Calculate he average over the given interval: f(x)=e^(-n*x), [-1,1].

I know that the average value = (1/b-a)integrate from a to b f(x)dx

You have stated the definition of the average value correctly. All you have to do is perform the integration and divide the integral by the integration interval, 2.

The indefinite integral of e^(-nx) is (-1/n)*e^(-nx).

The definite integral from -1 to 1 is
(-1/n)[e^-n - e^n]

Divide that by 2 for the average value.

Check my math. I am often sloppy.

thanks, but can we leave it in that form?

You can convert my answer to a hyperbolic sine of n, divided by 2, if you wish.

I meant "divided by n", not 2

What is hyperbolic sin of n? Never heard of that.

To calculate the average value of the function f(x) = e^(-n*x) over the interval [-1,1], you can follow these steps:

1. Plug in the values for a, b, and f(x) into the formula for calculating the average value:

Average = (1/(b - a)) * ∫(a to b) f(x) dx

2. Determine the values for a and b in the given interval. In this case, a = -1 and b = 1.

3. Substitute the function f(x) = e^(-n*x) into the integral:

Average = (1/(1 - (-1))) * ∫(-1 to 1) e^(-n*x) dx

4. Evaluate the integral ∫ e^(-n*x) dx. Since this is a standard integral, the antiderivative of e^(-n*x) is (-1/n) * e^(-n*x):

Average = (1/2) * [(-1/n) * e^(-n*x)] evaluated from -1 to 1

5. Evaluate the expression at the upper limit (1) and subtract the value evaluated at the lower limit (-1):

Average = (1/2) * [(-1/n) * e^(-n*1) - (-1/n) * e^(-n*(-1))]

6. Simplify the expression:

Average = (1/2) * [(-1/n) * e^(-n) - (-1/n) * e^n]

Average = (-1/2n) * [e^(-n) - e^n]

Therefore, the average value of the function f(x) = e^(-n*x) over the interval [-1,1] is (-1/2n) * [e^(-n) - e^n].