Find where the function equals its average value
An arch and its average height
The top edge of a stone arch has height \(f(x) = 16 - x^2\) meters at \(x\) meters from the center, for \(x\) from \(-3\) to 3.
The shaded region under the arch has area 78, and the orange rectangle has base 6. Drag its top. The areas are equal at height 13, because 6 times 13 is 78, so the average height of the arch is 13.
Press Set the height to the average. The two orange dots move to where the arch has its average height, 13 meters, at \(x = \pm\sqrt{3}\), about \(\pm 1.73\) meters.
A curve must cross its average
The orange line is at the average value of \(f\) on \([a, b]\), so the rectangle and the shaded region have equal area. Press Show hills and gaps.
The red hills are the parts of the region above the orange line. The green gaps are the parts of the rectangle above the curve. They have equal total area, because the rectangle and the region do. Both exist, so the curve is above the line in some places and below it in others.
Drag the point along the curve. The red segment shows how far \(f(c)\) is from the average, and it shrinks to zero where the curve meets the line.
A curve with no break cannot pass from above the line to below it without meeting it. At that place, \(f(c)\) equals the average value.
Notation
First find the average value of \(f\) on \([a, b]\). The orange line is at this height.
\[\cy{f_{\text{avg}}} = \frac{1}{b - a}\int_a^b f(x)\,dx\]Set \(f(c)\) equal to the average value and solve for \(c\). Each solution is an input where the curve meets the orange line.
\[f(c) = \cy{f_{\text{avg}}}\]Keep only the solutions in \([a, b]\). Solutions outside the interval do not count.
The Mean Value Theorem for integrals guarantees at least one solution. If \(f\) is continuous on \([a, b]\), then some \(c\) in \([a, b]\) has \(f(c)\) equal to the average value.
\[\begin{aligned} &\exists\, c \in [a, b]: \\ &\cb{\int_a^b f(x)\,dx} = f(c)\,(b - a) \end{aligned}\]This is \(f(c) = f_{\text{avg}}\) with both sides multiplied by \(b - a\). There may be more than one \(c\): this curve meets the line twice.
The condition is continuity on all of \([a, b]\). A graph with a break can jump over its average value and never take it.
When it fails
Press Put a jump in f. The graph is now two flat pieces of equal length, at heights 2 and 6, so its average value is 4.
The orange line at height 4 does not meet the graph, because the graph jumps from 2 to 6 and skips 4. No \(c\) has \(f(c) = 4\).
Example 1: A parabola
Find the value of \(c\) in \([0, 6]\) where \(f(x) = x^2\) equals its average value on \([0, 6]\).
Find the area under the curve.
\[\int_0^6 x^2\,dx = \Bigl[\frac{x^3}{3}\Bigr]_0^6 = \cb{72}\]Divide by the length, 6, to find the average value. The orange line is at height 12.
\[f_{\text{avg}} = \frac{72}{6} = \cy{12}\]Set \(f(c)\) equal to 12 and solve for \(c\). Both square roots of 12 solve the equation.
\[c^2 = 12 \quad\Longrightarrow\quad c = \pm 2\sqrt{3}\]Keep the solution in \([0, 6]\). The negative root \(-2\sqrt{3}\) is outside the interval, and \(2\sqrt{3}\) is about 3.46.
\[c = 2\sqrt{3} \approx 3.46\]
Example 2: Not at the middle
Find the value of \(c\) in \([0, 2]\) where \(f(x) = x^3\) equals its average value on \([0, 2]\).
The area under the curve is 4. The length of the interval is 2, so the average value is 2.
\[\begin{aligned} \int_0^2 x^3\,dx &= \Bigl[\frac{x^4}{4}\Bigr]_0^2 = 4 \\ f_{\text{avg}} &= \frac{4}{2} = \cy{2} \end{aligned}\]A tempting wrong move: Guess that \(f\) reaches its average value at the middle of the interval, \(c = 1\).
\[f(1) = \cr{1}\]At \(x = 1\) the curve is at height 1 and the orange line is at height 2, so the red segment shows a gap of 1. The curve is steeper on the right half, so it meets the line to the right of the middle.
Solve \(f(c) = 2\) instead. The number \(\sqrt[3]{2}\) is about 1.26, and it lies in \([0, 2]\).
\[c^3 = 2 \quad\Longrightarrow\quad c = \sqrt[3]{2} \approx 1.26\]
Practice
Find all values of \(c\) in \([0, 5]\) where \(f(x) = x^2 - 4x\) equals its average value on \([0, 5]\).
The parabola meets the orange line twice. Drag the point until the red segment shrinks to zero to estimate each c.
Solution
Integrate from 0 to 5. The red part of the region lies below the x-axis and counts as negative, so the integral is negative.
\[\begin{aligned} \int_0^5 (x^2 - 4x)\,dx &= \Bigl[\frac{x^3}{3} - 2x^2\Bigr]_0^5 \\ &= -\frac{25}{3} \end{aligned}\]Divide by the length 5. The orange line is at height \(-\tfrac53\).
\[f_{\text{avg}} = -\frac{5}{3}\]Set \(f(c)\) equal to \(-\tfrac53\) and move every term to one side. Then use the quadratic formula.
\[\begin{aligned} c^2 - 4c + \frac{5}{3} &= 0 \\ c &= \frac{4 \pm \sqrt{16 - \frac{20}{3}}}{2} \\ &= 2 \pm \sqrt{\frac{7}{3}} \end{aligned}\]Both solutions lie in \([0, 5]\). They are about 0.47 and 3.53.
\[c \approx 0.47 \quad\text{and}\quad c \approx 3.53\]
Questions
The figure shows \(f(x) = x^3 - 3x\) on \([-2, 3]\). The orange line is at the average value, 1.75, and the red hills and green gaps are marked.
How many values of \(c\) in \([-2, 3]\) have \(f(c)\) equal to the average value?
- 3
- 1
- 2
Answer
3
Why must the curve meet the orange line at least once?
- The curve has no break, and it is above the line in some places and below it in others.
- The interval is closed.
- The curve rises from left to right.
Answer
The curve has no break, and it is above the line in some places and below it in others.
A function \(g\) is continuous on \([1, 5]\) and \(\int_1^5 g(x)\,dx = 36\). Which value must \(g\) take somewhere in \([1, 5]\)?
- 9
- 36
- 7.2
Answer
9
Summary
You find the average value, set \(f(c)\) equal to it, solve for \(c\), and keep the solutions in \([a, b]\).
This works because a continuous curve is above its average in some places and below it in others, so somewhere it equals the average.
You use it when a question asks where a function takes its average value, and the guarantee needs a continuous function.
This is one of four skills in topic 8.1 of Which & Why: AP Calculus.