find the average value of the function (using integration) over the given interval `f(x)=e^(-3x) ` over `[0,12]`

Please provide exact answer, in improper fractions if necessary.

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To get the average value of a function in an interval [a,b], apply the formula:

`f_(avg)=1/(b-a)int_a^b f(x) dx`

So, plug-in `f(x)=e^(-3x)` .

`f_(avg)= 1/(b-a)int_a^b e^(-3x)dx`

Since the given interval is [0,12], then plug-in a=0 and b=12.

`f_(avg)=1/(12-0) int_0^12 e^(-3x)dx`

`f_(avg)=1/12int_0^12 e^(-3x) dx`

To integrate, apply the formula `int e^u du= e^u + C` .

`f_(avg)=1/12 * e^(-3x)/(-3)` `|_0^12`

`f_(avg)=-1/36e^(-3x)` `|_0^12`

`f_(avg)=-1/36(e^(-3*12)-e^(-3*0))`

`f_(avg)=-1/36 (e^(-36)-1)`

`f_(avg)=-e^(-36)/36 + 1/36`

`f_(avg)=1/36-e^(-36)/36`

**Hence, the average value of the given function for the interval [0,12] is `1/36-e^(-36)/36` .**

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