Question

Consider one arch of \(f(x)=\sin x .\) Show that the average value of \(f(x)\) over the middle third of the arch is twice the average value over the end thirds.

Short answer

The average value of \(f(x)\) over the middle third of the arch is twice the average value over the end thirds.

Step-by-step solution

- Determine the relevant intervals

The function \(f(x) = \sin x\) has an arch from \(x = 0\) to \(x = \pi\).

- Specify the middle third interval

The middle third of the interval \([0, \pi]\) is \(\left[ \frac{\pi}{3}, \frac{2\pi}{3} \right]\).

- Specify the end third intervals

The end thirds of the interval \([0, \pi]\) are \(\left[ 0, \frac{\pi}{3} \right]\) and \(\left[ \frac{2\pi}{3}, \pi \right]\).

- Calculate the average value over the middle third

The average value of \(f(x)\) over \(\left[ \frac{\pi}{3}, \frac{2\pi}{3} \right]\) is given by: \[ \frac{1}{\frac{2\pi}{3} - \frac{\pi}{3}} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \sin x \, dx \]Evaluating the integral: \[ = \frac{1}{\frac{\pi}{3}} \left[ -\cos x \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} = 3 \left( -\cos \frac{2\pi}{3} + \cos \frac{\pi}{3} \right) = 3 \left( -\left( -\frac{1}{2} \right) + \frac{1}{2} \right)\= 3 \left( \frac{1}{2} + \frac{1}{2} \right) = 3 \left( 1 \right) = 3\]So, the average value is \(1\).

- Calculate the average value over the first end third

The average value over \(\left[ 0, \frac{\pi}{3} \right]\) is given by: \[ \frac{1}{\frac{\pi}{3}} \int_{0}^{\frac{\pi}{3}} \sin x \, dx \= 3 \left[-\cos x\right]_{0}^{\frac{\pi}{3}} \= 3 \left( \cos 0 - \cos \frac{\pi}{3} \right)\= 3 \left(1 - \frac{1}{2}\right) \= 3 \cdot \frac{1}{2}\= \frac{3}{2}\]

- Calculate the average value over the second end third

The average value over \(\left[ \frac{2\pi}{3}, \pi \right]\) is given by: \[ \frac{1}{\frac{\pi}{3}} \int_{\frac{2\pi}{3}}^{\pi} \sin x \, dx \= 3 \left[-\cos x\right]_{\frac{2\pi}{3}}^{\pi}\= 3 \left( \cos \frac{2\pi}{3} - \cos \pi \right)\= 3 \left(-\frac{1}{2} - (-1)\right)\= 3 \left(-\frac{1}{2} + 1\right) \= 3 \cdot \frac{1}{2}\= \frac{3}{2}\]

- Compare the average values

Now combine the average values over the end intervals:\[ \frac{3}{2} + \frac{3}{2} = 3 \]The average value over the middle third (1) is half compared to their combination. Therefore, over an end third the average value is \(\frac{1}{2}\). The middle third's average value is twice the end thirds'.

Key concepts

integration

Understanding the concept of integration is fundamental to solving problems involving average values of functions. Integration helps us to find the total area under a curve. For example, when calculating the average value of a function over an interval, you need to integrate the function over that interval.
The formula to find the average value of a function \( f(x) \) over the interval \( [a, b] \) is given by: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] This formula essentially calculates the total area under the function between points \( a \) and \( b \), and then divides by the length of the interval to find the average height of the function over that range.
Breaking down the process:

• Identify the interval \( [a, b] \)
• Calculate the integral \[ \int_{a}^{b} f(x) \, dx \]
• Divide by the interval length \( (b - a) \)

This method is used in the exercise to find average values over specified sections of the sine function.

sin function

The sine function \( \sin x \) is one of the basic trigonometric functions. This function describes a wave-like pattern and is especially useful in modeling periodic phenomena like sound waves.
Key characteristics of the sine function:

• The sine function has a range of \(-1 \) to \( 1 \)
• It has a period of \( 2\pi \), meaning the function repeats every \( 2\pi \) units
• It reaches its maximum value of 1 at \( x = \frac{\pi}{2} + 2\pi k \) and its minimum value of \(-1 \) at \( x = \frac{3\pi}{2} + 2\pi k \)

When calculating the average value of \( \sin x \) over an interval, we are interested in the area under the sine curve within that interval. In the given problem, the different thirds of the interval \([0, \pi]\) represent different sections of the sine wave, each capturing parts of its rise and fall.
Understanding the behavior of the sine function in these segments helps us to conceptualize why the area (and thereby the average value) might differ between the middle and the end thirds.

average value

The essence of the exercise is to find and compare average values of a function over different sections of its domain. The concept of the average value of a function is like finding the mean value in a list of numbers but applies this idea to continuous functions graphed over an interval.
To calculate the average value of \( f(x) = \sin x \) over the middle third \(\frac{\pi}{3} \, \frac{2\pi}{3} \):

• Determine the length of the interval - in this case, \( \frac{\pi}{3} \)
• Find the integral of \( \sin x \) over \(\frac{\pi}{3} \, \frac{2\pi}{3} \)
• Divide the integral result by the length of the interval

Using the same steps, you find the average value over the first and second end thirds. The calculations show that the average value over the middle third (1) is twice as large as the average value over each of the end thirds (0.5 each). Thus, this demonstrates the relationship between these sections and illustrates how different parts of a function can vary in their average values. The comparison helps us understand that the specific shape and behavior of the sine wave cause the middle third to have a higher average value than the individual end thirds.