• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  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...


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


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









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.





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



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

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






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











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


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.


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.


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


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.



...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 (EE): Alhazen's Problem

    must equal � (angle of incidence = angle of reflection). Furthermore, if we extend lines OA, OC, and OB to the circle at A' and B' then we form 3 new angles. The angle between OA' and the x-axis is noted ?, the angle between OC and the x-axis is

  2. Math IA Type 1 In this task I will investigate the patterns in the ...

    Although I cannot formally prove this, I will try to justify this. From Vieta's theorem1 we know that: The sum of the roots of a cubic function = Where a and b are coefficients from . x1+x2+ x3 = where x1 ,x2, x3 are roots of the function Where x2,

  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. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    In Graph 1.3 has been flipped over and stretched by a factor of to give the new graph of . Overall, it can be seen that in the represents the 'height' of the graph. In mathematics this 'height' is known as the amplitude of the graph, meaning that in the equation above, represents the amplitude of the graph.

  1. 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)

  2. Investigating Slopes Assessment

    However, I will do the table normally and check my results. f(x)= 3X3 X=? Tangent Gradient X= -2 Y= 36x+44 36 X= -1 Y= 9x+6 9 X= 0 Y= 0 0 X= 1 Y= 9x+(-6) 9 X= 2 Y= 36x+(-44)

  1. In this investigation, I will be modeling the revenue (income) that a firm can ...

    of our demand curve, which we can substitute it into the straight-line equation, P= -0.75Q + c. We only have to find the value of c in order to complete our linear demand equation for the 1st quarter. To find the value of c, we simply have to substitute the

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

    added together. The area is approaching around the value of 3. However, more calculations with additional trapezoids are needed to determine a more precise value. Eight Trapezoids Trapezoid h a b Area 1 0.125 3 3.015625 0.375976562 2 0.125 3.015625 3.0625 0.379882812 3 0.125 3.0625 3.140625 0.387695312 4 0.125 3.140625

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