• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16

Extracts from this document...

Introduction

Standard Level

Name Anis Mebarek

Candidate Number:

We are able to find the area under the graph through calculating the area of the trapeziums under it. Through having numerous trapeziums under the curve the amount of uncertainty within the answer decreases this is due to the uncalculated area decreasing. I will be investigating and proving this theory from using the function  and to find out the area of the trapezium I will use this formula .

As seen from the graph the area between the curve and the trapezium is also being calculated when trying to find the area under the line  therefore we are expected to get a higher number then the area it’s self. But increasing the number of trapeziums on the graph from 1 to 2, the uncertainty for area decreases.  This can be seen from the graph below.

Therefore through increasing the number of trapeziums within the graph the uncertainty of the answer will decreases and so will the number for the area will decrease, as the trapeziums will over-estimate the number for the area. To prove this I will have to calculate the actual area under the curve through using integration.

Middle

T5

3.34

I have noticed there is a relationship between the number of trapeziums within the graph and the areas that they produce. The table below will show the differences in the areas that each number of trapeziums produces.

 Number of trapeziums Total area Total difference T1 3.5 0 T2 3.37 0.125 T4 3.34375 0.03125 T5 3.34 0.00375

When we take the total difference and square it, then add the two of the same squared differences.

Therefore by finding the potential “total difference” and have the total area be subtracted from the found total difference, we will be able to find the next total area.

Part 2

So the very basic idea behind finding the area underneath the line is through finding to how many trapeziums to separate the X axis into, which is deciding upon how much height each section will have. Then finding the trapeziums area through doing the . Therefore for example when we are trying to find the area with 5 trapeziums the formula will be:

But when we have this equation in terms of x we get:

Therefore it’s basically just plugging in the numbers into the equation above, for example for  you will have to find that length therefore you plug the 0.2 into the given equation, for this example it’s,

when we find the number of that X co-ordinate we can then plug it into the equation, and repeat the process with the other lengths. But his is not a general formula, where I will have to change the  to  and the different x axis sides, where the  will be used, which will give me this formula:

To simplify this equation I will simplify one part of it as it correlates towards the rest, as to use that part as an example:

Therefore we are able to factorize this equation as ½ and h as repeated throughout the equation which will give us this:

This is still not a general explanation to the formula, as its still focused on one example, therefore to change this example so that it can be used with any number of trapeziums, we have to change the height as it has to correlate with the length of the x-axis and the number of times that x-axis will be split. Simply we are able to change h to  where “b-a” will give us that length of the x-axis where they want to measure the area from, essentially the upper and the lower limits of their search for the area. “n” in that equation is the number of trapeziums that we want to use. Therefore together, “b-a” will be split by the number of trapeziums someone wants to use, therefore giving us height for the trapezium.

We can further simplify this equation into:

We are able to do this because, the numbers repeat twice throughout the equation apart from and  though by doing so, we are just making it simpler to use, but to make it more general we will have to replace  with therefore transforming the equation into this:

“…” in the formula is used as an uncertainty to how many trapeziums the user will have on his graph, therefore they have to continue depending adding until they reach the third last trapezium. Until reaching “n-1” where n is the number of trapeziums, this will allow the equation to substitute the uncertainty of the number of trapeziums within the equation. Though we are able to further simplify this equation to:

As is was simply

Therefore the most general formula is:

Consider the areas under the following three curves, from x=1 to x=3:

 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

, selected the lower limit to be 1, and the upper limit to be 3, then pressed enter to find the area between the limits.
 Area Y= Trapezium Integration Difference 0.0007 0.1335 0.0205

The trapezium rule came to a close proximity with its approximations, with only small differences between the actual area, and the estimation. Though increasing the number of trapeziums used within the equation would increases the accuracy of the answers.

The scope and limitations:

The way that the rule will be limited or even not work is through the sin curves, as they fluctuate between the x-axis, the areas calculated will be difficult to do, as the sides of the graph go from a positive number, to a negative, therefore canceling each other out.

Another limitation to the rule is when the graph is irregular, and then the uncertainty or the over/underestimations from the rule will increase as the irregular curves create difficulty in estimating the curving parts of the graphs. Even through increasing the number of trapeziums within the graph the uncertainty is still great.

Another limitation is an isotope, where the line does not touch the axis, therefore makes it difficult to calculate the lengths for the trapeziums. Which makes it impossible to do, but if it’s done then the uncertainty for the area found will be high.

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

= x2 + 3] ?ca f(x) dx --> ?10 x2 + 3 dx Where: c is equal to the end x value of the domain a is equal

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

is the height of the cuboid "h" Substituting (36-V) into equation (12) Y(36-v)=(36-v)2 + (36-v) = [(36-v)2-72(36-v)] = [(36-w)2- 2592+72v] = [1296-72v+v2-2592+72v] h =[1296 - v2] ------(13) Volume=width � height � length =2v �[(1296 - v2)] �150 = 150[(1296v - v3)] To get "v" for the maximum volume of the cuboid, the above equation should be differentiated with regard

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

= = =1 The values of sin? and cos? within the range of 0�???90� are analyzed in order to conjecture another relationship between sin? and cos? for any angle ?. A portion of Table 1 within the range of 0�???90� showing the relationship between sin? and cos? for any angle ?

2. ## Mathematics Higher Level Internal Assessment Investigating the Sin Curve

When is negative the graph is not only stretched by but it is also flipped over the x-axis. This is happening because all the values of the graph are multiplied by which causes there to be a shift in the amplitude of the graph without affecting the horizontal position of the graph.

1. ## The Sky Is the Limit Portfolio. In this assignment I will be building ...

I will define the parameters of each. For the linear equation: h is the winning height in centimetres, m is the slope of the function, t is the time in years, and b is the y-intercept. For the square root function: h is the winning height in centimetres and t is the time in years.

2. ## Investigating ratio of areas and volumes

and values for Area A and Area B were rounded to 3 d.p.) From the above data the following conjecture can be made: For the graph of y = xn in between the points x = 0 and x = 1 the ratio area A: area B is: n: 1.

1. ## The aim of this particular portfolio investigation is to find the area under a ...

+(g(0.4) +g(0.6) 0.2)+(g(0.6) +g(0.8) 0.2)+(g(0.8)+g(1) 0.2) 2 2 2 2 2 = (( 3+ 3.04)+ ( 3.04+ 3.16)+( 3.16+ 3.36) +( 3.36+ 3.64)+ ( 3.64+ 4)) ..........factorizing = or 3.34 The area under the curve approximated using five trapeziums was closer to the actual value than when using only two trapeziums.

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

this into account, the expression obtained would now become: A=12ng0+2gx1+2gx2+…+2g(xn-1+g(1)] From the previous analysis of the distance difference, where position of base ×âxn+x value of first base If the x value of first base = a, then the general expression of the area can be expressed by factorizing the pattern even more.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to