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

What is the average value of this function on the interval [โˆ’5, 4]?


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‘Ž to ๐‘. Itโ€™s one over ๐‘ minus ๐‘Ž times the integral of ๐‘“ of ๐‘ฅ evaluated between ๐‘Ž and ๐‘. In our case, the limits are four and negative five. So itโ€™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, weโ€™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 ๐‘ฅ-axis. We can therefore find the area between the graph of this function and the ๐‘ฅ-axis by splitting into subintervals and remembering that when we evaluate the area underneath the ๐‘ฅ-axis, weโ€™ll end up with a negative value.

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

The next shape weโ€™ll come across is a trapezium, though we couldโ€™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 thatโ€™s equal to the integral evaluated between negative five and four of ๐‘“ of ๐‘ฅ with respect to ๐‘ฅ. We see that the average value of our function is therefore a ninth times 15, which is five-thirds.

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