### 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 ๐ to ๐. Itโs one over ๐ minus ๐ times the integral of ๐ of ๐ฅ evaluated between ๐ and ๐. In our case, the limits are four and negative five. So itโs 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, weโre 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 ๐ฅ-axis. We can therefore find the area between the graph of this function and the ๐ฅ-axis by splitting into subintervals and remembering that when we evaluate the area underneath the ๐ฅ-axis, weโll end up with a negative value.

Weโll 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 ๐ฅ-axis, we give this a value of negative two in our integral. Our next triangle sits above the ๐ฅ-axis. Its area is a half times two times one, which is just one square unit. So we add one.

The next shape weโll come across is a trapezium, though we couldโve 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 thatโs equal to the integral evaluated between negative five and four of ๐ of ๐ฅ with respect to ๐ฅ. We see that the average value of our function is therefore a ninth times 15, which is five-thirds.