- #1

- 279

- 0

## Homework Statement

[tex]\int2e^-^7^xdx[/tex]

## Homework Equations

None

## The Attempt at a Solution

[tex](\frac{-2}{7})(\frac{e^-^7^x}{-7})+C[/tex]

This is as far as I can go, but the answer is:

[tex]\frac{-2e^-^7^x}{7}+C[/tex]

- Thread starter temaire
- Start date

- #1

- 279

- 0

[tex]\int2e^-^7^xdx[/tex]

None

[tex](\frac{-2}{7})(\frac{e^-^7^x}{-7})+C[/tex]

This is as far as I can go, but the answer is:

[tex]\frac{-2e^-^7^x}{7}+C[/tex]

- #2

- 1,752

- 1

How did you get two 1/7 terms?

- #3

- 279

- 0

I took the antiderivative of [tex]e^-^7^x[/tex].

- #4

- 1,752

- 1

So, why do you have 1/7 * 1/7? Just do it again from scratch and you'll probably see what you did wrong.I took the antiderivative of [tex]e^-^7^x[/tex].

- #5

- 1,752

- 1

[tex]-\frac 2 7\int -7e^{-7x}dx[/tex]

You don't need to divide by another -7. It's already in standard form!

- #6

- 279

- 0

I've got it now, thanks.

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