91Ó°ÊÓ

The average value of a function \(f(x)\) on an interval \([a, b]\) is defined as the following: $$ \text{Average}(f)=\frac{1}{b-a}\int_{a}^{b}f(x)dx $$ This means that in order to find the average value of a function, we need to integrate the given function over the interval \([a, b]\), and then divide the result by the length of the interval \((b - a)\).

Let's see an example to better understand the process. Suppose we have a function \(f(x) = x^2\) and we want to find its average value on the interval \([1, 3]\). 1. First, we need to calculate the integral of the function over the given interval: $$ \int_{1}^{3}x^2dx $$ 2. Next, we integrate the function: $$ \frac{x^3}{3}\Big|_1^3 $$ 3. We evaluate the integral at the limits of integration: $$ \frac{3^3}{3} - \frac{1^3}{3} = 8 - \frac{1}{3} = \frac{23}{3} $$ 4. Finally, divide the result by the length of the interval \((3 - 1)\): $$ \text{Average}(f) = \frac{\frac{23}{3}}{3 - 1} = \frac{23}{6} $$ So the average value of the function \(f(x) = x^2\) on the interval \([1, 3]\) is \(\frac{23}{6}\).

The definition of the average value of a function is similar to the definition of the average of a set of numbers in the sense that both involve dividing the sum (or integral) by the number of elements (or the length of the interval). For instance, let's consider a set of numbers \(S = \{x_1, x_2, \dots, x_n\}\). The average of this set is given by: $$ \text{Average}(S) = \frac{x_1 + x_2 + \dots + x_n}{n} $$ Comparing this definition with the definition of the average value of a function, we can see that both involve summing the values (integral for the function case), and dividing by the number of elements or the length of the interval: - For the set of numbers, we sum the values and divide by the number of elements (\(n\)). - For the function, we calculate the integral (which can be seen as a sum of infinitely many function values) and divide by the length of the interval (\(b - a\)). In both cases, we are essentially calculating a "weighted average" or "balance point" of the values.