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the average (mean) value of 1/x over the interval 1 less than or equal to x less than or equal to "e" is?
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(1+e)/2
The average of a function \(f(x)\) in a given interval is \([a,b]\) is given by \(\frac{1}{b-a} \int_a^b f(x)dx\).
In this case the average is \(\frac{1}{e-1} \int_1^e \frac{dx}{x}=\frac{1}{e-1} (\ln e-\ln 1)=\frac{1}{e-1}.\)
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