Integrate Square root of 3e^(2t) . The integral has a 5 at top and 1 at bottom

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Integrate Square root of 3e^(2t) . The integral has a 5 at top and 1 at bottom

Mathematics
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First of all, you need to simplify the equation: $$ \begin{align*} \sqrt{3e^{2t}} &= (3e^{2t})^{\frac{1}{2}} &= 3e^{2t*\frac{1}{2}} &= 3e^{t} \end{align*} $$ You can then integrate knowing that \(\int e^t dt = e^t\).
Ahh right right cause square root is ^(1/2) . Thank ya.
but wait, i've been working towards getting to the answer in the back of the book which is (square root of 3) times ((e^5) - e)) how'd they get that?

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Uh... I forgot to distribute the exponent. It should actually read: $$ (3e^{2t})^{\frac{1}{2}} = 3^{\frac{1}{2}}e^{2t*\frac{1}{2}} $$ That's how the \(\sqrt{3}\) comes out. The \(e^5-e\) results from integration within bounds: \[ \begin{align*} \int_{1}^{5} e^t dt &= e^t|_{1}^{5}\\ &= e^{5}-e^{1} \end{align*} \]
ahh ok, thanks.

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