calculus - how...?

Question

how do you find the average value for y = (some equation) on some interval [a,b]

Answer 1

I suppose you simply calculate the area and divide it by the length of the interval,

so you work out A = ∫ y.dx, then the average value for y is A/(b-a).

You are essentially changing the original area into a reactangle of width b - a and height "average y", which has exactly the same area as under the original curve.

Answer 2

for an equation
y=f(x)

The average value over the interval [a, b] is the same thing as the secant line from a to b.

This is expressed as (f(b)-f(a)) / (b-a). All you're doing is finding the slope of the line between point a and b

After you have this slope value you need to take the derivative of f(x)

Set f'(x) = to the slope of the secant line

Solve for x. This is known as Mean's Value theorem. Make sure that f(x) is continious and differentable on the interval (a, b)

Answer 3

The average value of y= f(x) over the interval a [f(b) -f(a)}/(b-a). It's that simple.

Answer 4

If y = f(x), then the expected value is ∫x*f(x)dx . Integrate over the interval [a,b].

Answer 5

You would find the area between the curve and the x-axis, and find some y for which y*(b-a)=area

Answer 6

integrate f(x) over [a, b], then divide by [b - a]