# Average Value

I forgot the formula for the average value of a function.
Is that $$f(b) - f(a) = f'(c)(b-a)$$ or am I thinking of something else?

Try:

$$\frac{\int_a^b f(x)\,dx}{b-a}}$$

Just remember that you're turning the area under the curve into a rectangle of base (b - a). The height is the average value.

okay, the actual problem now becomes
$$\frac{\int_{-1}^{1} e^{-x^2} \,dx}{-2}$$
any takers?

You've got an extra negative in the denominator.

The expression can be simplified at most to:

$$\int_0^1 e^{-x^2}\,dx$$

Which must be integrated numerically. Mathematica gives:

$$\int_0^1 e^{-x^2}\,dx = .746824$$