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Find the Value of the Integral โˆซ 1 0 X 2 1 + X 3 ๐’…๐’™ Using Trapezoidal Rule - Applied Mathematics 2

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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`

๐’™๐ŸŽ=0 ๐’™๐Ÿ=0.25 ๐’™๐Ÿ=0.50 ๐’™๐Ÿ‘=0.75 ๐’™๐Ÿ’=1.0
๐’š๐ŸŽ=0 ๐’š๐Ÿ=0.06153 ๐’š๐Ÿ=0.2222 ๐’š๐Ÿ‘=0.39560 ๐’š๐Ÿ’=๐ŸŽ.๐Ÿ“

Trapezoidal rule :

`"I"=h/2[X+2R]` -----------------(1)

๐‘ฟ=๐’”๐’–๐’Ž ๐’๐’‡ ๐’†๐’™๐’•๐’“๐’†๐’Ž๐’† ๐’๐’“๐’…๐’Š๐’๐’‚๐’•๐’†๐’”=๐’š๐ŸŽ+๐’š๐Ÿ’=๐ŸŽ+๐ŸŽ.๐Ÿ“=๐ŸŽ.๐Ÿ“ ๐‘น=๐’”๐’–๐’Ž ๐’๐’‡ ๐’“๐’†๐’Ž๐’‚๐’Š๐’๐’Š๐’๐’ˆ ๐’๐’“๐’…๐’Š๐’๐’‚๐’•๐’†๐’” = ๐’š๐Ÿ+๐’š๐Ÿ+๐’š๐Ÿ‘ =๐ŸŽ.๐ŸŽ๐Ÿ”๐Ÿ๐Ÿ“๐Ÿ‘+๐ŸŽ.๐Ÿ๐Ÿ๐Ÿ๐Ÿ+๐ŸŽ.๐Ÿ‘๐Ÿ—๐Ÿ“๐Ÿ”๐ŸŽ=๐ŸŽ.๐Ÿ”๐Ÿ•๐Ÿ—๐Ÿ‘๐Ÿ‘

`"I" = (0.25)/2`(๐ŸŽ.๐Ÿ“+๐Ÿ(๐ŸŽ.๐Ÿ‘๐Ÿ—๐Ÿ“๐Ÿ”๐ŸŽ))  ……………….(from 1)

∴ I = 0.2323

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Numerical Integrationโ€ by Trapezoidal
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2017-2018 (June) CBCGS
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