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?

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Replies to This Discussion

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. 

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