Find the average value of the function f over the interval..

1) Find the average value of the function f over the interval [0, 3].
f(x) = 5e^-x

2) Find the average value of the function f over the interval [-1, 2].
f(x) = 1-x^2

3) Find the area of the region under the graph of f on [-1, 12].
f(x) = e−x/2


I really don't get this. Someone help please.

To find the requested average, integrate f(x) over the interval; then divide
by the length of the interval. [You could think of the process as representing
the integral area as a rectangle.]

1) the integration gives 5[ (-1/e^3) +1] ≈4.75106; so the average over the
interval is ≈1.584 (from dividing by 3, the interval length).

2) your integral over [-1, 2] ends up zero. So your average function value is 0.

3) the integral over [-1, 12] is ≈ 3.2925; so the average over the interval is
abt 0.253268 [belatedly I notice you didN'T ask for an average on this one, just
the area; so you're after the 3.29 figure, not the quotient from dividing by 13.]

The average value of some function y = f(x) on the interval [a,b] is defined as:

1/(b - a) * ∫f(x) dx

1.) Avg. value = 1/3 * ∫5e^-x dx from 0 to 3

= 5/3*(1 - e^-3)

2.) Avg. value = 1/3 * (x - (x^3)/3 eval. from - 1 to 2)

= 0

1.
f(x) = 5e^-x
(f(3)-(0))/3
=(5e^-3 -5)/3
=(5/3)(e^-3 -5)

2.
(f(2)-f(1))/(2-1)
=(-3-0)/1
=-3