Aproxime $$$\int\limits_{0}^{1} \left(- 2 x^{2} + x\right)\, dx$$$ com $$$n = 3$$$ usando a regra trapezoidal

Aproxime a integral $$$\int\limits_{0}^{1} \left(- 2 x^{2} + x\right)\, dx$$$ com $$$n = 3$$$ usando a regra trapezoidal.

A regra trapezoidal usa trapézios para aproximar a área:

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

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

Temos que $$$f{\left(x \right)} = - 2 x^{2} + x$$$, $$$a = 0$$$, $$$b = 1$$$ e $$$n = 3$$$.

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

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

Agora, apenas avalie a função nesses pontos finais.

$$$f{\left(x_{0} \right)} = f{\left(0 \right)} = 0$$$

$$$2 f{\left(x_{1} \right)} = 2 f{\left(\frac{1}{3} \right)} = \frac{2}{9}\approx 0.222222222222222$$$

$$$2 f{\left(x_{2} \right)} = 2 f{\left(\frac{2}{3} \right)} = - \frac{4}{9}\approx -0.444444444444444$$$

$$$f{\left(x_{3} \right)} = f{\left(1 \right)} = -1$$$

Por fim, basta somar os valores acima e multiplicar por $$$\frac{\Delta x}{2} = \frac{1}{6}$$$: $$$\frac{1}{6} \left(0 + 0.222222222222222 - 0.444444444444444 - 1\right) = -0.203703703703704.$$$

$$$\int\limits_{0}^{1} \left(- 2 x^{2} + x\right)\, dx\approx -0.203703703703704$$$A