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Consider the curve y=f(x)=(2^x)-1 Find a value for a such that the average value of the function f(x) on the interval [0,a] is equal to 1.
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(1/a-0) ∫ [0,a] (2^x-1)dx = 1 I do not know how to solve for a
did you apply the value a
I found the antiderivative and I applied the first fundamental theorem of calculus. I'm getting stuck with simplification. 1/a[(2^a/ln(2))-a)-(2^0/ln(2))]
ok
I've gotten as far as: 2^a-aln(2)-1/aln(2)=1
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that looks right so far
2^a-1=-2aln(2)
we can really only approximate it
a=1.812647074
Yeah, that's what I got. So there's easy no way to simplify?
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no, not really.
unless you want to leave your answer in terms of the Lambert function
No, I think this is good. Thanks for the help.
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