Question Video: Finding the Average Value of a Function on a Given Interval from Its Graph

Video Transcript

What is the average value of this function on the closed interval negative five to four?

We begin by recalling the formula for the average value of a function in a closed interval 𝑎 to 𝑏. It’s one over 𝑏 minus 𝑎 times the integral of 𝑓 of 𝑥 evaluated between 𝑎 and 𝑏. In our case, the limits are four and negative five. So it’s one over four minus negative five times that definite integral. This graph, however, shows a piecewise function made up of a number of different functions.

Instead of working out the function at each point, we’re going to recall the most basic definition of the integral of a function. It allows us to find the net area between the graph of the function and the 𝑥-axis. We can therefore find the area between the graph of this function and the 𝑥-axis by splitting into subintervals and remembering that when we evaluate the area underneath the 𝑥-axis, we’ll end up with a negative value.

We’ll begin by finding the area of this triangle. The formula for area of a triangle is half times base times height. So the area of this triangle is a half times one times four, which is two square units. Since this triangle sits below the 𝑥-axis, we give this a value of negative two in our integral. Our next triangle sits above the 𝑥-axis. Its area is a half times two times one, which is just one square unit. So we add one.

The next shape we’ll come across is a trapezium, though we could’ve split this into a triangle and a square. Its area is a half times four plus three times three, which is 10.5 square units. So we add 10.5. We then find a second trapezium, which has an area of a half times three plus two times one, which is 2.5 square units. And we have one final trapezium, which has an area of a half times four plus two times two, which is three square units. Or we find the sum of these values, and that’s equal to the integral evaluated between negative five and four of 𝑓 of 𝑥 with respect to 𝑥. We see that the average value of our function is therefore a ninth times 15, which is five-thirds.