Problem 115 - Calculus Volume 1

Question

In the following exercises, find the average value \(f_{\text { ave }}\) of \(f\) between \(a\) and \(b,\) and find a point \(c,\) where \(f(c)=f_{\text { ave. }}\) \(f(x)=\cos x, a=0, b=2 \pi\)

Step-by-Step Solution

Answer

The average value of \(\cos x\) from 0 to \(2\pi\) is 0, occurring at \(c = \frac{\pi}{2}\) and \(c = \frac{3\pi}{2}\).

Step 1: Understand the Formula for Average Value

The average value of a function \(f\) over the interval \([a, b]\) is given by the formula:\[f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] For our problem, \(f(x) = \cos x\), \(a = 0\), and \(b = 2\pi\).

Step 2: Set Up the Integral

Substitute the function and limits into the formula for the average value:\[f_{\text{ave}} = \frac{1}{2\pi - 0} \int_{0}^{2\pi} \cos x \, dx \] which simplifies to:\[f_{\text{ave}} = \frac{1}{2\pi} \int_{0}^{2\pi} \cos x \, dx \]

Step 3: Compute the Integral

Find the integral of \(\cos x\):\[\int \cos x \, dx = \sin x + C\] Now, compute the definite integral from \(0\) to \(2\pi\):\[\int_{0}^{2\pi} \cos x \, dx = \left[ \sin x \right]_{0}^{2\pi} = \sin(2\pi) - \sin(0) = 0\]

Step 4: Calculate the Average Value of the Function

Substitute the result of the integral back into the equation for \(f_{\text{ave}}\):\[f_{\text{ave}} = \frac{1}{2\pi} \times 0 = 0\] Thus, the average value of \(\cos x\) from \(0\) to \(2\pi\) is 0.

Step 5: Find the Point \(c\) Where \(f(c) = f_{\text{ave}}\)

Set \(f(c) = \cos c = f_{\text{ave}} = 0\).Solve for \(c\):\[\cos c = 0\] The solutions on the interval \([0, 2\pi]\) are \(c = \frac{\pi}{2}\) and \(c = \frac{3\pi}{2}\).

Key Concepts

Definite Integrals

Definite integrals are a powerful mathematical tool used to calculate the accumulation of quantities over an interval. They are different from indefinite integrals because they provide a numerical value rather than a general function expression.

The definite integral of a function from a to b is represented as \[ \int_{a}^{b} f(x) \, dx \] and it measures the net area between the function and the x-axis over that interval. In our specific example with the function \( f(x) = \cos x \), we calculate its integral from 0 to \( 2\pi \) to find the average value.

Here, definite integrals help find this average by measuring the total "spread" of the cosine function over the interval \([0, 2\pi]\). Once integrated, this spread is scaled by dividing by the interval length, which provides the average value over that range.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, representing the relationships between the angles and sides of triangles. In the context of functions, they often model periodic phenomena, such as the cycles of the day or waves.

The cosine function, \( \cos x \), is a basic trig function and it is periodic, meaning it repeats its values at regular intervals.

• Its period is \( 2\pi \), which means that it completes one full cycle over this interval.
• It ranges between -1 and 1, providing smooth oscillations.

In average value problems like this one, the periodic nature of \( \cos x \) allows us to use integrals to find the mean of its values over a full cycle plus any additional cycles if needed. Understanding the properties of \( \cos x \) helps us interpret its average value accurately over different intervals.

Zero Values of Cosine

Finding where a trigonometric function like cosine equals zero is an important concept in trigonometry. These values indicate the points where the waveform crosses the x-axis.

For the cosine function \( \cos x \), the zeros occur at specific points where the angle \( x \) translates to the cosine value equalling zero. Within the interval \([0, 2\pi]\), cosine zeros are found at:

• \( x = \frac{\pi}{2} \)
• \( x = \frac{3\pi}{2} \)

These points are derived from the properties of cosine function, which shifts between positive and negative values about the zero line.

When determining where the function equals its average value, understanding the locations of zero values is crucial. In this example, it helped identify the points \( c = \frac{\pi}{2} \) and \( c = \frac{3\pi}{2} \) where the cosine function's value matches the calculated average of zero.