Aproxima $$$\int\limits_{0}^{1} e^{- \frac{x^{2}}{2}}\, dx$$$ con $$$n = 4$$$ usando la regla trapezoidal

Aproximar la integral $$$\int\limits_{0}^{1} e^{- \frac{x^{2}}{2}}\, dx$$$ con $$$n = 4$$$ usando la regla trapezoidal.

La regla trapezoidal usa trapezoides para aproximar el área:

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

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

Tenemos que $$$f{\left(x \right)} = e^{- \frac{x^{2}}{2}}$$$, $$$a = 0$$$, $$$b = 1$$$ y $$$n = 4$$$.

Por lo tanto, $$$\Delta x = \frac{1 - 0}{4} = \frac{1}{4}$$$.

Divida el intervalo $$$\left[0, 1\right]$$$ en $$$n = 4$$$ subintervalos de longitud $$$\Delta x = \frac{1}{4}$$$ con los siguientes puntos finales: $$$a = 0$$$, $$$\frac{1}{4}$$$, $$$\frac{1}{2}$$$, $$$\frac{3}{4}$$$, $$$1 = b$$$.

Ahora, simplemente evalúe la función en estos puntos finales.

$$$f{\left(x_{0} \right)} = f{\left(0 \right)} = 1$$$

$$$2 f{\left(x_{1} \right)} = 2 f{\left(\frac{1}{4} \right)} = \frac{2}{e^{\frac{1}{32}}}\approx 1.938466468952688$$$

$$$2 f{\left(x_{2} \right)} = 2 f{\left(\frac{1}{2} \right)} = \frac{2}{e^{\frac{1}{8}}}\approx 1.764993805169191$$$

$$$2 f{\left(x_{3} \right)} = 2 f{\left(\frac{3}{4} \right)} = \frac{2}{e^{\frac{9}{32}}}\approx 1.509679203978015$$$

$$$f{\left(x_{4} \right)} = f{\left(1 \right)} = e^{- \frac{1}{2}}\approx 0.606530659712633$$$

Finalmente, simplemente sume los valores anteriores y multiplíquelos por $$$\frac{\Delta x}{2} = \frac{1}{8}$$$: $$$\frac{1}{8} \left(1 + 1.938466468952688 + 1.764993805169191 + 1.509679203978015 + 0.606530659712633\right) = 0.852458767226566.$$$

$$$\int\limits_{0}^{1} e^{- \frac{x^{2}}{2}}\, dx\approx 0.852458767226566$$$A