Calculate an average value
A heater over ten hours
The power of a heater rises steadily from 0 to 8 kilowatts in the first 4 hours, then stays at 8 kilowatts until hour 10.
The shaded region is a triangle plus a rectangle with areas 16 and 48, so its area is 64 kilowatt-hours. One unit of area is one kilowatt times one hour.
Drag the top of the orange rectangle up and down. Its base is 10 hours, so its area is 10 times its height.
The two areas are equal when the height is 6.4, so the average power is 6.4 kilowatts.
Three ways to find the area
Without numbers, the region under \(f\) from \(a\) to \(b\) has some area \(A\). The orange rectangle has the same base, so when it has the same area, its height is \(A\) divided by \(b - a\). Press Fit the rectangle to move the top to that height.
Press the first three buttons. The area can come from a formula, from shapes such as triangles, or from a table of readings. In every case the last step is the same: you divide \(A\) by the length of the interval.
Notation
If you have a formula, find an antiderivative \(F\) and subtract its values at the two ends.
\[\cb{A} = \int_a^b f(x)\,dx = F(b) - F(a)\]If the graph is made of straight lines or arcs of circles, find the area with geometry.
If you have readings \(y_0, \dots, y_n\) at equal spacing \(h\), one trapezoid between each two readings estimates the area. Each has width \(h\), so its area is \(h\) times the mean of its two readings.
\[\begin{aligned} \cb{A} &\approx \frac{h}{2}(y_0 + y_1) + \frac{h}{2}(y_1 + y_2) \\ &\qquad + \dots + \frac{h}{2}(y_{n-1} + y_n) \end{aligned}\]Divide the area by the length of the interval. The result is the average value, the height of the orange rectangle.
\[\cy{f_{\text{avg}}} = \frac{\cb{A}}{b - a}\]The area has the unit of \(f\) times the unit of \(x\). Dividing by the length removes the unit of \(x\), so the average keeps the unit of \(f\).
The integral exists when \(f\) is continuous on \([a, b]\). The average lies between the smallest and the largest value of \(f\), because a level above the curve makes the rectangle larger than the region.
Example 1: A formula
A factory makes parts at the rate \(R(t) = t^2 - 4t + 7\) hundred parts per hour, with \(t\) in hours. Find the average value of \(R\) for \(0 \le t \le 6\).
Find an antiderivative of \(R\) by raising each power by one and dividing by the new power.
\[F(t) = \frac{t^3}{3} - 2t^2 + 7t\]Subtract \(F(0)\) from \(F(6)\). The result is the area, in hundreds of parts.
\[\begin{aligned} \cb{A} &= F(6) - F(0) \\ &= (72 - 72 + 42) - 0 = \cb{42} \end{aligned}\]Divide by the length, 6 hours.
\[\begin{aligned} R_{\text{avg}} &= \frac{42}{6} \\ &= \cy{7} \text{ hundred parts per hour} \end{aligned}\]As a check, 7 lies between the smallest value 3 and the largest value 19 of \(R\). The red parts of the region above the orange line fill the green gaps below it, so the two areas are equal.
Example 2: A table
A pump moves water at the rates in the table, in liters per minute. Find the average rate over the 20 minutes.\[\begin{array}{c|ccccc} t \text{ (min)} & 0 & 5 & 10 & 15 & 20 \\ \hline \text{rate (L/min)} & 0 & 6 & 10 & 12 & 12 \end{array}\]
Straight lines join the readings, and faint lines split the shaded region into four trapezoids of width 5 minutes. Each has area 5 times the mean of its two readings.
\[\begin{aligned} &5\cdot\frac{0+6}{2},\quad 5\cdot\frac{6+10}{2}, \\ &5\cdot\frac{10+12}{2},\quad 5\cdot\frac{12+12}{2} \end{aligned}\]The four areas are 15, 40, 55 and 60. Their sum is the area under the graph, in liters.
\[\cb{A} \approx 15 + 40 + 55 + 60 = \cb{170}\]A tempting wrong move: Divide the area by 5, the number of readings.
\[\frac{170}{5} = \cr{34}\]The red rectangle of height 34 rises off the top of the figure, far above the largest reading, 12. An average cannot be larger than the largest value, so 34 is not possible.
The base of the rectangle is the length of the interval, 20 minutes, not the number of readings.
\[\begin{aligned} \cy{f_{\text{avg}}} &\approx \frac{170}{20} \\ &= \cy{8.5} \text{ liters per minute} \end{aligned}\]
Practice
The temperature of a drink above room temperature is \(D(t) = 16e^{-t/4}\) degrees Celsius, with \(t\) in minutes. Find the exact average value of \(D\) for \(0 \le t \le 8\).
Drag the orange line to your answer. At the average, the red parts above the line have the same area as the green gaps below it.
Solution
An antiderivative of \(16e^{-t/4}\) is \(-64e^{-t/4}\). Evaluate it from 0 to 8.
\[\begin{aligned} &\int_0^8 16e^{-t/4}\,dt \\ &\quad = \Bigl[-64e^{-t/4}\Bigr]_0^8 \\ &\quad = -64e^{-2} + 64 \\ &\quad = 64\bigl(1 - e^{-2}\bigr) \end{aligned}\]Divide by the length, 8 minutes. The unit is degrees Celsius.
\[\begin{aligned} D_{\text{avg}} &= \frac{64(1 - e^{-2})}{8} \\ &= 8\bigl(1 - e^{-2}\bigr) \end{aligned}\]As a check, the average is about 6.92, between the largest value \(D(0) = 16\) and the smallest value \(D(8) \approx 2.17\).
Questions
The figure shows the speed of a boat, in kilometers per hour. It rises from 0 to 6 in 3 hours, then falls to 0 in 2 hours.
What is the area of the shaded region?
- 15 kilometers
- 30 kilometers
- 3 kilometers per hour
Answer
15 kilometers
What is the unit of the average value of the speed?
- Kilometers per hour
- Kilometers
- Hours
Answer
Kilometers per hour
Rain falls at the rate \(r(t) = 6\sqrt{t}\) millimeters per hour. What is the average value of \(r\) for \(0 \le t \le 4\)?
- 8 millimeters per hour
- 32 millimeters
- 12 millimeters per hour
- 6 millimeters per hour
Answer
8 millimeters per hour
Summary
For a formula, a graph or a table, you find the area under the graph and then divide it by the length of the interval.
This works because the average value is the height of a rectangle with the same base and the same area as the region.
The answer keeps the unit of the original quantity, here kilowatts, because the area in kilowatt-hours is divided by hours.
This is one of four skills in topic 8.1 of Which & Why: AP Calculus.