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Thread: Simple Integration

  1. #1
    Newbie AmberLamps's Avatar
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    Simple Integration

    I have an integrating factor of $\displaystyle e^(3lnt)$and have simplified it to $\displaystyle e^t^3$ near the end of the problem we are required to integrate $\displaystyle -3t^3 dt$ which comes to $\displaystyle -3/4t^4$ right?

    Now my friend has kept the integrating factor in the form $\displaystyle e^(3lnt)$ throughout the problem and thus has to integrate $\displaystyle -3e^(3lnt) dt$ at the end of the problem. She, however, thinks that this becomes $\displaystyle -te^(3lnt)$.

    Obviosuly one of us is wrong; but who?
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  2. #2
    Junior Member
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    Quote Originally Posted by AmberLamps View Post
    I have an integrating factor of $\displaystyle e^(3lnt)$and have simplified it to $\displaystyle e^{t^3}$ near the end of the problem we are required to integrate $\displaystyle -3t^3 dt$ which comes to $\displaystyle -3/4t^4$ right?

    Now my friend has kept the integrating factor in the form $\displaystyle e^(3lnt)$ throughout the problem and thus has to integrate $\displaystyle -3e^(3lnt) dt$ at the end of the problem. She, however, thinks that this becomes $\displaystyle -te^(3lnt)$.

    Obviosuly one of us is wrong; but who?
    If you post the problem, I could tell you for sure, but given an integrating factor of $\displaystyle e^{3ln(t)}$, I would simplify this to $\displaystyle e^{ln(t^3)}$. The log and the exponential then cancel out to give you an integrating factor of $\displaystyle t^3$. Is this the integrating factor that you actually got?

    Secondly, how did your friend get $\displaystyle -te^{3ln(t)}$? Anycase, he is wrong, if you take the super complicated and long way around, you still get the integration of $\displaystyle \frac{e^{4ln(t)}}{4}=\frac{t^4}{4}$.

    Out of interest, if you want to do the long way, let u = ln(t), hence t = exp(u) then let v =4u. Don't for get to change the dt in each case.
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