The average value of $f(x)=\dfrac{1}{x}$ from $x=1$ to $x=e$ is

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To find the average value:

$$\frac{1}{e-1}\int\limits_1^e \frac{1}{x}   dx$$ $$= \frac{1}{e-1}\Big[\ln x \Big]_1^e$$ $$= \frac{1}{e-1}(\ln e - \ln 1)$$ $$= \frac{1}{e-1}(1-0) = \boxed{\frac{1}{e-1}}$$