How to find the average value of $y = e^x$ between $x = e$ and $x = 2e$? [closed]

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What approach would be ideal in finding the average value of $y = e^x$ between $x = e$ and $x = 2e$?

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2 Answers

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Average value of a function is denoted by: $$\text{Average value of }f(x)=\frac{1}{b-a}\int_a^b f(x) \, dx$$

So applying $f(x)=e^x$ and the bounds $e$ and $2e$ yields \begin{align} \frac{1}{2e-e} \int_e^{2e}e^x \, dx &= \frac{1}{e} \int_e^{2e}e^x \, dx \\ &= \frac{1}{e}e^x \Bigg\vert_e^{2e} \\ &= \frac{1}{e} (e^{2e}-e^e) \\ &=e^{2e-1}-e^{e-1} \end{align}

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The average value of a (continuous) function on an interval $\lbrack a, b \rbrack$ is given by the following formula:

$$\frac{1}{b - a} \int_a^b f(x) \ \text{d}x$$

This is easy to remember: it's just a continuous version of your discrete arithmetic means, where you sum over the data set and divide by the size of the data set.

So the average value of your function on $\lbrack e, 2e \rbrack$ is:

$$\frac{1}{2e-e} \int_e^{2e} e^x \ \text{d}x$$

Can you go from here?

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