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Math Help - Integration - Simpsons Rule Problem

  1. #1
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    Integration - Simpsons Rule Problem

    Hi again guys,

    Working my way through some little problems and i'm totally baffled with this one.

    THE SIMPSONS RULE - PROBLEM CLICK HERE

    Could anybody help me out with it? Give me some sort of idea what the graph will look like? - Are there any online graph plotting programmes?

    Thanks in advance.
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  2. #2
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    Do you know how to calculate using Simpson's rule?.

    Code:
                        endpoint           f(x)         mult.         result      
    
    0                      0                   3          1                3
    1                     1/3                 1.65      4                6.605  
    2                     2/3                 1.235    2                 2.47
    3                       1                  1.406    4                5.624
    4                      4/3                1.986     2               3.972
    5                      5/3                2.885     4               11.54
    6                       2                  4.055     1               4.055
                                                                          -----------
                                                                         37.266


    Use \frac{b-a}{3n}=\frac{2-0}{3(6)}=\frac{1}{9}

    37.266/9=4.141

    Here's your graph:
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  3. #3
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    Quote Originally Posted by c00ky
    Hi again guys,

    Working my way through some little problems and i'm totally baffled with this one.

    THE SIMPSONS RULE - PROBLEM CLICK HERE

    Could anybody help me out with it? Give me some sort of idea what the graph will look like? - Are there any online graph plotting programmes?

    Thanks in advance.
    The number of partition needs to be an even number in this case n=6. On the interval of [2,0]. Therefore, the size of each partition is,
    \Delta x=\frac{2-0}{6}=\frac{1}{3}
    Thus, by Homer's Rule,
    \int_0^2 x^2+3e^{-2x}dx\approx  \frac{1}{3} [ f(0)+4f(\Delta x)+2f(2\Delta x)+ 4f(3\Delta x)+2(4\Delta x)+4(5\Delta x)+f(2)]\Delta x
    Since,
    \Delta x=\frac{1}{3}
    We can write,
    \int_0^2 x^2+3e^{-2x}dx\approx  \frac{1}{9} [ f(0)+4f(1/3)+2f(2/3)+ 4f(3/3)+2(4/3)+4(5/3)+f(2)]
    Now, create a table of values,
    \left\{ \begin{array}{cc}x&x^2-3e^{-2x}\\<br />
0/3&3.0000\\<br />
1/3&1.6514\\<br />
2/3&1.2352\\<br />
3/3&1.4060\\<br />
4/3&1.9862\\<br />
5/3&2.8848\\<br />
6/3&4.0549
    Now substitute their values into your Homer's approximation,
    \frac{1}{9}( 3+4(1.6514)+ 2(1.2352)+4(1.4060)+2(1.9862)+4(2.8848)+4.0549)
    Thus, calculate,
    \frac{1}{9}(3+6.6056+2.4704+5.624 +3.9724+11.5392+4.0549)
    Again,
    \frac{1}{9}(37.2665)
    Again,
    \approx 4.141
    Now the actual value is,
    \int_0^2 x^2+3e^{-2x}dx=\frac{1}{3}x^3-\frac{3}{2}e^{-2x} \big|^2_0
    =\frac{8}{3}+\frac{3}{2}e^{-4}+\frac{3}{2}e^0=4.1392....

    Note very accurate to the approximation.
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