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

1. anonymous

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 $$\int e^t dt = e^t$$.

2. anonymous

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

3. anonymous

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?

4. anonymous

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*}

5. anonymous

ahh ok, thanks.