Integrating \int_0^2 \sqrt{65e^{2t}} dt for Beginners

[7yler]

I am familiar with integration, but I'm stumped with this. How do you integrate something such as $$\int$$(from 0 to 2)$$\sqrt{65e(to the power of 2t}$$dt

That is supposed to be the square root of 65e to the power of 2t.

 

[zhermes]

you should simplify the argument first.

 

[foxjwill]

Here's how you'd write your integral using LaTeX:

\int_0^2 \sqrt{65 e^{2t}} dt

which, when enclosed by "TEX" and "/TEX" (with the quotation marks replaced by square brackets), gives

$$\int_0^2 \sqrt{65 e^{2t}} dt$$​

Anyway, try using the fact that $\sqrt{a}=a^{1/2}$.

 

[dextercioby]

Well, another to put it would be to see that

$$e^{2t} = (e^t)^2$$

which would fit nicely with the square root.

 

[7yler]

So I would get 65e^2 - 65?

 

[dextercioby]

Well, the 65 is still under a radical. Only e^t is squared under the square root.

 

[7yler]

So 65(e^2) - 65 is what I get when I follow through with the calculations, but I'm told that that answer is incorrect.

 

[dextercioby]

You didn't understand. The 65 must be under the radical sign, as in $\sqrt{65}$.

 

[7yler]

Oh okay. Thank you very much.