# Simply then integrate

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:
Code:
\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 $\sqrt{a}=a^{1/2}$.

Homework Helper

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?

Homework Helper

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

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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.

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