Question

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

Answer 1

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 \(\int e^t dt = e^t\).

Answer 2

Ahh right right cause square root is ^(1/2) . Thank ya.

Answer 3

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?

Answer 4

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:
\[
\begin{align*}
\int_{1}^{5} e^t dt &= e^t|_{1}^{5}\\
&= e^{5}-e^{1}
\end{align*}
\]

Answer 5

ahh ok, thanks.