Question

Find the mean value of the function on each of the indicated closed intervals. (a) \(f(x)=\sin x\) on \([0, \pi],[0,2 \pi],[\alpha, \alpha+2 \pi]\) \([\alpha, \alpha+\pi]\) (b) \(f(x)=\operatorname{sgn} x\) on \([-2,-1][-2,1],[-1,3],[-2,2]\), \([1,2]\). (c) Is the mean value of the function on each intervalone of the values of the function on that interval? Explain why in some cases the answer in positive and in the other negative.

Short answer

Answer: For the function \(f(x) = \sin x\), the mean value is one of the values of the function in each of the intervals \([0, \pi]\), \([0, 2\pi]\), \([\alpha, \alpha + 2\pi]\), and \([\alpha, \alpha + \pi]\). However, for the function \(f(x) = \operatorname{sgn}(x)\), the mean value is not one of the values of the function in the intervals \([-2, 1]\), \([-1, 3]\), and \([-2, 2]\) as the mean value 0 only occurs at single points; but it is one of the values of the function in intervals \([-2, -1]\) and \([1, 2]\) where the function is constant.

Step-by-step solution

Mean Value formulation

Mean value of a function \(f(x)\) on an interval \([a, b]\) can be calculated using the following formula: $$ \text{Mean value} = \frac{1}{b-a} \int_a^b f(x) dx $$ Now let's apply this formula for the given function \(f(x) = \sin x\) on each of the closed intervals.

Calculate Mean value on \([0, \pi]\)

Using the formula as mentioned earlier, we get $$ \text{Mean value} = \frac{1}{\pi - 0} \int_0^\pi \sin(x) dx = \frac{1}{\pi} \left[-\cos(\pi) + \cos(0)\right] = \frac{2}{\pi} $$

Calculate Mean value on \([0, 2\pi]\)

Now let's calculate the mean value on the interval \([0, 2\pi]\). $$ \text{Mean value} = \frac{1}{2\pi - 0} \int_0^{2\pi} \sin(x) dx = \frac{1}{2\pi} \left[-\cos(2\pi) + \cos(0)\right] = 0 $$

Calculate Mean value on \([\alpha, \alpha + 2\pi]\) and \([\alpha, \alpha + \pi]\)

Now let's calculating the mean value on the interval \([\alpha, \alpha + 2\pi]\) and \([\alpha, \alpha + \pi]\). $$ \text{Mean value on } [\alpha, \alpha + 2\pi] = \frac{1}{2\pi} \left[-\cos(\alpha + 2\pi) + \cos(\alpha)\right] = 0 $$ $$ \text{Mean value on } [\alpha, \alpha + \pi] = \frac{1}{\pi} \left[-\cos(\alpha + \pi) + \cos(\alpha)\right] = \frac{2\cos(\alpha)}{\pi} $$ (b) The function \(f(x) = \operatorname{sgn}(x)\) has only three values, -1, 0, and 1. We can apply the mean value formula directly as the function is constant in each interval.

Mean value of \(f(x) = \operatorname{sgn}(x)\) for all the intervals

We can directly write the mean values of the function for given intervals: \([-2,-1]\): Mean value = -1 \([-2,1]\): Mean value = 0 \([-1,3]\): Mean value = 0 \([-2,2]\): Mean value = 0 \([1,2]\): Mean value = 1 (c) Now we need to discuss if the mean value of the functions on the given interval, is one of the values of the function on that interval or not. For the function \(f(x)=\sin x\), the mean value is one of the values of the function in each of the intervals as we can find some point in the interval where the function takes this value. However, for the function \(f(x)=\operatorname{sgn}(x)\), in the intervals \([-2,1]\), \([-1,3]\), and \([-2,2]\) the mean value is not one of the values of the function because the function only takes values from the set \{-1,0,1\}, and the mean value 0 only occurs at a single point in each interval and does not represent the true average behavior of the function within that interval. Therefore, for these intervals, the mean value is not one of the values of the function on that interval. On the other hand, in the intervals \([-2,-1]\) and \([1,2]\), the mean value is one of the values of the function as the function is constant in each interval, and the mean value is equal to that constant value.

Key concepts

Integral Calculus

Integral calculus, a core component of mathematical analysis, is essential for computing areas, volumes, and for solving problems involving the accumulation of quantities such as distance or mass. In the context of finding the mean value of a function over an interval, integral calculus plays a pivotal role.

The mean value is essentially an average of the function's values over a range. It's expressed as the integral of the function over the interval divided by the length of the interval, which is precisely what the formulas in the textbook solution illustrate. As an example, when evaluating the mean value of the sine function over various intervals, one must integrate the function and then divide the result by the interval's length to find the average value. This process showcases the power of integral calculus in summarizing the behavior of a function across a domain into a single representative number.

Mean Value Theorem

The Mean Value Theorem for integrals states that if a function is continuous on a closed interval, there is at least one point in this interval at which the function takes its average (mean) value. The theorem essentially guarantees that the average height of the graph of the function over an interval is the height of the graph at some point in the interval.

This theorem gives a strong justification for the observations made in part (c) of the exercise. For the continuous sine function, the theorem assures us that the mean value calculated using the integral is actually a value that the function takes at some point in each specified interval. However, the theorem does not apply to functions with discontinuities within the interval like the 'sign function'. As a result, the mean value might not always be a value that the function takes on the interval, especially when the function's profile varies significantly.

Sign Function

The sign function, often denoted as \( f(x) = \operatorname{sgn}(x) \), is a mathematical function that extracts the sign of a real number. Its values are -1, 0, or 1 depending on whether x is negative, zero, or positive. Unlike continuous functions, which smoothly transition through their values, the sign function has abrupt jumps from one value to another, creating discontinuities at the points where it changes value.

In part (b) of the solution, it's noted that for certain intervals, the mean value of the sign function is not necessarily a value that the function takes within that interval. This is attributed to the discontinuous nature of the sign function, which causes it to behave differently from continuous functions when considering mean values. The distinct characteristic of the sign function—being piecewise constant with abrupt changes—makes it an interesting case study in the application of integral calculus and the Mean Value Theorem.