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I am in an online Calculus 2 class, and before my professor gets back to me, I was wondering if you guys could help. I was reading through an example:

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How was 1/27 pulled out from the coefficient next to u^2? I am probably missing something dumb. Thanks.

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  • $\begingroup$ The $27$ is not coming from the radical but rather from substituting the exponential in the numerator: $e^{3t} \rightarrow u^3 / 27$ $\endgroup$
    – zahbaz
    Jun 19, 2016 at 1:16

2 Answers 2

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Note that $e^{3t} = \frac{1}{27} u^3, 9e^{2t} = u^2, dt = \frac{du}{3e^t} = \frac{du}{u}$. Now substitute and see what you get.

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  • $\begingroup$ Ahhh yes makes sense, thank you! $\endgroup$
    – Brandex
    Jun 19, 2016 at 1:29
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There's an error in the problem-the numerator in the u-substitution should be $u^3$, not $u^2$. The numerator here is $(1/3 u)^3$ (compare to u). So a factor of 1/27 drops out into the denominator. The rest is easy to see by careful substitution.

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  • $\begingroup$ I don't think there's an error, a factor of $u$ gets cancelled by the $dt$ substitution $\endgroup$
    – MCT
    Jun 19, 2016 at 1:25

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