Set up the average-value calculation
Averaging many readings
A sensor records the depth of water in a tide pool, and the depth is \(d(t) = t(6 - t)\) centimeters at time \(t\) hours, from \(t = 0\) to \(t = 6\).
The interval is cut into equal parts, and each chalk dot is a reading of the depth in the middle of one part. The orange line is at the mean of the readings, their sum divided by how many there are.
With 3 readings, the dots are at \(t = 1\), 3 and 5, so the depths are 5, 9 and 5 centimeters and their mean is about 6.33.
Press Use 6 readings, then Use 12 readings. The mean falls to about 6.08, then to about 6.02, so more readings give a mean closer to 6.
Any function on any interval
Without numbers, the curve is any function \(f\) on an interval from \(a\) to \(b\). The dots are \(n\) readings at equal spacing, and the orange line is their mean.
Drag the slider to add readings. The orange line moves less and less, because it approaches one height that depends only on \(f\), \(a\) and \(b\).
Press Pour the region into the rectangle, which uses 60 readings. The red parts are where the curve is above the orange line, and the green parts are where the orange line is above the curve. The red parts fill the green parts, so the rectangle with base \(b - a\) has the same area as the region.
The height of this rectangle, the orange line, is the average value of \(f\) on the interval.
Notation
Split the interval into \(n\) equal parts. Each part has width \(\Delta x\), so the number of parts is \(n = \frac{b - a}{\Delta x}\).
\[\cq{\Delta x} = \frac{b - a}{n}\]The readings are \(f(x_1), \ldots, f(x_n)\), one in the middle of each part, and their mean is their sum divided by \(n\). Because \(n = \frac{b - a}{\Delta x}\), you can replace \(\frac1n\) by \(\frac{\Delta x}{b - a}\).
\[\begin{aligned} &\frac{1}{n}\sum f(x_i) \\ &= \frac{1}{\cy{b - a}}\sum f(x_i)\,\Delta x \end{aligned}\]The divisor is now the length of the interval, not the number of readings.
The sum on the right is a Riemann sum, which adds height times width for each part. As \(n\) grows, the sum approaches the definite integral, the area of the shaded region.
\[\frac{1}{b - a}\,\cb{\int_a^b f(x)\,dx}\]The average value is the area divided by the length of the interval, so the area equals the base \(b - a\) times the orange height.
\[\begin{aligned} \cy{f_{\text{avg}}} &= \frac{1}{b - a}\int_a^b f(x)\,dx \\ \cb{\int_a^b f(x)\,dx} &= \cy{f_{\text{avg}}}\,(b - a) \end{aligned}\]The formula needs \(a < b\), so that the length \(b - a\) is positive. It also needs \(f\) to be continuous on \([a, b]\), so that the integral exists.
Example 1: A machine's power
A machine uses power \(P(t) = 3 + 2t\) kilowatts, with \(t\) in hours. Find its average power from \(t = 1\) to \(t = 5\).
The interval runs from 1 to 5, so the length to divide by is 4 hours.
\[b - a = 5 - 1 = \cy{4}\]Write the formula with these numbers, where the integral is the area of the shaded region.
\[P_{\text{avg}} = \frac{1}{4}\int_1^5 (3 + 2t)\,dt\]An antiderivative of \(3 + 2t\) is \(3t + t^2\), so you evaluate it at 5 and at 1 and subtract.
\[\begin{aligned} \int_1^5 (3 + 2t)\,dt &= \Bigl[\,3t + t^2\,\Bigr]_1^5 \\ &= 40 - 4 = \cb{36} \end{aligned}\]Divide the area by the length. The area is in kilowatt-hours and the length in hours, so the average power is 9 kilowatts.
\[P_{\text{avg}} = \frac{36}{4} = \cy{9}\]The orange rectangle has base 4 and height 9, so its area is also 36. The red triangle above the orange line fills the green triangle below it.
Example 2: The divisor is the length
Find the average value of \(f(x) = x^2\) on \([2, 5]\).
Find the area under the curve from 2 to 5 first.
\[\begin{aligned} \int_2^5 x^2\,dx &= \Bigl[\frac{x^3}{3}\Bigr]_2^5 \\ &= \frac{125}{3} - \frac{8}{3} = \cb{39} \end{aligned}\]A tempting wrong move: Divide the area by 5, because the interval ends at 5.
\[\frac{39}{5} = \cr{7.8}\]The red rectangle has base 3 and height 7.8, so its area is \(7.8 \cdot 3 = 23.4\), less than 39. The curve is above 7.8 on most of the interval.
The base of the rectangle is the interval from 2 to 5, so its length is 3, not 5.
\[b - a = 5 - 2 = \cy{3}\]Divide the area by 3. The result lies between the smallest height \(f(2) = 4\) and the largest \(f(5) = 25\), as an average must.
\[f_{\text{avg}} = \frac{39}{3} = \cy{13}\]
Practice
Set up and find the average value of \(f(x) = 3x^2 + 1\) on \([-1, 2]\).
The base runs from \(-1\) to 2. Count its length on the number line, not from 0.
Solution
The interval runs from \(-1\) to 2, so its length is 3.
\[b - a = 2 - (-1) = 3\]Integrate with the antiderivative \(x^3 + x\).
\[\begin{aligned} \int_{-1}^{2} (3x^2 + 1)\,dx &= \Bigl[\,x^3 + x\,\Bigr]_{-1}^{2} \\ &= 10 - (-2) \\ &= 12 \end{aligned}\]Divide the area by the length. The average value is 4.
\[f_{\text{avg}} = \frac{12}{3} = 4\]
Questions
The figure shows \(y = x^2\) on \([0, 3]\). The shaded region has area 9, and the orange rectangle has the same base and the same area.
How tall is the orange rectangle?
- 3
- 9
- 4.5
Answer
3
Which part of the figure gives the number that you divide the area by?
- The width of the interval along the x-axis.
- The height of the curve at \(x = 3\).
- The area of the shaded region.
Answer
The width of the interval along the x-axis.
What is the average value of \(g(x) = 6x^2\) on \([1, 3]\)?
- 26
- 52
- \(\tfrac{52}{3}\)
- 30
Answer
26
Summary
The average value is the definite integral divided by the length of the interval, so the pool's average depth is \(36 \div 6 = 6\) centimeters.
This works because the mean of \(n\) readings is a Riemann sum divided by \(b - a\), and as \(n\) grows the sum becomes the integral.
You use it when a question asks for the average of a quantity that changes continuously over an interval.
This is one of four skills in topic 8.1 of Which & Why: AP Calculus.