Evaluate $\int\limits^1_{-1}\dfrac {dx}{1+x^2}$ by

1views

written 9.5 years ago by

teamques10 ★ 70k

Let $a = -1, b = -1$ and $n = 6$

Let $y=\dfrac 1{1+x^2}\\ \therefore h=\dfrac {b-a}n=\dfrac {1-(-1)}6=\dfrac 13$

enter image description here

1) Trapezoidal method

$\int\limits_a^bf(x)dx=\dfrac h2[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)]\\ \therefore \int\limits_{-1}^1\dfrac {dx}{1+x^2}=\dfrac {\frac 13}2\Bigg[\Bigg(\dfrac 12+\dfrac12\Bigg)+2\Bigg(\dfrac 9{13}+\dfrac 9{10}+1+\dfrac 9{10}+\dfrac 9{13}\Bigg)\Bigg] \\ =\dfrac 16\Bigg[1+2\times \dfrac {272}{65}\Bigg] \\ =1.5615$

2) Simpsons 1/3rd method

$\int\limits_a^bf(x)dx=\dfrac h3[(y_0+y_6)+4(y_1+y_2+y_5)+2(y_2+y_4)]\\ =\dfrac {\frac 13}2\Bigg[\Bigg(\dfrac 12+\dfrac12\Bigg)+2\Bigg(\dfrac 9{13}+1+\dfrac 9{13}\Bigg) +2\Bigg(\dfrac 9{10}+\dfrac 9{10}\Bigg)\Bigg] \\ =\dfrac 19\Bigg[1+4\times \dfrac {31}{13}+\dfrac {2\times 9}5 \Bigg] \\ =1.5709$

3) Simpsons 3/8th method.

$\int\limits_a^bf(x)dx=\dfrac {3h}8[(y_0+y_6)+3(y_1+y_2+y_4+y_5)+2(y_3)]\\ =\dfrac {3\times\frac 13}8\Bigg[\Bigg(\dfrac 12+\dfrac12\Bigg)+3\Bigg(\dfrac 9{13}+\dfrac 9{10} +\dfrac 9{10}+\dfrac 9{13}\Bigg)+2(1)\Bigg] \\ =\dfrac 18\Bigg[1+3\times \dfrac {207}{65}+2 \Bigg] \\ =1.5692$

Exact solution

$\int\limits^1_{-1}\dfrac 1{1+x^2}dx=[\tan^{-1}x]^1_{-1}\\ =\tan^{-1}1-\tan^{-1}(-1)\\ =\dfrac \pi4+\dfrac \pi4 =\dfrac \pi2\\ =1.57079$

ADD COMMENT EDIT