Where does the 1/3 come from in ∫₀¹ e^{-3x} dx?

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Ok I know that the $$\int e^x = e^x + C$$

now I don't understand this problem.

$$\int_0^1 e^{-3x} dx = -(1/3)e^{-3(0)} - -(1/3)e^{-3(1)}$$
$$= 1/3(1- e^{-3} )$$

Where does the 1/3 come from?

 

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anyone?? have to study for a test any help would be greatly appreciated.

I just want to know where the constant -1/3 came from.

 

[JasonRox]

Because...

*I'm using -> to point to the derivative.

e^x -> e^x

e^3x -> 3e^3x

e^-3x -> -3e^-3x

See how the 3 came up? The -1/3 is used to get rid of the -3 because in the integral we don't have a -3.

-1/3e^-3x -> e^-3x

Note: I know it's sloppy, but you should get the idea.

 

[digink]

Because...

*I'm using -> to point to the derivative.

e^x -> e^x

e^3x -> 3e^3x

e^-3x -> -3e^-3x

See how the 3 came up? The -1/3 is used to get rid of the -3 because in the integral we don't have a -3.

-1/3e^-3x -> e^-3x

Note: I know it's sloppy, but you should get the idea.

wow I feel really stupid to forget that lol, I forgot that when $$e^x$$ is raised to something like 3x you use the derivative of that times the original $$e^x$$ function.

thanks :D

 

[JasonRox]

I forget it sometimes to because your so used to copying e^x down as the derivative.

Good Luck!