Problem 66 - Essential Calculus

Question

Find the average value of the function over the given interval. $$ f(x)=\sec \frac{\pi x}{6}, \quad[0,2] $$

Step-by-Step Solution

Answer

The average value of the function \(f(x)=\sec \frac{\pi x}{6}\) over the interval [0, 2] is approximately 1.4545.

Step 1: Rewrite the secant function

In order to handle the integration more easily, we rewrite the secant function in terms of cosine, which yields \(f(x)=\frac{1}{\cos(\frac{\pi x}{6})}\)

Step 2: Calculate the integral

Apply the formula for the average value of a function on an interval: \(\frac{1}{b - a} \int_a^b f(x) dx\). Plugging in our function and the limits of [0, 2] we get: \(\frac{1}{2-0} \int_0^2 \frac{1}{\cos(\frac{\pi x}{6})} dx\). The result of this integral is \(2 \times 6/\pi\times [2 \tan^{-1}(2\sqrt{2+\sqrt{3}}\tan(\pi x/12))]|_0^2 \)

Step 3: Evaluate the result

Replace x in the result of the integral with the limits 0 and 2, then make the subtraction: \(2 \times 6/\pi\times [2 \tan^{-1}(2\sqrt{2+\sqrt{3}}\tan(\pi x/12))]|_0^2 = 2 \times 6/\pi\times [2 \tan^{-1}(2\sqrt{2+\sqrt{3}}\tan(\pi \times 2/12)) - 2 \tan^{-1}(2\sqrt{2+\sqrt{3}}\tan(\pi \times 0/12))]. Simplify the expression to get the final result, which is approximately 1.4545