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Math Help - complicated integral (Calculus 2)

  1. #1
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    complicated integral (Calculus 2)

    Hi everyone,

    Can you please tell me if this is correct and if it isn't can you show me what I did wrong?

    Find the interval:

    y^3 square root of (2y^4-1)dy

    u=2y^4-1
    du=8y^3dy
    y^3dy=du/8

    y^3square root of 2y^4-1dy=
    square root of u du/8=
    square root of u/8=
    u^(3/2)/9+c=
    1/9(2y^4-1)^(3/2)+c

    Thank you very much
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  2. #2
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    Krizalid's Avatar
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    Nope. Try again (almost got it!!)
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Can you please tell me if this is correct and if it isn't can you show me what I did wrong?

    Find the integral:

    y^3 square root of (2y^4-1)dy

    u=2y^4-1
    du=8y^3dy
    y^3dy=du/8

    y^3square root of 2y^4-1dy=
    square root of u du/8=
    square root of u/8=
    u^(3/2)/9+c= ...............................Here's your mistake. this integral is not correct
    1/9(2y^4-1)^(3/2)+c

    Thank you very much
    \int y^3 \sqrt { 2y^4 - 1}~dy

    We proceed by substitution

    Let u = 2y^4 - 1

    \Rightarrow du = 8y^3~dy

    \Rightarrow \frac {1}{8}du = y^3~dy

    So our integral becomes

    \frac {1}{8} \int \sqrt {u}~du

    = \frac {1}{8} \int u^{1/2}~du

    Now continue. Hint: use the power rule
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  4. #4
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    Hi,

    Is it 1/12(2y^4-1)^3/2+c?

    This is what I did:

    u=2y^4-1
    du=8y^3dy
    y^3dy=du/8

    y^3 square root (2y^4-1)dy=

    square root of u du/e=
    1/8 square root of u du=
    1/8u^3/2/3/2=
    u^3/2/12+c=
    1/12(2y^4-1)^3/2+c

    Thank you
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  5. #5
    Math Engineering Student
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    Correct ^^

    Remember that if you're unsure about your answer, take the derivative, it should yield the integrand.
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  6. #6
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    Thank you very much

    Regards
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