• 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

Shady Areas. In this investigation you will attempt to find a rule to approximate the area under a curve (i.e. between the curve and the x-axis) using trapeziums (trapezoids). Consider the function g(x) = x2 + 3

Extracts from this document...


In this investigation you will attempt to find a rule to approximate the area under a curve (i.e. between the curve and the x-axis) using trapeziums (trapezoids).

Consider the function g(x) = x2 + 3

The diagram below shows the graph of g.  The area under this curve from x = 0 to x = 1 is approximated by the sum of the area of two trapeziums.  Find this approximation.

In this investigation I will attempt to find the approximate area under a curve (i.e. between the curve and the x-axis) using trapeziums (trapezoids).

The area of a trapezoid can be expressed as:

A= (1/2)(b)(h1+h2)


                b is the length of the base of the trapezium along the x-axis

        h1 is the vertical length (height) of one side on the trapzium, the equivalent y- value for one of the endpoints on the base

h2 is the other height of the trapezium for which it is the corresponding y-value for the second endpoint of the base.

A is the area of a specific trapezium in units squared.

The total area underneath the curve can be expressed as:



TA is the total area (or sum) of the two trapeziums, which is equivalent to the approximated area underneath the curve, in units squared.

AA is the estimated, calculated area of the trapezium labeled as A, in units squared.


...read more.







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.

The smaller the trapeziums, the more precise the area is because a smaller unit of measurement was utilized. It is possible to go on to infinity trapeziums, in which case the uncertainty would decrease more and more resulting in the relative error heading towards zero. This is because with more trapeziums it is possible to get a better approximation of the area since the line is going to a microscopic level, nearer to the original line. It gets nearer to the real value.

Use the diagram below to find a general expression for the area under the curve of g, from x = 0 to x = 1, using n trapeziums.         

Based on the pattern observed from the diagrams above, a general equation can be written for the area under the curve of g, from x=0 to x=1, using n trapeziums.  [g(x) = x2 + 3]

ca   f(x) dx  10   x2 + 3 dx


        c is equal to the end x value of the domain

        a is equal to the beginning x value for the domain

        Y= f(x), the equation of the curve

        dx indicates that everything is being taken in respects to x

The above equation in its expanded form is:  

10 x2 + 3 dx =(1/2)

...read more.


2 + 4) 2nd      TRACE      7     ENTER   graph displayed   0     ENTER   1 S  ENTER S

f(x)dx = 0.463647609

  1. y = (2x+1)

ca f(x) dx = [(xn – x0)/(2n)] (g(x0) + g(xn) + 2[∑ g(xi)])

x0: g(x) = (2x+1)

g(0) = (2(0)+1)

               = 1

xn: g(x) = (2x+1)

g(1) = (2(1)+1)

               = 1.732050808

10 ((2x+1)) dx = (1/8) (1 + 1.73 + 2[1.224744871 + 1.414213562 + 1.58113883]

= 1.396530667


math-> 9. fnInt (  enter   2/(x2 + 4)      ,X     ,1      ,3    )   ENTER  OR Y=  enter y = 2/(x2 + 4) 2nd      TRACE      7     ENTER   graph displayed   0     ENTER   1 S  ENTER S

f(x)dx = 1.398717474


The manual calculations derived from the general statement appeared to be very close to the exact answers as computed by the calculator. Based on the results, it appears that no matter the shape, using the trapezium rule provides with a fairly accurate and precise approximation.

However, it can be determined in two ways whether the approximation is an overestimate or an underestimate. The first is to sketch a graph and draw the trapeziums. If the tops of the trapeziums are above the curve, there is an overestimate, and if the trapeziums are below the curve, it is an underestimate. The second method is to examine the second derivative of the graph. If it is negative, indicating concave down, then the curve will have a lesser gradient at any given interval in the positive x-direction, and therefore the trapeziums will be underneath the curve. If the second derivative is positive, indicating upwards concavity, trapeziums will extrude, and thus give an overestimate.

Overall, to achieve the most precise and accurate approximation for an area under the curve, n, the number of trapeziums, needs to be divided into smaller subintervals.

-  -

...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

    �"a�_][,�a�B�@0G�-0`���/(c)��qÉA��H�� X�MÕ¢8c'� �"k�Lm3@@w M��>K�JM��v("q���+D�v��FzH-lkTBC��`�����I"&_�1/2'G@\2b"A| =z�j�g�!0r�HL"��(r)E1/2� Ò¸"�K�3UQ Y��-5(c)ñ­¦1/4�Z�s�,Z����Z�U��H-� 'cÆ�� P|�1/4�k~�����e.z�Er�9ß�۷�b�-R\�2�'M�]"�Êu7n��P�R���� �$Lm��=����`a�'���#�\i! I�� L=�q��%6�o"1/4"S'c�:w�L�N�-g�z�%+P�""ݺuS�dh'1/4Õ¥K|er�4�1_q��%���{�...QR411/4��� 2��"Bg$É"'�������"S�Ke�99*6,'(c)�\�ܬ�y��;v�PnXB�,Y ,(tm)6D�E-�'���'#F���ذH"Ó�]`'4>��3e��Z���%� �U"V!Ψ��'�"�}���/GI��E�,����U�X\w�%��$�o���d����XW(c) �������2 3� E� 7��'I��شiS�M��'�E'L���;O���HFW��P-Y�$��m-$NB�ʥH3W<�rs'G6yo��-~�V�Zq�&���V�!@"/Pd�D� ��gÍ�";�M��jP"&Mv��M=:��n�r$K�� 8m.��r�H6%s Tr�ܹs-(tm)OVÓ�I��<P�E27Dot 'p�����J-��3/4��_|Q�����MQ�<G'�HQ�`iH�1/4Ùµk�3/4}��CAof"Y�sR^�� �EZ��a���gΜiÕ¬Ò ï¿½o���aX$�:�"�x�4�-���%��"mR(r)"�PÝ��J* ��+�k�(���#���E"x8 �*$�ki�G�(r)>}��#��*.fÏ1/2aÆÅ�ա�`���;,21/47�1/4e�-���S[�$�",��c�`1t���'U�� @� D$��c�ÍN���$�'<�d]4���--[�X�["�;-�8q"Gl��k�(r)...4Ó§O�p�j�$c��H"'"#X�Q��� C �1/2Û¶m#�/Lc��K�w�}� -�e�G-B&����(c) �g�6R$�Vj'" 3<�3d���� ��Ñ~T����ߧg� �ڴi��O"]�"�H��~ߧ�5 �^�"�L�,'_"�cE�"@� �S�(r)�C6���""=�� &|��G�{��7���[d+4n���x�j-���W-�yJ��N�*� �^SzW/u?��a��E'�2��Xh���)�x��{.��@ $Ϫ�|�:)ز5jT�����ذHTkV�(r)�_�'u-��IS...J��Ok t<n�A�...���v!�CL&��NI��[���iÒ¡8��N; #,�_ï¿½ï¿½× R^��@1/2���o% ]t�E�sL��1/2{���#��g\�� 8H�1/4��׿���^{�1/2Ħ �(s�%gÌ=�P*�(c)�"@ �~'�8v�X �yJ�E�h��3&9�I,�51/2�a�+�1/4Ò~�Yg��O*�'/1/4���ÍKfH�r.I��O��O<��q��={�p�QbÍ5"v����\�B��H�y�GL1O>�d��za%��>��c�R�?!h�Jq)��y�'d��...uv'��K-.%Zn�O'O`�Ð~�)~�'���:��ñ\�wQ�])i��$C�Ü"'qWҦ�y"%IzÓ±6U?�?�jpI�,�O3��L�O\R�7�Æ�� �vxQ�8#*��8ɣ�t���v��'G�%R' �2�i��{�z表�%o{��W�����w%m*�Y'wz|a���5z1/44s$N��WVX!�!E���;x.���gr1/2=��m &���-�J�X8��*tE�����Í�'Üh!8):��E� �.r��6m�1/2J�V�fZ�Üjf�L�&�-'�/�5�(tm)A�jf���Ok5�7�Q `"8R~� �a��}-w�u��tK�Þ�~�"��r 3/4u��0��+���"���d��agJ"���"##d�ذ"<�,�/viV�PF�;�ÒÅ(tm)Ûo3/4y�QG������G�:Û°y ;�04��K-D�į-rs�� �⤣�v-(�"dF��"vgeX$��a' DY��,'�9�Þ1/4��6!s'*'S7-A�%�n�N�z�D��-�i��Ia;w�"R~L#G�$5�Cd'�`�`�rK�7nQ/�"" �FU"F�*Û³"1/4�%$[�h!��=;��(tm)

  2. Mathematics (EE): Alhazen's Problem

    They emphasize the fact that for any given ellipse, if a ball is placed in each focus then any point on the rim would be a solution (such is the nature of the foci points).

  1. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    Moving on, we saw that the variables and correspond to the horizontal and vertical translation of the sine curve respectively. The curve is translated to the right horizontally and units vertically upwards from the original sine curve. Since we know the general rule of the variables, it can help predict

  2. Logarithm Bases - 3 sequences and their expression in the mth term has been ...

    Now, lets consider the pattern of the second column. This time, not the base of the logarithm, but the value of the logarithm will be taken. Thus for the second column after calculating the values of the terms, we obtain the values: 3 (log864), 1(log4949), -1(Log 125) and 9 (log2512)

  1. Parabola investigation. In this task, we will investigate the patterns in the intersections of ...

    -4 -4 1 -2 12 -10 0.5 2 -10 5 0.5 Proof: Find the two intersections between parabola f(x) and g(x): or Find the two intersections between parabola f(x) and h(x): or Hence, the conjecture about the values of D, for all real values of a, .

  2. Maths Portfolio Shady Areas

    "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,

  1. Analysis of Functions. The factors of decreasing and decreasing intervals (in the y ...

    Power functions are not periodic because a periodic function repeats function values after regular intervals. It is defined as a function for which f(x+a) = f(x), where T is the period of the function. In the case of power functions, it can't be a periodic function is that there is no definite or constant period like "a".

  2. Logarithm Bases

    The next two terms are log 243 81 and log 729 81 General expression for the nth term of this sequence is log 3n 81 = 4/n. View the work below to understand. * log 5 25, log 25 25, log 125 25, log 625 25,...

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