Question

In Exercises \(43-48\), find the average value of the function over the given interval and all values of \(x\) in the interval for which the function equals its average value. $$ f(x)=4-x^{2}, \quad[-2,2] $$

Short answer

The average value of the function \(f(x) = 4 - x^{2}\) over the interval \([-2,2]\) is 2. The values of \(x\) in this interval for which the function equals its average value are \(\sqrt{2}\) and \(-\sqrt{2}\).

Step-by-step solution

Calculate the Average Value

First, we'll find the average value of \(f(x) = 4 - x^{2}\) over the interval \([-2,2]\). The average value of a function \(f(x)\) over the interval \([a,b]\) is given by \( \frac{1}{b - a} \int_{a}^{b} f(x) dx \). So here, the average value of \(f(x) = 4 - x^{2}\) over \([-2,2]\) is \( \frac{1}{2 - (-2)} \int_{-2}^{2} (4 - x^{2}) dx = \frac{1}{4} \int_{-2}^{2} (4 - x^{2}) dx = \frac{1}{4} [4x - \frac{x^{3}}{3} ]_{-2}^{2} = 2 \).

Find \(x\) Values for which \(f(x)\) Equals its Average Value

Next, we'll find all \(x\) values in the \([-2,2]\) interval for which \(f(x) = 4 - x^{2}\) equals its average value. We set \(f(x) = \) Average Value and solve for \(x\). So \(4 - x^{2} = 2 \Rightarrow x^{2} = 2 \Rightarrow x = \sqrt{2}, -\sqrt{2}\). Both of these solutions are in the interval \([-2,2]\), so they are the values of \(x\) for which \(f(x)\) equals its average value in this interval.

Key concepts

Definite Integrals

The calculation of definite integrals is central to understanding how to find the average value of a function over an interval, which involves integrating the function between two specified limits. A definite integral of a function on a closed interval \[a, b\] is represented by \( \int_{a}^{b} f(x) dx \). It essentially provides the accumulated value—often interpreted as the area under the curve—of the function \(f(x)\) between \(x=a\) and \(x=b\).

To calculate this integral, we must find the antiderivative of \(f(x)\), also known as an indefinite integral, and then evaluate this antiderivative at the upper and lower bounds of the interval. Subtracting these two values gives us the value of the definite integral. For instance, in our exercise \(f(x) = 4 - x^{2}\), after computing the antiderivative, we substitute the limits \(x=2\) and \(x=-2\) to get \( \frac{1}{4} [4x - \frac{x^{3}}{3} ]_{-2}^{2}\) which simplifies to the result of the integral over the interval.

Integral Calculus

Integral calculus, comprising the study of integrals and their properties, is a cornerstone of function analysis and plays a vital role in numerous fields of mathematics and applied sciences. It is employed not only for finding areas under curves but also for solving problems related to volumes, lengths, work, and pressure among others.

The process of integration, which is opposite to differentiation, can often be understood as the 'reversal' of finding derivatives. If a function \(F(x)\) is the antiderivative of \(f(x)\), then the integral of \(f(x)\) with respect to \(x\) over an interval \[a,b\] can be given by \(F(b) - F(a)\). For example, the integration of our function \(f(x) = 4 - x^{2}\) over the interval \[a,b\] determines the average value by incorporating the limits of integration, thereby incorporating the entire behavior of the function within the mentioned bounds.

Function Analysis

Function analysis is a part of calculus that involves examining various aspects of functions, such as their behavior, extremities, continuity, and in this context, their average values over a specific interval. When we explore the average value of a function, we are essentially seeking a single representative value that summarizes the behavior of the function over the interval.

Average value, a concept related to the arithmetic mean, is often visualized as the 'height' that would create a rectangle with the same area under the function's curve over the specified interval. To determine where the function attains this average value, we can set \(f(x)\) equal to the average value and solve for \(x\). This procedure can reveal how often and where exactly the function's output matches the computed average, as seen in \(4 - x^{2} = 2\), solving which gives us the precise points \(\sqrt{2}\) and \(\-\sqrt{2}\) where the function's value equals its average over the interval \[\-2,2\].