- #1

- 31

- 0

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

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- Thread starter 7yler
- Start date

- #1

- 31

- 0

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

- #2

- 1,254

- 3

you should simplify the argument first.

- #3

- 354

- 0

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

Code:

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

[tex]\int_0^2 \sqrt{65 e^{2t}} dt[/tex]

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

- #4

- 13,046

- 594

Well, another to put it would be to see that

[tex] e^{2t} = (e^t)^2 [/tex]

which would fit nicely with the square root.

- #5

- 31

- 0

So I would get 65e^2 - 65?

- #6

- 13,046

- 594

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

Last edited:

- #7

- 31

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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.

- #8

- 13,046

- 594

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

- #9

- 31

- 0

Oh okay. Thank you very much.

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