.36397023 0.363970234 220 -0.6427876 -0.76604 0.83909963 0.839099631 240 -0.8660254 -0.5 1.73205081 1.732050808 260 -0.9848078 -0.17365 5.67128182 5.67128182 270 -1 -1.8E-16 undefined 5.44152E+15 280 -0.9848078 0.173648 -5.6712818 -5.67128182 300 -0.8660254 0.5 -1.7320508 -1.73205081 320 -0.6427876 0.766044 -0.8390996 -0.83909963

1. ## Math Portfolio Type II

Part 6 To initiate an annual harvest of 5000 fishes, let us first find out at what point the fish population stabilizes when the initial growth rate, r = 1.5. The following table shows the growth in the population over the first 20 years taking r = 1.5 and thus

2. ## Stellar Numbers math portfolio

quadratic equation, the only values that now need to be solved for are the b-values and c-values. Therefore the equation as it stands is f= 0.5n2+bn+c. Now, by substituting in 2 ordered pairs in the equation, two equations are created with only b and c as the unsolved variable in each.

1. ## Artificial Intelligence &amp;amp; Math

This is a great benefit as the automated system saves police's time, money, and human resources as the computer does the majority of the police's work for them and allows police to do other work that requires human influence. Explained and analysed.

2. ## The investigation given asks for the attempt in finding a rule which allows us ...

Based on Wolfram Mathematica online integrator, the integration formula is x223=3x535×223. to find the accurate value, x =3 minus x = 1 to get the approximate area. Hence, 33535×223-31535×223 2.358667255-0.377976315=1.98069094≈1.981 The answer from the general statement is almost identical to the accurate measured value. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 