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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First 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 \(\int e^t dt = e^t\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ahh right right cause square root is ^(1/2) . Thank ya.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uh... 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 \(e^5e\) results from integration within bounds: \[ \begin{align*} \int_{1}^{5} e^t dt &= e^t_{1}^{5}\\ &= e^{5}e^{1} \end{align*} \]
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