# Math Help - Integration - Simpsons Rule Problem

1. ## 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?

2. 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

3. 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?

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,
$\left\{ \begin{array}{cc}x&x^2-3e^{-2x}\\
0/3&3.0000\\
1/3&1.6514\\
2/3&1.2352\\
3/3&1.4060\\
4/3&1.9862\\
5/3&2.8848\\
6/3&4.0549$

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.