Problem 3 - Calculus: Early Transcendentals

Question

Find the average value of the function on the given interval. \( g(x) = 3 \cos x \) , \( [\frac{-\pi}{2}, \frac{\pi}{2}] \)

Step-by-Step Solution

Answer

The average value is \( \frac{6}{\pi} \).

Step 1: Understand the Average Value Formula

The average value of a function \( g(x) \) on the interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \). For our function \( g(x) = 3 \cos x \) and interval \( \left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \), we identify \( a = \frac{-\pi}{2} \) and \( b = \frac{\pi}{2} \).

Step 2: Set up the Integral

We need to set up the integral \( \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 3 \cos x \, dx \). This will help us find the total area under the curve from \( a \) to \( b \).

Step 3: Calculate the Integral

To calculate \( \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 3 \cos x \, dx \), we use the antiderivative of \( 3 \cos x \), which is \( 3 \sin x \). Thus, we have:\[ \left. 3 \sin x \right|_{-\pi/2}^{\pi/2} = 3 \sin \frac{\pi}{2} - 3 \sin \left(\frac{-\pi}{2}\right) \]\[ = 3(1) - 3(-1) = 3 + 3 = 6 \]

Step 4: Apply the Average Value Formula

Now that we have calculated the integral to be \( 6 \), we apply the average value formula. The length of the interval \( [\frac{-\pi}{2}, \frac{\pi}{2}] \) is \( \pi \). Therefore, the average value is:\[ \frac{1}{\pi} \times 6 = \frac{6}{\pi} \]

Step 5: Conclusion

The average value of the function \( g(x) = 3 \cos x \) on the interval \([\frac{-\pi}{2}, \frac{\pi}{2}]\) is \( \frac{6}{\pi} \).

Key Concepts

Understanding Integrals

Integrals are a fundamental concept in calculus, often used to find areas under curves. \( \int g(x) \, dx \) symbolizes the process of integration. When calculating the average value of a function over a specific interval, integrals help us compute the total accumulated value of the function within that range.
The integral \( \int_{a}^{b} g(x) \, dx \) gives the total area beneath the function \( g(x) \) from \( x=a \) to \( x=b \). In the context of the exercise, integrating \( 3 \cos x \) across the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) allows us to determine the sum of all values of the function within this interval. This integral evaluates to 6, indicating the total area under \( 3 \cos x \) from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
It is crucial to set up the integral correctly based on the function and given limits to ensure accurate calculations.

Exploring Trigonometric Functions

Trigonometric functions, such as \( \cos x \), play a vital role in mathematics, particularly in analyzing periodic phenomena. Functions like sine and cosine are periodic, meaning they repeat their values in regular intervals.
In this exercise, \( g(x) = 3 \cos x \), acts as the function under investigation. The factor of 3 stretches the cosine wave vertically, meaning it reaches three times as high or as low as the standard cosine wave.
The cosine function itself varies between -1 and 1. Within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\), cosine starts at 0, rises to 1, and returns to 0. The adjustment \(3 \cos x\) means it peaks at 3 and descends to -3. Understanding these characteristics is key when visualizing how the function behaves on a specific interval and its impact on calculus operations like integration.

Discovering Antiderivatives

Antiderivatives, or indefinite integrals, reverse the process of differentiation, helping find a function whose derivative is the given function. In the exercise, finding the antiderivative of \( 3 \cos x \) is crucial to solving the integral \( \int \).
The antiderivative of \( \cos x \) is \( \sin x \), leading us to determine that the antiderivative of \( 3 \cos x \) is \( 3 \sin x \). This antiderivative is written with a constant of integration typically considered in indefinite integrals, but here it's not needed due to definite limits.
Calculating the definite integral, we evaluate \( 3 \sin x \) at the upper and lower bounds \( \frac{\pi}{2} \) and \(-\frac{\pi}{2} \), respectively, resulting in the value 6. This step is essential for analyzing the behavior of functions over intervals, which supports determining the average value.