Distinguish average value from average rate
Two questions about one flight
A drone climbs so that its height is \(H(t) = 8t - t^2\) meters, for \(t\) from 0 to 4 seconds.
Question A asks for the average height. Press Show the average height. The orange line is the top of a rectangle with the same area as the shaded region, about 10.67 meters high.
Question B asks for the average rate of climb. Press Show the average rate of climb. The straight line through the two ends rises 16 meters in 4 seconds, so its slope is 4 meters per second.
The two answers have different units: meters and meters per second.
A height and a slope
Without numbers, the curve is any function \(f\) on an interval from \(a\) to \(b\).
Press Show the orange rectangle. Its top is the average value of \(f\), which comes from the area under the curve.
Press Show the straight line through the ends. This is a secant line. Its slope is the average rate of change of \(f\), which uses the two end values only.
Drag the right end \(b\). Both change, but the orange height still comes from the area and the slope from the two ends.
Notation
The average value of \(f\), a typical value of \(f\) on the interval, comes from the area under the curve.
\[\cy{f_{\text{avg}}} = \frac{1}{b - a}\int_a^b f(x)\,dx\]The average rate of change of \(f\) is how fast \(f\) changes from \(a\) to \(b\). It is the slope of the secant line, from the two end values only.
\[\frac{\cg{f(b) - f(a)}}{b - a}\]The units tell the two apart. The average value has the unit of \(f\), and the average rate has the unit of \(f\) per unit of \(x\).
Read the words of the question. "Average height" asks for an average value, and "average rate" asks for an average rate of change.
A rate is also a function with an average value. By the Fundamental Theorem of Calculus, the integral of the rate \(f'\) is the change \(f(b) - f(a)\). This means the average value of \(f'\) is the average rate of change of \(f\).
\[\begin{aligned} &\frac{1}{b - a}\cg{\int_a^b f'(x)\,dx} \\ &= \frac{\cg{f(b) - f(a)}}{b - a} \end{aligned}\]This holds when \(f'\) is continuous on \([a, b]\).
When it fails
Press Bend the curve between the ends. The end heights do not change, so the secant line and the average rate of change stay the same.
The area under the curve changes, so the orange line moves. The average value depends on the whole curve, the average rate of change on the two ends only.
Example 1: Both averages of one function
The mass of a bacteria culture is \(m(t) = t^2 + 1\) grams, with \(t\) in hours. For \(1 \le t \le 3\), find the average mass and the average rate at which the mass changes.
"Average mass" asks for an average value, so integrate the mass.
\[\begin{aligned} \int_1^3 (t^2 + 1)\,dt &= \Bigl[\frac{t^3}{3} + t\Bigr]_1^3 \\ &= 12 - \frac{4}{3} = \cb{\frac{32}{3}} \end{aligned}\]Divide by the length of the interval, 2 hours. The unit is grams.
\[m_{\text{avg}} = \frac{32/3}{2} = \cy{\frac{16}{3}} \approx 5.33\]"Average rate" asks for a slope, so use the end values \(m(1) = 2\) and \(m(3) = 10\). The unit is grams per hour.
\[\frac{m(3) - m(1)}{3 - 1} = \frac{10 - 2}{2} = 4\]
Example 2: The average speed of a round trip
A car drives 150 miles out at 30 miles per hour. It returns the same 150 miles at 50 miles per hour. Find its average speed for the whole trip.
The graph shows the speed against time. The trip out takes 5 hours, because \(150 \div 30 = 5\), and the return takes 3 hours, because \(150 \div 50 = 3\). Each shaded part has area 150 miles, the distance driven.
\[5 + 3 = 8\]A tempting wrong move: Average the two speeds.
\[\frac{30 + 50}{2} = \cr{40}\]The red rectangle, 40 high over 8 hours, has area 320 miles, but the car traveled 300. It spends 5 hours at 30 and only 3 at 50, so the average is closer to 30.
The area under the speed graph is the 300 miles traveled, so the average value of the speed is 300 divided by 8 hours. It is also the average rate of change of the distance traveled.
\[\begin{aligned} \frac{150 + 150}{8} &= \frac{300}{8} \\ &= \cy{37.5} \text{ miles per hour} \end{aligned}\]
Practice
The volume of water in a tank is \(V(t) = 3t^2 + 10\) liters after \(t\) minutes. For \(0 \le t \le 4\), find the average volume and the average rate of change of the volume.
The average volume is the height of the orange line, and the average rate of change is the slope of the straight line.
Solution
"Average volume" asks for an average value, so divide the integral by the 4 minutes. The unit is liters.
\[\begin{aligned} &\frac{1}{4}\int_0^4 (3t^2 + 10)\,dt \\ &= \frac{1}{4}\Bigl[t^3 + 10t\Bigr]_0^4 \\ &= \frac{104}{4} = 26 \end{aligned}\]The average rate of change uses the end values \(V(0) = 10\) and \(V(4) = 58\). The unit is liters per minute.
\[\frac{V(4) - V(0)}{4 - 0} = \frac{48}{4} = 12\]
Questions
The figure shows \(f(x) = 2 + 4x - x^2\) on \([0, 4]\), the straight line through its ends, and the orange line at about 4.67.
What is the average rate of change of \(f\) from \(x = 0\) to \(x = 4\)?
- 0
- About 4.67
- 2
Answer
0
What does the height of the orange line, about 4.67, tell you?
- The average value of \(f\) on \([0, 4]\).
- The average rate of change of \(f\) on \([0, 4]\).
- The largest value of \(f\) on \([0, 4]\).
Answer
The average value of \(f\) on \([0, 4]\).
A continuous function \(f\) has \(\int_2^6 f(x)\,dx = 20\), \(f(2) = 3\) and \(f(6) = 9\). What is its average value on \([2, 6]\)?
- 5
- 1.5
- 20
- 6
Answer
5
Summary
The drone's average height, 10.67 meters, is the height of the orange rectangle with the same area as the shaded region.
Its average rate of climb, 4 meters per second, is the slope of the line through the two ends only.
An average amount is the integral divided by the length of the interval, and an average rate is the change divided by that length.
This is one of four skills in topic 8.1 of Which & Why: AP Calculus.