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Math Help - difficult antiderivative

  1. #16
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    I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?

    Can someone please explain this to me?

    Thank you very much

    Regards
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?

    Thank you very much

    Regards
    if u = mx then du = m~dx \implies \frac {1}{m}du = dx ........there's your factor of \frac {1}{m}

    Also, we have to change the x^2
    if u = mx then x = \frac {1}{m}u \implies x^2 = \frac {1}{m^2}u^2 .........there's your factor of \frac {1}{m^2}

    multiplying these factors together, you get \frac {1}{m^3}

    when Krizalid was doing the integral, he left off this factor and so he said you should remember to multiply through by it, as his final equation suggested
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  3. #18
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    Thank you very much

    So this is the final answer, right?

    ;
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Thank you very much

    So this is the final answer, right?

    ;
    yes. but Krizalid suggested to multiply through by the \frac {1}{m^3}, that way, you got the answer in the form he had it in on the first page. this is not absolutely necessary though. i think it's fine as it is, it's up to you and your preferences
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