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Question
Find the value of the integral `int_0^1 x^2/(1+x^3`๐ ๐ using Trapezoidal rule
Sum
Solution
Let I = `int_0^1 x^2/(1+x^3)dx`
a=0 , b=1
Dividing limits into 4 parts i.e n = 4
`thereforeh=(b-a)/n=1/4=0.25`
๐๐=0 | ๐๐=0.25 | ๐๐=0.50 | ๐๐=0.75 | ๐๐=1.0 |
๐๐=0 | ๐๐=0.06153 | ๐๐=0.2222 | ๐๐=0.39560 | ๐๐=๐.๐ |
Trapezoidal rule :
`"I"=h/2[X+2R]` -----------------(1)
๐ฟ=๐๐๐ ๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐=๐๐+๐๐=๐+๐.๐=๐.๐ ๐น=๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐ = ๐๐+๐๐+๐๐ =๐.๐๐๐๐๐+๐.๐๐๐๐+๐.๐๐๐๐๐=๐.๐๐๐๐๐
`"I" = (0.25)/2`(๐.๐+๐(๐.๐๐๐๐๐)) ……………….(from 1)
∴ I = 0.2323
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Numerical Integrationโ by Trapezoidal
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