Calculadora da regra 3/8 de Simpson para uma função

Aproxime a integral $$$\int\limits_{0}^{3} \sqrt{x^{3} + 5}\, dx$$$ com $$$n = 6$$$ usando a regra 3/8 de Simpson.

A regra 3/8 de Simpson usa polinômios cúbicos para aproximar a área:

$$$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \frac{3 \Delta x}{8} \left(f{\left(x_{0} \right)} + 3 f{\left(x_{1} \right)} + 3 f{\left(x_{2} \right)} + 2 f{\left(x_{3} \right)} + 3 f{\left(x_{4} \right)} + 3 f{\left(x_{5} \right)} + 2 f{\left(x_{6} \right)}+\dots+3 f{\left(x_{n-5} \right)} + 3 f{\left(x_{n-4} \right)} + 2 f{\left(x_{n-3} \right)} + 3 f{\left(x_{n-2} \right)} + 3 f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right)$$$

onde $$$\Delta x = \frac{b - a}{n}$$$.

Temos que $$$f{\left(x \right)} = \sqrt{x^{3} + 5}$$$, $$$a = 0$$$, $$$b = 3$$$ e $$$n = 6$$$.

Portanto, $$$\Delta x = \frac{3 - 0}{6} = \frac{1}{2}$$$.

Divida o intervalo $$$\left[0, 3\right]$$$ em $$$n = 6$$$ subintervalos de comprimento $$$\Delta x = \frac{1}{2}$$$ com os seguintes pontos finais: $$$a = 0$$$, $$$\frac{1}{2}$$$, $$$1$$$, $$$\frac{3}{2}$$$, $$$2$$$, $$$\frac{5}{2}$$$, $$$3 = b$$$.

Agora, apenas avalie a função nesses pontos finais.

$$$f{\left(x_{0} \right)} = f{\left(0 \right)} = \sqrt{5}\approx 2.23606797749979$$$

$$$3 f{\left(x_{1} \right)} = 3 f{\left(\frac{1}{2} \right)} = \frac{3 \sqrt{82}}{4}\approx 6.791538853603062$$$

$$$3 f{\left(x_{2} \right)} = 3 f{\left(1 \right)} = 3 \sqrt{6}\approx 7.348469228349534$$$

$$$2 f{\left(x_{3} \right)} = 2 f{\left(\frac{3}{2} \right)} = \frac{\sqrt{134}}{2}\approx 5.787918451395113$$$

$$$3 f{\left(x_{4} \right)} = 3 f{\left(2 \right)} = 3 \sqrt{13}\approx 10.816653826391968$$$

$$$3 f{\left(x_{5} \right)} = 3 f{\left(\frac{5}{2} \right)} = \frac{3 \sqrt{330}}{4}\approx 13.624426593438712$$$

$$$f{\left(x_{6} \right)} = f{\left(3 \right)} = 4 \sqrt{2}\approx 5.65685424949238$$$

Por fim, basta somar os valores acima e multiplicar por $$$\frac{3 \Delta x}{8} = \frac{3}{16}$$$: $$$\frac{3}{16} \left(2.23606797749979 + 6.791538853603062 + 7.348469228349534 + 5.787918451395113 + 10.816653826391968 + 13.624426593438712 + 5.65685424949238\right) = 9.79911172128198.$$$

$$$\int\limits_{0}^{3} \sqrt{x^{3} + 5}\, dx\approx 9.79911172128198$$$A