Soluzione La regola di Simpson dei 3/8 utilizza polinomi cubici per approssimare l'area:
∫ a b f ( x ) d x ≈ 3 Δ x 8 ( f ( x 0 ) + 3 f ( x 1 ) + 3 f ( x 2 ) + 2 f ( x 3 ) + 3 f ( x 4 ) + 3 f ( x 5 ) + 2 f ( x 6 ) + ⋯ + 3 f ( x n − 5 ) + 3 f ( x n − 4 ) + 2 f ( x n − 3 ) + 3 f ( x n − 2 ) + 3 f ( x n − 1 ) + f ( x n ) ) \int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \frac{3 \Delta x}{8} \left(f{\left(x_{0} \right)} + 3 f{\left(x_{1} \right)} + 3 f{\left(x_{2} \right)} + 2 f{\left(x_{3} \right)} + 3 f{\left(x_{4} \right)} + 3 f{\left(x_{5} \right)} + 2 f{\left(x_{6} \right)}+\dots+3 f{\left(x_{n-5} \right)} + 3 f{\left(x_{n-4} \right)} + 2 f{\left(x_{n-3} \right)} + 3 f{\left(x_{n-2} \right)} + 3 f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right) a ∫ b f ( x ) d x ≈ 8 3Δ x ( f ( x 0 ) + 3 f ( x 1 ) + 3 f ( x 2 ) + 2 f ( x 3 ) + 3 f ( x 4 ) + 3 f ( x 5 ) + 2 f ( x 6 ) + ⋯ + 3 f ( x n − 5 ) + 3 f ( x n − 4 ) + 2 f ( x n − 3 ) + 3 f ( x n − 2 ) + 3 f ( x n − 1 ) + f ( x n ) )
dove Δ x = b − a n \Delta x = \frac{b - a}{n} Δ x = n b − a
Si ha che f ( x ) = x 3 + 5 f{\left(x \right)} = \sqrt{x^{3} + 5} f ( x ) = x 3 + 5 a = 0 a = 0 a = 0 b = 3 b = 3 b = 3 n = 6 n = 6 n = 6
Pertanto, Δ x = 3 − 0 6 = 1 2 \Delta x = \frac{3 - 0}{6} = \frac{1}{2} Δ x = 6 3 − 0 = 2 1
Dividere l'intervallo [ 0 , 3 ] \left[0, 3\right] [ 0 , 3 ] n = 6 n = 6 n = 6 Δ x = 1 2 \Delta x = \frac{1}{2} Δ x = 2 1 a = 0 a = 0 a = 0 1 2 \frac{1}{2} 2 1 1 1 1 3 2 \frac{3}{2} 2 3 2 2 2 5 2 \frac{5}{2} 2 5 3 = b 3 = b 3 = b
Ora, è sufficiente valutare la funzione su questi punti finali.
f ( x 0 ) = f ( 0 ) = 5 ≈ 2.23606797749979 f{\left(x_{0} \right)} = f{\left(0 \right)} = \sqrt{5}\approx 2.23606797749979 f ( x 0 ) = f ( 0 ) = 5 ≈ 2.23606797749979
3 f ( x 1 ) = 3 f ( 1 2 ) = 3 82 4 ≈ 6.791538853603062 3 f{\left(x_{1} \right)} = 3 f{\left(\frac{1}{2} \right)} = \frac{3 \sqrt{82}}{4}\approx 6.791538853603062 3 f ( x 1 ) = 3 f ( 2 1 ) = 4 3 82 ≈ 6.791538853603062
3 f ( x 2 ) = 3 f ( 1 ) = 3 6 ≈ 7.348469228349534 3 f{\left(x_{2} \right)} = 3 f{\left(1 \right)} = 3 \sqrt{6}\approx 7.348469228349534 3 f ( x 2 ) = 3 f ( 1 ) = 3 6 ≈ 7.348469228349534
2 f ( x 3 ) = 2 f ( 3 2 ) = 134 2 ≈ 5.787918451395113 2 f{\left(x_{3} \right)} = 2 f{\left(\frac{3}{2} \right)} = \frac{\sqrt{134}}{2}\approx 5.787918451395113 2 f ( x 3 ) = 2 f ( 2 3 ) = 2 134 ≈ 5.787918451395113
3 f ( x 4 ) = 3 f ( 2 ) = 3 13 ≈ 10.816653826391968 3 f{\left(x_{4} \right)} = 3 f{\left(2 \right)} = 3 \sqrt{13}\approx 10.816653826391968 3 f ( x 4 ) = 3 f ( 2 ) = 3 13 ≈ 10.816653826391968
3 f ( x 5 ) = 3 f ( 5 2 ) = 3 330 4 ≈ 13.624426593438712 3 f{\left(x_{5} \right)} = 3 f{\left(\frac{5}{2} \right)} = \frac{3 \sqrt{330}}{4}\approx 13.624426593438712 3 f ( x 5 ) = 3 f ( 2 5 ) = 4 3 330 ≈ 13.624426593438712
f ( x 6 ) = f ( 3 ) = 4 2 ≈ 5.65685424949238 f{\left(x_{6} \right)} = f{\left(3 \right)} = 4 \sqrt{2}\approx 5.65685424949238 f ( x 6 ) = f ( 3 ) = 4 2 ≈ 5.65685424949238
Infine, basta sommare i valori precedenti e moltiplicarli per 3 Δ x 8 = 3 16 \frac{3 \Delta x}{8} = \frac{3}{16} 8 3Δ x = 16 3 3 16 ( 2.23606797749979 + 6.791538853603062 + 7.348469228349534 + 5.787918451395113 + 10.816653826391968 + 13.624426593438712 + 5.65685424949238 ) = 9.79911172128198. \frac{3}{16} \left(2.23606797749979 + 6.791538853603062 + 7.348469228349534 + 5.787918451395113 + 10.816653826391968 + 13.624426593438712 + 5.65685424949238\right) = 9.79911172128198. 16 3 ( 2.23606797749979 + 6.791538853603062 + 7.348469228349534 + 5.787918451395113 + 10.816653826391968 + 13.624426593438712 + 5.65685424949238 ) = 9.79911172128198.
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