Question

Find the average value of the function on the given interval. Lise equation (4.8) if it applies. If an average value is zero, you may be able to determine this from a sketch. \(\sin 2 x\) on \(\left(\frac{\pi}{6}, \frac{7 \pi}{6}\right)\)

Short answer

-\frac{1}{2\pi}

Step-by-step solution

- Identify the interval

The given interval is \(\frac{\pi}{6}, \frac{7 \pi}{6}\).

- Recall the formula for average value of a function

The average value of a function \(f(x)\) on the interval \([a, b]\) is given by \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \].

- Set up the integral

For the function \(f(x) = \sin 2x\) on the interval \(\frac{\pi}{6}, \frac{7\pi}{6}\), we have \(\text{Average value} = \frac{1}{\frac{7\pi}{6} - \frac{\pi}{6}} \int_{\frac{\pi}{6}}^{\frac{7\pi}{6}} \sin 2x \, dx\).

- Simplify the interval length

The length of the interval is \[ \frac{7\pi}{6} - \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]. So, the formula now becomes \[ \text{Average value} = \frac{1}{\pi} \int_{\frac{\pi}{6}}^{\frac{7\pi}{6}} \sin 2x \, dx \].

- Integrate \(\sin 2x\)

To solve the integral \(\int \sin 2x \, dx\), use the substitution method. Let \(u = 2x\), so \(du = 2 dx\) or \(dx = \frac{du}{2}\). The integral becomes \[ \int \sin 2x \, dx = \frac{1}{2} \int \sin u \, du = \frac{-1}{2} \cos u + C = \frac{-1}{2} \cos (2x) + C \].

- Evaluate the definite integral

Now evaluate \(\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}} \sin 2x \, dx\) by substituting the limits: \[ \left[ \frac{-1}{2} \cos (2x) \right]_{\frac{\pi}{6}}^{\frac{7\pi}{6}} = \frac{-1}{2} \left( \cos (2(\frac{7\pi}{6})) - \cos (2(\frac{\pi}{6})) \right) \]. Simplifying further: \[ = \frac{-1}{2} \left( \cos \left( \frac{7\pi}{3} \right) - \cos \left( \frac{2\pi}{3} \right) \right) = \frac{-1}{2} \left( \cos \left( \frac{\pi}{3} \right) - \cos \left( \frac{2\pi}{3} \right) \right) \].

- Simplify using trig values

Recall that \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}\) and \(\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}\). Substitute these values in: \[ \frac{-1}{2} \left( \frac{1}{2} - ( -\frac{1}{2}) \right) = \frac{-1}{2} \left( \frac{1}{2} + \frac{1}{2} \right) = \frac{-1}{2} \cdot 1 = -\frac{1}{2}. \].

- Calculate the average value

Finally, compute the average value using the formula:\[ \text{Average value} = \frac{1}{\pi} \cdot -\frac{1}{2} = -\frac{1}{2\pi}. \]

Key concepts

average value formula

To find the average value of a function over an interval, we use the average value formula. This formula is essential for integrating functions over a given range. The formula can be written as: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] Here, \( a \) and \( b \) are the beginning and end of the interval, respectively. \( f(x) \) is the function you are calculating the average for. Breaking down the formula:

• \( \frac{1}{b-a} \) is the factor that ensures we average over the whole interval length.
• \( \int_{a}^{b} f(x) \, dx \) represents the definite integral of the function from \( a \) to \( b \).

This approach gives you a simple yet powerful tool to find out the overall behavior of a function over any given interval.

integration

Integration is a fundamental concept in calculus that is used to find areas under curves, among other things. In our specific case, we needed to integrate the function \( \sin 2x \) over the interval \( \left( \frac{\pi}{6}, \frac{7 \pi}{6} \right) \). The integral of a function like \( \sin 2x \) is quite straightforward but can involve certain techniques to simplify the process. To solve the integral \( \int \sin 2x \, dx \), you follow these steps:

• Set up the integral with the given limits \( \left( \frac{\pi}{6}, \frac{7 \pi}{6} \right) \).
• Use substitution to simplify the integral if needed.
• Evaluate the integral over the limits.

These steps help you methodically break down the problem and find the definite integral value, which is essential for applying the average value formula.

trigonometric functions

Trigonometric functions like sine, cosine, and tangent are vital in many areas of mathematics, including calculus. In our exercise, the function we are dealing with is \( \sin 2x \). Trigonometric functions often involve specific properties and values that come in handy during integration. For instance, in our example, we utilized the specific values of the cosine function:

• \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \)
• \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \)

Knowing these values helps simplify the integral evaluation and ensures that the calculations are accurate. Becoming familiar with the key properties of trig functions can significantly ease solving such problems.

substitution method

The substitution method is a powerful technique for simplifying integrals by transforming them into an easier form to integrate. In our exercise, we used substitution to handle the integral \( \int \sin 2x \, dx \). By substituting \( u = 2x \), we could transform the integral: \[ dx = \frac{du}{2} \] \[ \frac{1}{2} \int \sin u \, du \] The substitution simplifies the integration process because it converts it into a simpler integral form. Once the integral is solved, you can then substitute back to the original variable to conclude the solution. This method is particularly useful when dealing with more complicated functions or when a direct approach to integration is challenging.