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Chapter 4: Problem 66

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

Short Answer

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

Step by step solution

01

Rewrite the secant function

In order to handle the integration more easily, we rewrite the secant function in terms of cosine, which yields \(f(x)=\frac{1}{\cos(\frac{\pi x}{6})}\)
02

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 \(2 \times 6/\pi\times [2 \tan^{-1}(2\sqrt{2+\sqrt{3}}\tan(\pi x/12))]|_0^2 \)
03

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

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