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Math Help - integral with square root

  1. #1
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    integral with square root

    Hi everyone,

    In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

    Int. 0-1 r^3/sq. 4+r^2 dr

    t=r^3
    tu=3r^2dr
    r^2dr=tu/3

    Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

    Thank you
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  2. #2
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    Krizalid's Avatar
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    Set u=\sqrt{4+r^2}
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  3. #3
    MHF Contributor red_dog's Avatar
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    You better use the substitution
    \sqrt{4+r^2}=u\Rightarrow 4+r^2=u^2\Rightarrow rdr=udu
    r=0\Rightarrow u=2
    r=1\Rightarrow u=\sqrt{5}
    Now
    \displaystyle\int_0^1r^2\sqrt{4+r^2}\cdot rdr=\int_2^{\sqrt{5}}(u^2-4)u^2du
    and you can continue.
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  4. #4
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    Hi,

    Can you explain how you got "" please?

    Also, your final answer would be -1 right?

    This is what I did:

    u squared -4=(2 squared-4)-(sq. 5) sq. -4=-1

    Thank you
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  5. #5
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    u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr, so u\,du=r\,dr

    Does that make sense?
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  6. #6
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    Yes, thank you very much
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr, so u\,du=r\,dr

    Does that make sense?
    how about root (4-2t)
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  8. #8
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

    Int. 0-1 r^3/sq. 4+r^2 dr

    t=r^3
    tu=3r^2dr
    r^2dr=tu/3

    Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

    Thank you
    sssss
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