Find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the \(x\) axis are the same. $$x-\cos ^{2} 6 x \text { on }\left(0, \frac{\pi}{6}\right)$$

The function given is \(f(x) = x - \cos^2(6x)\). The interval provided is \(\left(0, \frac{\pi}{6}\right)\).

The formula for the average value of a function \(f(x)\) on the interval \([a, b]\) is \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. In this particular problem, \(a = 0\) and \(b = \frac{\pi}{6}\).

Substitute \(f(x) = x - \cos^2(6x)\), \(a = 0\), and \(b = \frac{\pi}{6}\) into the average value formula: \[ f_{avg} = \frac{1}{\frac{\pi}{6} - 0} \int_{0}^{\frac{\pi}{6}} \left(x - \cos^2(6x)\right) \, dx \]

First simplify the denominator: \[ \frac{1}{\frac{\pi}{6}} = \frac{6}{\pi} \]The new equation becomes: \[ f_{avg} = \frac{6}{\pi} \int_{0}^{\frac{\pi}{6}} \left(x - \cos^2(6x)\right) \, dx \]

Now, evaluate the integral: \[ \int_{0}^{\frac{\pi}{6}} \left(x - \cos^2(6x)\right) \, dx = \int_{0}^{\frac{\pi}{6}} x \, dx - \int_{0}^{\frac{\pi}{6}} \cos^2(6x) \, dx \] The first integral is straightforward: \[ \int_{0}^{\frac{\pi}{6}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{6}} = \frac{\left(\frac{\pi}{6}\right)^2}{2} = \frac{\pi^2}{72} \] Solving the second integral requires the trigonometric identity \(\cos^2(6x) = \frac{1 + \cos(12x)}{2}\): \[ \int_{0}^{\frac{\pi}{6}} \cos^2(6x) \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{6}} (1 + \cos(12x)) \, dx = \frac{1}{2} \left[ x + \frac{\sin(12x)}{12} \right]_{0}^{\frac{\pi}{6}} = \frac{1}{2} \left[ \frac{\pi}{6} + \frac{\sin(2\pi)}{12} \right] = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12} \] Putting it all together, we have: \[ \int_{0}^{\frac{\pi}{6}} (x - \cos^2(6x)) \, dx = \frac{\pi^2}{72} - \frac{\pi}{12} = \frac{\pi(\pi - 6)}{72} \]

Finally, substitute the evaluated integral back into the average value formula: \[ f_{avg} = \frac{6}{\pi} \cdot \frac{\pi(\pi - 6)}{72} = \frac{6(\pi - 6)}{72} = \frac{\pi - 6}{12} \]