1) Trapezoidal method 2) Simpsons 1/3 rd method 3) Simpsons 3/8 th method

Compare result with exact values

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : DEC 2013

Let $a = 0, b = b$ and $n = 6$

$Let \space y=\dfrac 1{1+x^2}\\ \therefore h=\dfrac {b-a}n=\dfrac {6-0}6=1$

1) Trapezoidal method

$\int\limits_a^bf(x)dx=\dfrac h2[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)]\\ \therefore \int\limits_0^6 \dfrac {dx}{1+x}=\dfrac 12\Bigg[\Bigg(1+\dfrac 17\Bigg)+2\Bigg(\dfrac 12+\dfrac 13+\dfrac 14+\dfrac 15+\dfrac 16\Bigg)\Bigg]\\ =\dfrac 12\Bigg[\dfrac 87+2\times \dfrac {29}{20}\Bigg]\\ =12.0214$

2) Simpsons 1/3 rd method

$\int\limits_a^b f(x)dx=\dfrac h3[(y_0+y_6)+4(y_1+y_2+y_5)+2(y_2+y_4)]\\ =\dfrac 13\Bigg[\Bigg(1+\dfrac 17\Bigg)+4\Bigg(\dfrac 12+\dfrac 14+\dfrac 16\Bigg)+2\Bigg(\dfrac 13+\dfrac 15\Bigg)\Bigg]\\ =\dfrac 13\Bigg[\dfrac 87+4\times \dfrac {11}{12} +\dfrac {2\times 8}{15}\Bigg]\\ =1.9587$

3) Simpsons 3/8 th method

$\int\limits_a^b f(x)dx=\dfrac {3h}8[(y_0+y_6)+3(y_1+y_2+y_4+y_5)+2(y_3)]\\ =\dfrac 38\Bigg[\Bigg(1+\dfrac 17\Bigg)+3\Bigg(\dfrac 12+\dfrac 13+\dfrac 15+\dfrac 16\Bigg)+2\Bigg(\dfrac 14\Bigg)\Bigg]\\ =\dfrac 18\Bigg[\dfrac 87+3\times \dfrac 65 +\dfrac 12\Bigg]\\ =1.9661$

Exact solution

$\int\limits^1_{-1}\dfrac {dx}{1+x}=[\log(1+x)]^6_0\\ =\log(1+6)-\log(1+0)\\ =\log 7\\ =1.9459$