Por Simpson, calcule o valor de a partir da relação com 4 subintervalos.... Pergunta de ideia deDanJR

January 2020 1 211 Report

Por Simpson, calcule o valor de $\mathsf{\pi}$ a partir da relação
$\displaystyle{\mathsf{\frac{\pi}{4} = \int_{0}^{1} \frac{dx}{1 + x^2}}}$
com 4 subintervalos.

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Niiya

Verified answer

Vamos estimar a integral

$\displaystyle\int_{a}^{b}f(x)\,dx$

pela Regra de Simpson com 4 subintervalos, onde $a=0,~b=1~~$ e $f(x)=\dfrac{1}{1+x^{2}}$
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Para termos pontos igualmente espaçados, definimos

$h=\dfrac{b-a}{n}=\dfrac{1-0}{4}=0,25=\frac{1}{4}$

Com isso, geramos os pontos

$x_{0}:=a=0\\\\\\\begin{cases}x_{1}=x_{0}+h=0,25=\frac{1}{4}\\\\x_{2}=x_{1}+h=0,5=\frac{1}{2}\\\\x_{3}=x_{2}+h=0,75=\frac{3}{4}\end{cases}\\\\\\x_{4}:=b=1$

Pela Regra de Simpson, aproximamos a integral da seguinte forma:

$\displaystyle\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx\dfrac{h}{3}\bigg[f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+f(x_{4})\bigg]\\\\\\\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx\dfrac{\big(\frac{1}{4}\big)}{3}\bigg[f(0)+4f(0,25)+2f(0,5)+4f(0,75)+f(1)\bigg]$

Encontrando o valor de $f$ nesses pontos:

$f(0)=\dfrac{1}{1+0^{2}}=1\\\\\\f\big(\frac{1}{4}\big)=\dfrac{1}{1+\big(\frac{1}{4}\big)^{2}}=\dfrac{1}{\big(\frac{17}{16}\big)}=\dfrac{16}{17}\\\\\\f\big(\frac{1}{2}\big)=\dfrac{1}{1+\big(\frac{1}{2}\big)^{2}}=\dfrac{1}{\big(\frac{5}{4}\big)}=\dfrac{4}{5}\\\\\\f\big(\frac{3}{4}\big)=\dfrac{1}{1+\big(\frac{3}{4}\big)^{2}}=\dfrac{1}{\big(\frac{25}{16})}=\dfrac{16}{25}\\\\\\f(1)=\dfrac{1}{1+1^{2}}=\dfrac{1}{2}$

Para facilitar, vamos colocar essas frações no mesmo denominador:

$f(0)=\dfrac{1}{1+0^{2}}=1=\dfrac{850}{850}\\\\\\f\big(\frac{1}{4}\big)=\dfrac{1}{1+\big(\frac{1}{4}\big)^{2}}=\dfrac{1}{\big(\frac{17}{16}\big)}=\dfrac{16}{17}=\dfrac{800}{850}\\\\\\f\big(\frac{1}{2}\big)=\dfrac{1}{1+\big(\frac{1}{2}\big)^{2}}=\dfrac{1}{\big(\frac{5}{4}\big)}=\dfrac{4}{5}=\dfrac{680}{850}\\\\\\f\big(\frac{3}{4}\big)=\dfrac{1}{1+\big(\frac{3}{4}\big)^{2}}=\dfrac{1}{\big(\frac{25}{16})}=\dfrac{16}{25}=\dfrac{544}{850}\\\\\\f(1)=\dfrac{1}{1+1^{2}}=\dfrac{1}{2}=\dfrac{425}{850}$

Então, por Simpson, temos

$\displaystyle\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx\dfrac{1}{12}\bigg[\dfrac{850}{850}+\dfrac{4\cdot800}{850}+\dfrac{2\cdot680}{850}+\dfrac{4\cdot544}{850}+\dfrac{425}{850}\bigg]\\\\\\\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx\dfrac{1}{12}\bigg[\dfrac{850+3200+1360+2176+425}{850}\bigg]\\\\\\\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx\dfrac{1}{12}\cdot\dfrac{8011}{850}\\\\\\\boxed{\boxed{\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx0,78539216}}$

Como o valor exato dessa integral é $\frac{\pi}{4}$ , temos a seguinte aproximação:

$\dfrac{\pi}{4}=\displaystyle\int_{0}^{1}\dfrac{dx}{1+x^{2}}\approx0,78539216$

Daí, geramos uma aproximação para $\pi$ :

$\dfrac{\pi}{4}\approx0,78539216~\Leftrightarrow~\pi\approx4\cdot0,78539216~\Leftrightarrow~\boxed{\boxed{\pi\approx3,14156864}}$

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