Question

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. $$x-\cos ^{2} 6 x \text { on }\left(0, \frac{\pi}{6}\right)$$

Short answer

\( \frac{\pi - 6}{12} \)

Step-by-step solution

Identify the function and interval

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

Set up the average value formula

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 the function and interval into the formula

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

Simplify the expression

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

Solve the integral

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: \[ \int_{0}^{\frac{\pi}{6}} (x - \cos^2(6x)) \, dx = \frac{\pi^2}{72} - \frac{\pi}{12} = \frac{\pi(\pi - 6)}{72} \]

Calculate the average value

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

Key concepts

Integral

Finally, we combine both integrals with:
$$\frac{\theta}\theta

Interval

Intervals are essential in various practical applications. They help in calculating distances, areas, and even averages in contexts ranging from physics to economics.
In calculus, intervals delimit the precise scope of our calculations, whether we are finding areas under curves or solving differential equations.
Understanding intervals is fundamental to solving real-world mathematical problems accurately.

Trigonometric Identity

Using the identity we rewrote the integral:
$$\1\cos(12x) = \frac{1 + \cos(12x)}{2}$$
They then broke it down to manageable parts:
$$frac{1}{2}\1\textit(dx$$
Integrating, simplified our task significantly