The Central Idea
For finitely many numbers
\[ y_1,y_2,\ldots,y_n, \]the arithmetic mean is
\[ \frac{y_1+y_2+\cdots+y_n}{n}. \]For a continuously varying function \(f(x)\) on \([a,b]\), the corresponding average value is
\[ \boxed{ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\,dx }. \]Think:
\[ \boxed{ \text{average value} = \frac{\text{total accumulation}}{\text{interval length}} }. \]Why We Divide by \(b-a\)
The integral
\[ \int_a^b f(x)\,dx \]measures the total signed accumulation of \(f\) across the interval.
The interval length is
\[ b-a. \]Dividing the total by the length converts total accumulation back into an average level of the function.
The Continuous Analogue of an Arithmetic Mean
For equally spaced sample points, an ordinary average looks like
\[ \frac{1}{n} \sum_{i=1}^{n}f(x_i). \]Since
\[ \Delta x=\frac{b-a}{n}, \]we can rewrite
\[ \frac1n = \frac{\Delta x}{b-a}. \]Therefore,
\[ \frac1n \sum_{i=1}^{n}f(x_i) = \frac{1}{b-a} \sum_{i=1}^{n}f(x_i)\Delta x. \]As the partition becomes finer, the sum approaches the definite integral.
Deriving the Formula from Riemann Sums
Starting with
\[ \frac{1}{b-a} \sum_{i=1}^{n} f(x_i^*)\Delta x, \]take the limit as \(n\to\infty\):
\[ \lim_{n\to\infty} \frac{1}{b-a} \sum_{i=1}^{n} f(x_i^*)\Delta x. \]Pull the constant factor outside:
\[ \frac{1}{b-a} \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x. \]The limit is the definite integral:
\[ \boxed{ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\,dx }. \]Geometric Interpretation
Suppose \(f(x)\geq0\) on \([a,b]\).
The integral
\[ \int_a^b f(x)\,dx \]is the area under the graph.
The average value is the height of a rectangle having:
- width \(b-a\);
- the same area as the region under \(f\).
Therefore,
\[ (b-a)f_{\text{avg}} = \int_a^b f(x)\,dx. \]The Equal-Area Rectangle
Imagine replacing a varying graph with one constant horizontal level.
We choose the height so that the rectangle preserves the same total accumulation.
That height is
\[ \boxed{f_{\text{avg}}}. \]Average value therefore converts a varying quantity into an equivalent constant quantity over the same interval.
Example: Average Value of \(f(x)=x^2\)
Find the average value of
\[ f(x)=x^2 \]on
\[ [0,2]. \]Use
\[ f_{\text{avg}} = \frac1{2-0} \int_0^2x^2\,dx. \]Since
\[ \int_0^2x^2\,dx = \frac83, \]we get
\[ f_{\text{avg}} = \frac12\cdot\frac83. \]Therefore,
\[ \boxed{ f_{\text{avg}}=\frac43 }. \]Average Value of a Constant Function
If
\[ f(x)=c, \]then
\[ f_{\text{avg}} = \frac1{b-a} \int_a^bc\,dx. \]Since
\[ \int_a^bc\,dx=c(b-a), \]we obtain
\[ \boxed{f_{\text{avg}}=c}. \]A constant function is already at its own average level everywhere.
Average Value of a Linear Function
For
\[ f(x)=mx+c \]on \([a,b]\), the average value is
\[ \boxed{ f_{\text{avg}} = \frac{f(a)+f(b)}{2} }. \]Geometrically, the area under a line forms a trapezoid, whose average height is the mean of the endpoint heights.
Example: Average Value of a Line
Let
\[ f(x)=3x+1 \]on \([0,4]\).
Endpoint values are
\[ f(0)=1 \]and
\[ f(4)=13. \]Therefore,
\[ f_{\text{avg}} = \frac{1+13}{2}. \]Hence,
\[ \boxed{f_{\text{avg}}=7}. \]Checking the Linear Example with Integration
\[ f_{\text{avg}} = \frac14 \int_0^4(3x+1)\,dx. \] \[ = \frac14 \left[ \frac32x^2+x \right]_0^4. \] \[ = \frac14(24+4). \]Therefore,
\[ \boxed{f_{\text{avg}}=7}. \]Average Value Can Be Negative
Average value is based on signed accumulation.
If a function spends enough of the interval below the axis, then
\[ \int_a^b f(x)\,dx \]can be negative.
Therefore,
\[ f_{\text{avg}} \]can also be negative.
Average Value Can Be Zero
If positive and negative accumulation cancel exactly, then
\[ \int_a^b f(x)\,dx=0. \]Therefore,
\[ \boxed{f_{\text{avg}}=0}. \]This does not mean the function itself is zero throughout the interval.
Odd Symmetry
If \(f\) is odd and the interval is symmetric,
\[ [-a,a], \]then
\[ \int_{-a}^{a}f(x)\,dx=0. \]Therefore,
\[ \boxed{f_{\text{avg}}=0}. \]Even Symmetry
If \(f\) is even, then
\[ \int_{-a}^{a}f(x)\,dx = 2\int_0^a f(x)\,dx. \]Since the interval length is \(2a\),
\[ f_{\text{avg}} = \frac{1}{2a} 2\int_0^a f(x)\,dx. \]Thus,
\[ \boxed{ f_{\text{avg}} = \frac1a \int_0^a f(x)\,dx }. \]Example: Average Value of \(\sin x\)
Find the average value of
\[ \sin x \]on
\[ [0,\pi]. \]The interval length is \(\pi\), so
\[ f_{\text{avg}} = \frac1\pi \int_0^\pi\sin x\,dx. \]Since
\[ \int_0^\pi\sin x\,dx=2, \]we get
\[ \boxed{ f_{\text{avg}}=\frac2\pi }. \]Example: Average over a Full Period
Consider
\[ f(x)=\cos x \]over one full period:
\[ [0,2\pi]. \]Since positive and negative lobes cancel,
\[ \int_0^{2\pi}\cos x\,dx=0. \]Therefore,
\[ \boxed{f_{\text{avg}}=0}. \]Mean Value Theorem for Integrals
If \(f\) is continuous on \([a,b]\), then there exists at least one point \(c\in[a,b]\) such that
\[ \boxed{ f(c) = \frac{1}{b-a} \int_a^b f(x)\,dx }. \]In other words,
\[ \boxed{f(c)=f_{\text{avg}}}. \]What the Mean Value Theorem Says
A continuous function actually reaches its average value somewhere on the interval.
The average value is not merely an abstract computed number.
There is at least one input \(c\) for which the function's instantaneous value equals that continuous average.
Example: Finding a Mean-Value Point
For
\[ f(x)=x^2 \]on \([0,2]\), we found
\[ f_{\text{avg}}=\frac43. \]To find a point where the function equals its average, solve
\[ x^2=\frac43. \]On \([0,2]\),
\[ \boxed{ c=\frac{2}{\sqrt3} }. \]There May Be More Than One Mean-Value Point
The theorem guarantees at least one point \(c\), not exactly one.
A function that oscillates around its average level may cross that level many times.
Every such crossing satisfies
\[ f(c)=f_{\text{avg}}. \]Units of Average Value
Suppose \(f(x)\) has units \(U\), while \(x\) has units \(X\).
Then
\[ \int_a^b f(x)\,dx \]has units
\[ U\cdot X. \]Dividing by interval length \(b-a\), which has units \(X\), gives
\[ \boxed{U}. \]Average value therefore has the same units as the original function.
Example: Average Temperature
Suppose temperature over a \(24\)-hour period is modeled by \(T(t)\), where \(t\) is measured in hours.
The average temperature is
\[ \boxed{ T_{\text{avg}} = \frac1{24} \int_0^{24}T(t)\,dt }. \]The result has the same temperature units as \(T\).
Average Velocity as an Average Value
If velocity is \(v(t)\), then its average value over \([a,b]\) is
\[ \boxed{ v_{\text{avg}} = \frac1{b-a} \int_a^b v(t)\,dt }. \]Since
\[ \int_a^bv(t)\,dt \]is displacement, this becomes
\[ \boxed{ v_{\text{avg}} = \frac{\text{displacement}}{\text{elapsed time}} }. \]Connection to the Familiar Average-Velocity Formula
Since
\[ \int_a^b v(t)\,dt = s(b)-s(a), \]we obtain
\[ v_{\text{avg}} = \frac{s(b)-s(a)}{b-a}. \]Therefore,
\[ \boxed{ \text{average value of velocity} = \text{average rate of change of position} }. \]Average Velocity versus Average Speed
Average velocity uses signed velocity:
\[ \frac1{b-a} \int_a^bv(t)\,dt. \]Average speed uses the magnitude:
\[ \boxed{ \text{average speed} = \frac1{b-a} \int_a^b|v(t)|\,dt }. \]This mirrors the distinction between displacement and total distance.
Average Flow Rate
If \(r(t)\) is a flow rate, then
\[ \frac1{b-a} \int_a^b r(t)\,dt \]gives the average flow rate over that time interval.
Since the integral gives total volume change,
\[ \boxed{ \text{average flow rate} = \frac{\text{total volume change}}{\text{elapsed time}} }. \]Average Density
Suppose \(\rho(x)\) is linear density along a rod of length \(L=b-a\).
The total mass is
\[ M= \int_a^b\rho(x)\,dx. \]Therefore, the average density is
\[ \boxed{ \rho_{\text{avg}} = \frac{M}{b-a} = \frac1{b-a} \int_a^b\rho(x)\,dx }. \]Average of a Continuous Cost Rate
If \(c(t)\) represents cost per unit time, then
\[ \int_a^b c(t)\,dt \]gives accumulated cost.
Dividing by elapsed time gives
\[ \boxed{ c_{\text{avg}} = \frac1{b-a} \int_a^b c(t)\,dt }. \]Average Power
If \(P(t)\) is instantaneous power, then energy accumulated over \([a,b]\) is
\[ E= \int_a^bP(t)\,dt. \]The average power is therefore
\[ \boxed{ P_{\text{avg}} = \frac{E}{b-a} = \frac1{b-a} \int_a^bP(t)\,dt }. \]Equivalent Constant Force
If a variable force \(F(x)\) acts over \([a,b]\), then the work is
\[ W=\int_a^bF(x)\,dx. \]The constant force that would do the same work over the same displacement is
\[ \boxed{ F_{\text{avg}} = \frac1{b-a} \int_a^bF(x)\,dx }. \]Average value therefore has a direct physical interpretation as an equivalent constant level.
Average Height of a Graph
When \(f(x)\geq0\), the average value is sometimes called the average height of the graph over the interval.
The equal-area relation is
\[ \boxed{ (\text{average height})(\text{interval width}) = \text{area under the curve} }. \]Average Value Lies Between the Minimum and Maximum
If \(f\) is continuous on \([a,b]\) and
\[ m\leq f(x)\leq M, \]then
\[ m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a). \]Dividing by \(b-a>0\),
\[ \boxed{ m\leq f_{\text{avg}}\leq M }. \]Adding a Constant Shifts the Average
Suppose
\[ g(x)=f(x)+c. \]Then
\[ g_{\text{avg}} = \frac1{b-a} \int_a^b[f(x)+c]\,dx. \]By linearity,
\[ g_{\text{avg}} = f_{\text{avg}}+c. \]Therefore,
\[ \boxed{ (f+c)_{\text{avg}} = f_{\text{avg}}+c }. \]Scaling a Function Scales Its Average
If
\[ g(x)=cf(x), \]then
\[ \boxed{ g_{\text{avg}} = c f_{\text{avg}} }. \]This follows directly from the linearity of the definite integral.
Linearity of Average Value
For constants \(\alpha\) and \(\beta\),
\[ \boxed{ (\alpha f+\beta g)_{\text{avg}} = \alpha f_{\text{avg}} + \beta g_{\text{avg}} }. \]The averaging operation is linear.
Combining Average Values across Subintervals
Suppose \(a
The average over the full interval is not usually the simple arithmetic
mean of the two subinterval averages.
Instead, the averages must be weighted by interval length:
Example: Combining Two Time Periods
Suppose a process averages \(10\) units per hour for \(2\) hours and \(16\) units per hour for \(4\) hours.
The overall average is
\[ \frac{2(10)+4(16)}{6}. \]Therefore,
\[ \boxed{ \frac{84}{6}=14 }. \]The longer interval receives greater weight.
Average Value from Numerical Integration
If the integral cannot be evaluated exactly, approximate it numerically.
If
\[ \int_a^b f(x)\,dx\approx I, \]then
\[ \boxed{ f_{\text{avg}} \approx \frac{I}{b-a} }. \]Trapezoidal or Simpson estimates can therefore be converted directly into average-value estimates.
Example: Average Value from a Table
Suppose a rate is measured every hour:
| \(t\) | \(r(t)\) |
|---|---|
| 0 | 4 |
| 1 | 6 |
| 2 | 8 |
| 3 | 7 |
| 4 | 5 |
Using the trapezoidal rule,
\[ \int_0^4 r(t)\,dt \approx \frac12[4+2(6)+2(8)+2(7)+5]. \]This equals
\[ \frac12(51)=25.5. \]Therefore,
\[ r_{\text{avg}} \approx \frac{25.5}{4}. \]Hence,
\[ \boxed{r_{\text{avg}}\approx6.375}. \]Average over a Periodic Cycle
If \(f\) has period \(T\), a natural long-run cycle average is
\[ \boxed{ f_{\text{avg}} = \frac1T \int_a^{a+T}f(x)\,dx }. \]For a periodic function, this average is independent of the chosen starting point \(a\), provided the interval spans one complete period.
Example: Average of an Offset Sinusoid
Consider
\[ f(t)=12+3\sin t. \]Over one full period \([0,2\pi]\),
\[ f_{\text{avg}} = \frac1{2\pi} \int_0^{2\pi} (12+3\sin t)\,dt. \]The sine contribution averages to zero, so
\[ \boxed{f_{\text{avg}}=12}. \]Average Value Is Not the Same as Average Rate of Change
For a function \(f\), average value is
\[ \boxed{ \frac1{b-a} \int_a^bf(x)\,dx }. \]Average rate of change is
\[ \boxed{ \frac{f(b)-f(a)}{b-a} }. \]These are generally different quantities.
A Special Connection through Derivatives
If we average \(F'(x)\) over \([a,b]\), then
\[ \frac1{b-a} \int_a^bF'(x)\,dx. \]By the Fundamental Theorem of Calculus,
\[ \int_a^bF'(x)\,dx = F(b)-F(a). \]Therefore,
\[ \boxed{ (F')_{\text{avg}} = \frac{F(b)-F(a)}{b-a} }. \]So the average value of a derivative equals the average rate of change of the original function.
Common Error: Forgetting to Divide by Interval Length
The integral itself gives total accumulation:
\[ \int_a^bf(x)\,dx. \]Average value requires
\[ \boxed{ \frac{1}{b-a} \int_a^bf(x)\,dx }. \]Do not confuse the total with the average.
Common Error: Dividing by the Wrong Quantity
The denominator is the interval length
\[ \boxed{b-a}, \]not \(a+b\), not \(b\), and not the number of algebraic terms in the function.
Common Error: Taking Absolute Values Automatically
Average value is based on the signed integral.
If the problem asks for average value of \(f\), use
\[ \frac1{b-a}\int_a^bf(x)\,dx. \]Use
\[ \frac1{b-a}\int_a^b|f(x)|\,dx \]only when an average magnitude, such as average speed, is intended.
Common Error: Using the Midpoint Function Value as the Average
In general,
\[ f\left(\frac{a+b}{2}\right) \]is not equal to
\[ f_{\text{avg}}. \]They happen to agree for every linear function, but not for arbitrary nonlinear functions.
Common Error: Averaging Subinterval Averages Without Weights
If subintervals have different lengths, their averages must be weighted.
A two-hour average and an eight-hour average do not contribute equally to a ten-hour overall average.
A Reliable Average-Value Procedure
- Identify the function and interval \([a,b]\).
- Compute the interval length \(b-a\).
- Evaluate or approximate \(\int_a^bf(x)\,dx\).
- Divide by \(b-a\).
- Attach the same units as the original function.
- If asked for a point where the average occurs, solve \(f(c)=f_{\text{avg}}\).
- Check that the average lies between the function's minimum and maximum when appropriate.
Example: Full Average-Value Workflow
Find the average value of
\[ f(x)=x^3+1 \]on \([0,2]\).
The interval length is
\[ 2. \]Therefore,
\[ f_{\text{avg}} = \frac12 \int_0^2(x^3+1)\,dx. \]An antiderivative is
\[ \frac{x^4}{4}+x. \]Thus,
\[ f_{\text{avg}} = \frac12 \left[ \frac{x^4}{4}+x \right]_0^2. \] \[ = \frac12(4+2). \]Hence,
\[ \boxed{f_{\text{avg}}=3}. \]Example: Find Where the Average Occurs
For the same function,
\[ f(x)=x^3+1, \]we found
\[ f_{\text{avg}}=3. \]Solve
\[ x^3+1=3. \] \[ x^3=2. \]Therefore,
\[ \boxed{ c=\sqrt[3]{2} }. \]Practice 1: Constant Function
Find the average value of
\[ f(x)=7 \]on \([2,10]\).
Practice 2: Linear Function
Find the average value of
\[ f(x)=2x+3 \]on \([1,5]\).
Practice 3: Quadratic Function
Find the average value of
\[ f(x)=x^2 \]on \([1,4]\).
Practice 4: Trigonometric Function
Find the average value of
\[ f(x)=\cos x \]on \([0,\pi]\).
Practice 5: Odd Symmetry
Find the average value of
\[ f(x)=x^5 \]on \([-3,3]\).
Practice 6: Mean-Value Point
For
\[ f(x)=x^2 \]on \([0,3]\), find \(f_{\text{avg}}\), then find \(c\in[0,3]\) such that
\[ f(c)=f_{\text{avg}}. \]Practice 7: Average Velocity
A particle has velocity
\[ v(t)=4t-2 \]on \([0,3]\). Find its average velocity.
Practice 8: Average Speed Setup
Write the integral formula for average speed if velocity is \(v(t)\) on \([a,b]\).
Practice 9: Weighted Subinterval Averages
A process averages \(5\) units for \(3\) hours and \(11\) units for \(1\) hour. Find the overall average.
Practice 10: Units
If \(T(t)\) is measured in degrees Celsius, what are the units of
\[ \frac1{b-a}\int_a^bT(t)\,dt? \]Practice 11: Average Value versus Average Rate of Change
State the formula for:
- the average value of \(f\) on \([a,b]\);
- the average rate of change of \(f\) on \([a,b]\).
Practice 12: Identify the Error
A student computes
\[ \int_0^4x\,dx=8 \]and says the average value of \(f(x)=x\) on \([0,4]\) is \(8\).
Explain and correct the error.
Practice Answers
Practice 1
A constant function has the same average value as its constant level:
\[ \boxed{7}. \]Practice 2
Since the function is linear, average the endpoint values:
\[ f(1)=5,\qquad f(5)=13. \] \[ f_{\text{avg}} = \frac{5+13}{2}. \]Therefore,
\[ \boxed{9}. \]Practice 3
\[ f_{\text{avg}} = \frac1{4-1} \int_1^4x^2\,dx. \] \[ = \frac13 \left[ \frac{x^3}{3} \right]_1^4. \] \[ = \frac13 \cdot \frac{64-1}{3}. \] \[ = \frac{63}{9}. \]Hence,
\[ \boxed{7}. \]Practice 4
\[ f_{\text{avg}} = \frac1\pi \int_0^\pi\cos x\,dx. \] \[ = \frac1\pi[\sin x]_0^\pi. \] \[ =0. \]Therefore,
\[ \boxed{0}. \]Practice 5
Since \(x^5\) is odd and the interval is symmetric,
\[ \int_{-3}^{3}x^5\,dx=0. \]Therefore,
\[ \boxed{f_{\text{avg}}=0}. \]Practice 6
\[ f_{\text{avg}} = \frac13 \int_0^3x^2\,dx. \] \[ = \frac13 \left[ \frac{x^3}{3} \right]_0^3. \] \[ = \frac13(9). \]Thus,
\[ \boxed{f_{\text{avg}}=3}. \]Solve
\[ c^2=3. \]On \([0,3]\),
\[ \boxed{c=\sqrt3}. \]Practice 7
\[ v_{\text{avg}} = \frac13 \int_0^3(4t-2)\,dt. \] \[ = \frac13 [2t^2-2t]_0^3. \] \[ = \frac13(18-6). \] \[ = \boxed{4}. \]Practice 8
\[ \boxed{ \text{average speed} = \frac1{b-a} \int_a^b|v(t)|\,dt }. \]Practice 9
Weight by time:
\[ \frac{3(5)+1(11)}{4} = \frac{26}{4}. \]Therefore,
\[ \boxed{6.5}. \]Practice 10
Average value has the same units as the original function.
Therefore,
\[ \boxed{\text{degrees Celsius}}. \]Practice 11
Average value:
\[ \boxed{ \frac1{b-a} \int_a^bf(x)\,dx }. \]Average rate of change:
\[ \boxed{ \frac{f(b)-f(a)}{b-a} }. \]Practice 12
The integral \(8\) is total accumulation, not average value.
Divide by the interval length:
\[ f_{\text{avg}} = \frac8{4-0}. \]Therefore,
\[ \boxed{f_{\text{avg}}=2}. \]What to Remember
The average value of \(f\) on \([a,b]\) is
\[ \boxed{ f_{\text{avg}} = \frac1{b-a} \int_a^bf(x)\,dx }. \]It is the continuous counterpart of an arithmetic mean:
\[ \boxed{ \text{average} = \frac{\text{total}}{\text{amount of input}} }. \]Geometrically, \(f_{\text{avg}}\) is the height of an equal-area rectangle.
For continuous \(f\), the Mean Value Theorem for Integrals guarantees at least one point \(c\) such that
\[ \boxed{ f(c)=f_{\text{avg}} }. \]Average value has the same units as the original function and often represents an equivalent constant rate, density, temperature, force, or other continuously varying quantity.
Where We Go Next
Average values summarize continuous variation with one equivalent constant level. The next lesson, Accumulation Functions, returns to integrals with a variable endpoint and studies how an accumulated total itself behaves as a function.