Concept:
A) Simpson’s one-third rule:
For applying this rule, the number of subintervals must be a multiple of 2.
\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)
B) Simpson’s three-eighths rule:
For applying this rule, the number of subintervals must be a multiple of 3.
\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)
Calculation:
x0 = 0, xn = 1 = x0 + nh
h = 1/n
h = 1/2 = 0.5
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x
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0
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0.5
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1
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\(y = f\left( x \right) = \frac{{3{x^2}}}{5} + \frac{9}{5}\)
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\(\frac{9}{5}\)
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\(\frac{{39}}{{20}}\)
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\(\frac{{12}}{5}\)
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\(\mathop \smallint \nolimits_{{0}}^{{1} } f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_2}} \right) + 4y_1 } \right]\)
\(\mathop \smallint \limits_0^1 ydx = \frac{{0.5}}{3}\left[ {\left( {\frac{9}{5} + \frac{{12}}{5}} \right) + 4\left( {\frac{{39}}{{20}}} \right)} \right]=2\)
