Question

In Exercises \(31 - 36 ,\) find the average value of the function on the interval, using antiderivatives to compute the integral. $$y = \frac { 1 } { x } , \quad [ e , 2 e ]$$

Short answer

The average value of the function \( y = \frac { 1 } { x } \) on the interval \([ e , 2 e ]\) is \( ln|2| \)

Step-by-step solution

Write Down the Average Value Formula

The formula for the average value of the function on the interval \([a, b]\) is \( \frac{1}{b-a} \int_{a}^{b} f(x) dx \). On applying this formula to the problem, the formula becomes: \( \frac{1}{2e - e} \int_{e}^{2e} \frac {1}{x} dx \) which simplifies to \( \int_{e}^{2e} \frac {dx}{x} \)

Use the Antiderivative to Compute the Integral

The antiderivative of \( \frac {1}{x} \) is \( ln|x| \). So, the integral can be rewritten as: \( \int_{e}^{2e} d(ln|x|) \). Evaluate the integral using the limits of integration, which gives us: \( ln|2e| - ln|e| \)

Simplify the Result

Simplify the result from the previous step: \( ln|2| + ln|e| - ln|e| \). This simplifies to \( ln|2| \)