# Average value of a curve

1. Jan 22, 2006

### perryben

If I had a sinusoid, how would I find the average value of it over a given interval. Say -pi/5 to pi/5 for instance. Thanks everybody.

2. Jan 22, 2006

### benorin

The average value of a function, say $$f(x),$$ over the interval [a,b] is given by the the formula

$$f_{\mbox{ave}}=\frac{1}{b-a}\int_{x=a}^{b}f(x)dx$$

where I have assumed that $$f(x),$$ is properly integrable over [a,b].

3. Jan 23, 2006

### HallsofIvy

The point is: if you had a constant function, f(x)= c, the "area under the curve" from a to b would f(x)(b-a)= c(b-a). With a variable function, that area is $$\int_a^b f(x)dx$$.

If fave is the average of the function we must have
$$\int_a^b f(x)dx= f_{ave}(b-a)$$

4. Jan 23, 2006

### mathman

The sine function is odd. Therefore the average over an interval symmetric around 0 will be 0.