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Mathematics 9 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

(1/a-0) ∫ [0,a] (2^x-1)dx = 1 I do not know how to solve for a

OpenStudy (perl):

did you apply the value a

OpenStudy (anonymous):

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))]

OpenStudy (perl):

ok

OpenStudy (anonymous):

I've gotten as far as: 2^a-aln(2)-1/aln(2)=1

OpenStudy (perl):

that looks right so far

OpenStudy (anonymous):

2^a-1=-2aln(2)

OpenStudy (perl):

we can really only approximate it

OpenStudy (perl):

a=1.812647074

OpenStudy (anonymous):

Yeah, that's what I got. So there's easy no way to simplify?

OpenStudy (perl):

no, not really.

OpenStudy (perl):

unless you want to leave your answer in terms of the Lambert function

OpenStudy (anonymous):

No, I think this is good. Thanks for the help.

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