Welcome back to Professor Baker's Math Class! Tonight, we are diving into Section 6-5: Average Value of a Function. This is a fascinating application of integration that connects standard statistics concepts with calculus.
Moving Beyond Discrete Averages
We all know how to take the average of a finite list of numbers (like averaging your quiz grades). You simply add them up and divide by the number of items ($n$). But how do we compute the average temperature during a day? Because temperature changes continuously, there are infinitely many readings possible.
To solve this, we use integration. Just as a Riemann sum adds up infinite rectangles to find area, we can use a definite integral to sum up values to find an average. The definition of the Average Value of a function $f$ on the interval $[a, b]$ is given by:
$$f_{ave} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx$$Think of the integral $\int_{a}^{b} f(x) \, dx$ as the sum of all the values, and $(b-a)$ as the width of the interval we are dividing by.
Example: Applying the Formula
Let's look at Example 1 from the lecture notes. We want to find the average value of the function $f(x) = 1 + x^2$ on the interval $[-1, 2]$.
- Identify the interval: $a = -1$ and $b = 2$. The width is $2 - (-1) = 3$.
- Set up the integral:
$$f_{ave} = \frac{1}{3} \int_{-1}^{2} (1 + x^2) \, dx$$ - Evaluate:
$$= \frac{1}{3} \left[ x + \frac{x^3}{3} \right]_{-1}^{2}$$
$$= \frac{1}{3} \left[ \left(2 + \frac{8}{3}\right) - \left(-1 - \frac{1}{3}\right) \right]$$
$$= \frac{1}{3} [6] = 2$$
So, the average value of the function over that specific interval is 2.
The Mean Value Theorem for Integrals
A key concept in this section is the Mean Value Theorem for Integrals. It states that if $f$ is continuous on $[a, b]$, there exists at least one number $c$ in that interval such that:
$$f(c) = f_{ave}$$In terms of our temperature analogy, this means there is at least one specific moment in the day when the actual temperature is exactly equal to the average temperature.
Geometric Interpretation
Visualizing this concept can be very helpful. Geometrically, the area under the curve $y=f(x)$ is equal to the area of a rectangle with base $(b-a)$ and height $f_{ave}$.
As described in the class notes: "You can always chop off the top of a (two-dimensional) mountain at a certain height and use it to fill in the valleys so that the mountain becomes completely flat." That flattened height is your average value.
Be sure to review the attached PDF for more examples, including how this applies to velocity and other continuous variables. Keep practicing those integrals!