I have a problem that seems really basic, but for whatever reason I keep messing up...
f(r) = 7/(1+r)^2
Find the average value of the function on the interval [1,5]
I let u = 1+r
so du = dx
and integral 7u^-2 du
integrated I get -7u^(-1) and the new interval is [2,6]
-7(6)^-1+7(2)^-1 = 7/3
So the answer I keep coming up with is 7/3 which is wrong evidently........ Can anyone help?
ah, I didn't know I had to multiply it by 1/(b-a) for the average value of a function. So 7/3 * 1/(5-1) = 7/12.....
This problem was probably so basic that you guys didn't bother with it, lol. Ugh, I get stuck on the dumbest things. At least now I'll remember that.