• 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
  17. 17
    17
  18. 18
    18

Approximations of areas The following graph is a curve, the area of this curve is an approximation of two trapeziums from x=1 to x=0.

Extracts from this document...

Introduction

Internal Assessment 1             (05/11/09)    

Approximations of areas

The following graph is a curve, the area of this curve is an approximation of two trapeziums from x=1 to x=0.

Graph 1

image00.png

The Y-values (N) are found through the function F(x) = image01.pngimage01.pngwhere (x) is (M)

F(x) = image01.pngimage01.png                F(x) = image01.pngimage01.png                        F(x) = image01.pngimage01.png

F(0) = image02.pngimage02.png                F(.5) = .52 + 3                        F(1) = image09.pngimage09.png

F(0) = 3                        F(.5) = 3.25                        F(1) = 4

Table 1

M

N

0

3

0.5

3.25

1

4

I used Microsoft Excel and my Ti-84 calculator to retrieve this data. I graphed my M-values against my N-values in the graph as shown and I drew a line to represent the two trapezoids that are present in my graph. To find the area I will calculate The area of each trapezoid and then add them to retrieve the area of the graph.

M is obtained depending on the number of trapeziums in a graph. I will show a function to obtain this method during my general statement where you will be able to observe this data.

The area of a trapezium is as shown.image19.pngimage19.png where Y1 and Y2 are the parallel sides and ‘x’ ‘in this case’ is the base of width. For the first set of trapeziums I will not round unless the Ti-84 does it, so that I can keep my answer a little more precise.

...read more.

Middle

image12.pngimage12.pngimage13.pngimage13.pngimage14.pngimage14.png

image15.pngimage15.pngimage16.pngimage16.pngimage17.pngimage17.pngimage18.pngimage18.png

image20.pngimage20.pngimage21.pngimage21.pngimage22.pngimage22.pngimage23.pngimage23.png

Trapezoid (5)                        Trapezoid (6)                         Trapezoid (7)

image24.pngimage24.pngimage25.pngimage25.pngimage26.pngimage26.png

image27.pngimage27.pngimage28.pngimage28.png

image29.pngimage29.png.320                        image31.pngimage31.pngimage32.pngimage32.png

Trapezoid (8)                        Trapezoid (9)                        Trapezoid (10)

image33.pngimage33.pngimage34.pngimage34.pngimage35.pngimage35.png

image36.pngimage36.pngimage37.pngimage37.pngimage38.pngimage38.png

image39.pngimage39.pngimage40.pngimage40.pngimage42.pngimage42.png

image43.pngimage43.png372+.390

image44.pngimage44.png

The approximation under the curve using 10 trapezoids is approximately 3.335 units2

So far the approximations we have are 3.375 for 2 trapezoids, 3.34 for 5 trapezoids, 3.336 for 8 trapezoids and 3.335 for 10 trapeziums.

Table 6

T

A

2

3.375

5

3.34

8

3.336

10

3.335

We notice that although they don’t differ by much, they do decrease as the number of trapeziums increase. And this makes sense. Let’s go back to the first graph. Now as you can see we are actually trying to find the area of the graph under the curve(the black line). But using trapeziums, we end up calculating a little over it as well, though the approximation is close, it cannot be exact.

Graph 5

image45.png

Now if we take four trapeziums instead of 2,less of the outer area is calculated because the lines drawn from one f(x) value to another is not as high, and is therefore more minimal.

image46.png

The trapezoids here are much more fit for the curve and will therefore give you a much better estimation. Using this theory we can conclude that as the base width gets smaller, or the number of trapeziums gets higher, we will receive a better estimation, and to get an exact answer we must have w width of basically 0.

...read more.

Conclusion

It is wrong to get a negative integral since we are calculating the area.

Appendix

Microsoft Word XP and Microsoft Excel was used to type this assignment. Microsoft office’s “Equation” option was used to type the equations. The “Area” graphs were used in Excel to draw them. The tables and graphs were all formed using Microsoft Excel. I also used it’s “function” option to give the answers for the tables.

 All calculations were done using Ti-84.

The following will describe how the Ti-84 was used to perform the functions and find the area from section ‘Equations vs. Ti-84’. I will show it according to the first problem, Y1.

I went first to ‘Y=’ and typed in my equation, image111.pngimage111.pngAfter it was drawn I went to ‘WINDOW’ and set it accordingly, I changed my ‘Xmin’ to 1 and ‘xmax’ to 3.

After that I pressed ‘2ND’ and ‘Trace’ to get to ‘CALC’. I went to option ‘7’ which is the integration function. After I clicked that it took me back to the graph where I typed in my end points.

Lower limit ‘1’ ‘ENTER’

Upper limit ‘3’ ‘ENTER’

And on the bottom it then showed me approximation which in this case was A image82.pngimage82.png8.2597312.

I used Paint to create the following screen shots of the process described above.

image112.pngimage114.png

image115.pngimage116.png

...read more.

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. 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.

  2. Lacsap's Fractions : Internal Assessment

    = ... E7 (6) = The seventh row comes out as shown below: Again, putting the "1" back into the beginning and end of the row: In order to test the validity of the general statement, I will use the statement to find additional rows.

  1. Moss's Egg. Task -1- Find the area of the shaded region inside the two ...

    formula above: Area of triangle ABC = bh = 6 3 = 9 cm2 b) The properties of this triangle tell us that the perpendicular line of an isosceles triangle meets the base at angles of 900 and thus we have two right angles triangles.

  2. The purpose of this paper is to investigate an infinite summation patter where Ln(a) ...

    Sn approaches a horizontal asymptote when y=9. There is a y-intercept at (0,1). tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 6.238325 7.238325 2.000000 2.162039 9.400363 3.000000 0.499537 9.899900 4.000000 0.086563 9.986463 5.000000 0.012000 9.998464 6.000000 0.001386 9.999850 7.000000 0.000137 9.999987 8.000000 0.000012 9.999999 9.000000 0.000001 10.000000 10.000000 0.000000 10.000000 As n � +?, Sn �

  1. Mathematics (EE): Alhazen's Problem

    From looking at the changing "chord a - chord b" we can see that solutions should be at point 4, next to point 7, between points 10 and 11, and between points 11 and 12. However there appears to be an apparent paradox as although our results suggest that there

  2. Investigating the Ratios of Areas and Volumes around a Curve

    The following table demonstrates the results obtained using this program for several values of n, a, and b. Results which do not correlate with the observed pattern (and cannot be put down to the calculator's internal rounding errors) are shown in bold.

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

    Using the equation (13)to find out the height: h =[1296 - (20.78)2] h =0.67h The above relation is used to find the dimensions and the volume of the cuboid for different heights of the structure. The length of the cuboid= 150 meters The width of the cuboid =41.56meters The width

  2. Mathematics Internal Assessment: Finding area under a curve

    The figure, once graphed in totality: Here, the two trapeziums present are ADEB and BCFE. The area of trapezium ADEB is 1.56 cm2 and the area of trapezium BCFE is 1.81 cm2. Therefore, one can conclude that the approximate area present underneath the curve f(x)

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