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Math Help - integral substitution

  1. #1
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    integral substitution

    evaluate integral for interval between x=0 and x=1

    INT: (x^2)(1+x)^0.5 dx

    thanks
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by daaavo View Post
    evaluate integral for interval between x=0 and x=1

    INT: (x^2)(1+x)^0.5 dx

    thanks
    Try a u sub

    let u=x+1 then du=dx and u-1=x

    using the above and substition
    \int x^2(x+1)^{\frac{1}{2}}dx=\int(u-1)^2 u^{\frac{1}{2}}du

    expanding the binomial we get...

    \int (u^2-2u+1)u^{\frac{1}{2}}du=\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}du

    From here it is off to the races...
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by daaavo View Post

    evaluate integral for interval between x=0 and x=1

    INT: (x^2)(1+x)^0.5 dx
    After substitution u^2  = 1 + x the integral becomes 2\int_1^{\sqrt 2 } {u^2 \left( {u^2  - 1} \right)^2 \,du}.

    Now this is routine. After some straightforward calculations your integral is equal to \frac{4}<br />
{{105}}\left( {11\sqrt 2  - 4} \right).
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