• 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

Shady areas; math portfolio type 1

Extracts from this document...

Introduction

Shady areas

This investigation is carried out in order to find a rule to approximate the area under a curve using trapeziums (trapezoids).

From calculus it is known that by the help of integration, the definite area under a curve could be calculated. In this case a geometrical method will be used in order to approximate the area under a curve.

The function g(x) = x2+3 is considered. The graph of g is shown below. The area under the curve from x = 0 to x = 1 is approximated by the sum of the area of two trapeziums.

image00.png

In order to approximate the area under this curve, the composite trapezoidal rule is used. The composite trapezoidal rule is a numerical approach for approximating a definite integral.

image01.pngg(x) dx = Area

image01.pngx2 + 3 dx = Area

According to composite trapezoidal rule:

Area = ½ (width of strip) [first height + 2(sum of all middle heights) + last height]

Since there are two trapeziums from 0 to 1 in fig 1.1 hence the width of each strip: 1/2 = 0.5

x

g(x) or height

0

02 + 3 = 3

1/2

(1/2)2 + 3 = 3.25

1

12 + 3 = 4

So: Area = ½ (0.5) [3 + 2( 3.25) + 4]

     Area = 3.37 area units (a.u.)

Increasing the number of trapeziums to five:

image17.png

Approximating the area in this case with five trapeziums:

Width of every strip: 1/5 = 0.2

x

g(x) or height

0

3

1/5

3.04

2/5

3.16

3/5

3.36

4/5

3.64

1

4

Area = ½ (1/5) [3 + 2(3.04 + 3.16 + 3.36 + 3.64) + 4]

Area = 3.34 a.u.

...read more.

Middle

3.34

1.3

3.3359375

1.4

3.334490741

1.5

3.333984375

It is clearly observed that as the number of trapeziums increase, the uncertainty in approximating the area decreases since increasing number of segments of area under the curve are being taken into account. Thereof the value gets closer and closer to the actual value of the integral which is: 3.333333333 (from integration).

In order to find a general expression for the area under the curve of g(x) = x2 + 3 from x = 0 to x = 1, the following diagram is considered with n trapeziums:

image21.png

As mentioned earlier, area according to composite trapezoidal rule is:

Area = ½ (width of strips) [first height + 2(sum of all middle heights) + last height]

So the general expression in this case for n trapeziums is:

Area = ½ (x) [g (0) + 2(g (x1) + g (x2) +…………….+ g (xn-1)) + g (1)]

[OBS! Width of every strip is x. g (0) = g (x0) and g (1) = g (xn)

Using the results, the general statement that will estimate the area under any curve y = f (x), from x = a  to x = b, using n trapeziums could be developed.

Assuming that the following graph is y = f (x):

image22.png

Composite trapezoidal rule: Area = ½ (width of strips) [first height + 2(sum of all middle heights) + last height]

So the general statement for approximating the area under any curve according to fig 1.7:

Area = ½ (c) [f (a) + 2( f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]

[OBS! Width of every strip is c. f (a) = f (x) since (a = x) and f (b) = f (x+nc) since (b = x+nc)]

Using the general statement with eight trapeziums the area under the following three curves from x = 1 x = 3 could be found:

  1. y= image03.png

image02.png

  1. y= 4x3 – 23x2 + 40x – 18
  1. y= image03.png

image04.png

Area = ½ (c) [f (a) + 2( f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]

Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]

x values

f(x)

1

0.6299605249

1+1/4 = 5/4

0.7310044346

1 +2( 1/4) = 3/2

0.8254818122

1 +3( 1/4) = 7/4

0.9148264275

1 +4( 1/4) = 2

1

1 +5( 1/4) = 9/4

1.081687178

1 +6( 1/4) = 5/2

1.160397208

1+7( 1/4)  = 11/4

1.236521861

1 +8( 1/4) = 3

1.310370697

Area = ½ (1/4) [0.6299605249 +2 (0.7310044346 + 0.8254818122 + 0.9148264275 + 1 + 1.081687178 + 1.160397208 + 1.236521861) + 1.310370697]

Area = 1.980021133 a.u.

image02.png

image05.png

Area = ½ (c) [f (a) + 2(f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]

Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]

x values

f(x)

1

2.846049894

1+1/4 = 5/4

3.474792043

1 +2( 1/4) = 3/2

3.837612894

1 +3( 1/4) = 7/4

4.156356486

1 +4( 1/4) = 2

4.365641251

1 +5( 1/4) = 9/4

4.484455912

1 +6( 1/4) = 5/2

4.534134497

1+7( 1/4)  = 11/4

4.534086944

1 +8( 1/4) = 3

4.5

Area = ½ (1/4) [2.846049894 + 2(3.474792043 + 3.837612894 + 4.156356486 + 4.365641251 + 4.484455912 + 4.534134497 + 4.534086944) + 4.5]

Area = 8.265026244 a.u.

  1. y = 4x3 – 23x2 + 40x – 18

image06.png

Area = ½ (c) [f (a) + 2(f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]

Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]

x values

f(x)

1

3

1+1/4 = 5/4

3.875

1 +2( 1/4) = 3/2

3.75

1 +3( 1/4) = 7/4

3

1 +4( 1/4) = 2

2

1 +5( 1/4) = 9/4

1.125

1 +6( 1/4) = 5/2

0.75

1+7( 1/4)  = 11/4

1.25

1 +8( 1/4) = 3

3

...read more.

Conclusion

fig 3.1), for instance, is more straight and geometrical compared to the graph of a quadratic equation(fig3.2) and the graph of latter is more geometrical compared to the graph of a fifth degree polynomial(fig 3.3). Hence the shape of a graph influences the approximation to a great extent and therefore the more geometrical and linear the shape of a graph the more accurate will be the approximation. The relationship between the shape of a graph and the accuracy of approximation is also evident in figures: 2.1, 2.2, 2.3 and the data related to them. The accuracy decreases in b (fig 2.2) compared to a (fig 2.1) as the shape of the graph becomes less linear and more curved. Similarly the accuracy decreases in c (fig 2.3) compared to b (fig 2.2) as the shape of the graph becomes more haphazard and less linear.

The accuracy of approximation also depends upon the number of trapeziums used. This is evident in figures (1.1 – 1.7) and the data related to them. As the number of trapeziums increases, the approximate value gets closer and closer to the true value of the area.

It could be concluded that the composite trapezoidal rule is exact for polynomials of degree less than or equal to one.

Math Portfolio SL type 1

Shiba Younus IB 2

                                                                                Date: 23-04-09

                                                                                Katedralskolan

...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. Extended Essay- Math

    �^f-x��p1Ñ�""� �>����ÔS1/4�...�B,�9 ��)j(r)�(tm)M�� �=U'^w�9�&�"Q��H�D��l�"��'G#...�!{5���-��q&�Zax�3/4�m��cR��-ï¿½É 1\�r�^yB,�i�m3��������"0��n�(���b� ÓR-�p1��H...I� �"`W��"�5-Y���p�D��e�EM�Q��mͰ�-�����K,Re����PM�21 �{��u]s1�!�x �-&0T-�cTc�� ���8-[�hx+Dj�...���&�"8Ppc$C|�Þ�4J���n��I'��Ôga����4Z�> � X(�4J]Õ´K����(c)b4L1/2�5�����bB$�V ��}��i8�� nW�֤p�]4��(_�E��.�i�l1/2~� ���O�2X��8��~����u6��k]U��s?\ Cڹ�^�� `@��{����Z�J�O}e��;�ä³(c)�U �Ï..."g-��1/2 1�1/4hLEF '`����KT/o...O��3/4=a�9���n � S.�G�"��S��`.��lFB�J� ���E�"Z�...E�.�dW|�~�]OU�z��c(c)�KSuz-^�t.�çªï¿½j���5-�* �r�]�9�h����a%�8m ���h< �AN� WM��� �H��1/4i".(c)��$-~Ac-�:��t8��1:����� ld6�A�"@CTw�6�Fl03/4��9��*�2 ��><�-T�3/4"�M���`���x�CM�1/2B�-"�$�-p�� 1/4HwX�(c)��%:� �-��o\�2]��|�o��6"��x&��, �n���, ���h?vc"��?��/���6���4Å��S��.RæY����H�Pz-�_*~'���;<���" M�%5"w� �t�_ "�"$v���,1/2Ee�o���8��S(}�Ml8Q�1k�Øa�� 3/4�1/2'F� P�dN�,h(4 k!d��"f�mmq��?��=P��"ѧM�U'�z��8�0"�:4]���1/4(c)B>���-"q8�U3Ø°+��*P�1/4�Ѧ���`�>�� {�z��Ø�|q^"I�Y<�� �F#�bÐ�ÆG� V��F ����re�����a]�dR�E��1/2B8�- z��F"n�R8B/]T���G(c)��z�BP"1/2� o�"e"3/4 d"�o�W�ÇH��<��4Z�hf��� �%�^��W�(r)Û´Pi!��N��"� ki�c\w3�d�V�zl��W\ XO[��W`�i��"�'d4z"ÐhqM�E��$(tm)���V{'��"��c��"�J�R"�dÝ-Ew

  2. Maths IA Type 2 Modelling a Functional Building. The independent variable in ...

    volume of the multiple cuboids, we must determine the efficiency by calculating the ratio of wasted space volume to the volume of the cuboids. By implementing Method 5, the ratio was found to be: This value tells us that the wasted space volume is equivalent to 1.79% of the volume of the multiple cuboids.

  1. Ib math HL portfolio parabola investigation

    (x2 r2x r3x + r2r3) = a(x-r1) (x2 (r2+r3)x + r2r3) = a(x3 - (r2+r3)x2 + r2r3x r1 x2 + r1(r2+r3)x r1r2r3) = a(x3 - (r1+ r2+r3)x2 + (r1+r2 + r1r3 + r2r3)x - r1r2r3) = ax3 - a(r1+ r2+r3)x2 + a(r1+r2 + r1r3 + r2r3)x - a(r1r2r3)

  2. Math Portfolio: trigonometry investigation (circle trig)

    The value of y equals 0 and the value of x equals a positive number on the positive x axis. Therefore, when the value of y is divided by the value of x, 0 is divided by a positive number, resulting to 0.

  1. Maths Portfolio Shady Areas

    And as stated, it is due to the over-estimation of the area with the trapeziums, but adding more within the graph reduces that over estimation. Number of trapeziums Total area T1 3.5 T2 3.37 T4 3.34375 T5 3.34 I have noticed there is a relationship between the number of trapeziums within the graph and the areas that they produce.

  2. Gold Medal heights IB IA- score 15

    However, the height of 208 in 2016 would be approximately correct since the data indicates a change from upward concavity to downward concavity. This would mean that in the near future from 1980 the height of 2016 would be less than 236.

  1. IB Math Methods SL: Internal Assessment on Gold Medal Heights

    Let us attempt to discover what the height will be for the record in 1984; assuming that the functions remains true. Take a look at the graph below. Graph 6: Functions showing future heights based upon trends: 1984 However, there are flaws in this hypothetical argument.

  2. Gold Medal Heights Maths Portfolio.

    Therefore, the height which may have been achieved in 1940 is 200cm and 198cm in 1944. Since 1984 is close to the data set, one can again use the sin function to solve for its predicted value since it has the lowest RMSE.

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