From Graph or Table to Average Value
A graph or table can provide information about a function without giving a formula for it. To find the function’s average value, the key question is whether that information determines—or only helps estimate—the function’s integral over the interval.
A graph may show enough shape to compute signed area exactly. A table may supply integral values directly. But a list of function values at selected inputs is not, by itself, enough to determine an exact integral: many different functions can pass through the same points and have different areas between them.
The average value has the same units as \(f\). The integral has units of \(f\) multiplied by the units of \(x\); dividing by the interval width returns the units of \(f\).
Why the Integral Determines the Average
The definition can be understood by comparing the graph’s signed area with a rectangle. The rectangle has width \(b-a\) and constant height \(f_{\text{avg}}\). Its signed area is \((b-a)f_{\text{avg}}\). Substituting the definition gives the following derivation:
Thus the constant-height rectangle and the graph have the same signed integral over the interval. This does not mean the graph itself is a rectangle; it means that its accumulated signed area can be represented by a rectangle of the same width and average height.
Reading Graphs: Exact Area or Estimate?
When a graph is explicitly made of straight-line segments, its integral can be found exactly by adding the signed areas of trapezoids or triangles. The points where the segments meet give the heights needed for those calculations. If a graph is merely drawn as a curve, its appearance may support an estimate, but a sketch without exact geometric information usually does not specify an exact integral.
Worked Example: Average Value from a Piecewise-Linear Graph
Suppose \(f\) is drawn as straight-line segments through \((0,1)\), \((2,3)\), \((4,3)\), and \((6,1)\). Find its average value on \([0,6]\).
Because the graph consists of straight-line segments and stays above the \(x\)-axis, the integral is the sum of three trapezoid areas:
Therefore, \(\int_0^6 f(x)\,dx=4+6+4=14\). The interval width is \(6-0=6\), so
Check: A rectangle of width \(6\) and height \(7/3\) has area \(6(7/3)=14\), matching the integral computed from the graph.
If some parts of a graph lie below the \(x\)-axis, their signed contributions are negative. The formula uses the signed integral, not the sum of all geometric areas taken as positive. For an exact answer, the graph or accompanying information must specify enough to calculate those signed contributions.
Tables That Give Integral Information
A table can provide the accumulated integral over each subinterval, rather than just the function’s value at selected inputs. These are different kinds of information. If adjacent subinterval integrals are supplied, add them to obtain the integral over the full interval, then divide by the full interval width.
Worked Example: Average Value from Subinterval Integrals
A table gives the following signed integral values for a function \(f\):
| Subinterval | Supplied integral |
|---|---|
| \([1,3]\) | \(\int_1^3 f(x)\,dx=8\) |
| \([3,5]\) | \(\int_3^5 f(x)\,dx=-2\) |
Find the average value on \([1,5]\). Add the two contributions:
The full interval has width \(5-1=4\). Therefore,
Check: A constant height of \(3/2\) across width \(4\) gives signed area \(4(3/2)=6\), the supplied total integral. The negative subinterval contribution is included with its sign.
Some tables instead give values of \(f(x)\) at particular inputs. Those entries are point samples, not subinterval integrals. They may support a numerical estimate if an estimation method is requested, but they do not automatically determine the exact average.
Using Point Samples to Estimate
One common estimate joins neighboring data points with straight segments and adds the resulting trapezoid areas. This is the trapezoidal estimate of the integral. It is exact if the function really is linear on each subinterval; otherwise, it is an estimate based on that straight-segment assumption. Unless the problem states that the graph is piecewise linear, do not present the estimate as an exact integral.
Worked Example: A Trapezoidal Estimate from a Table
Suppose a table lists \(f(0)=2\), \(f(1)=4\), \(f(3)=2\), and \(f(4)=0\). Estimate the average value on \([0,4]\) by treating the graph between successive samples as linear.
The widths of the three subintervals are \(1\), \(2\), and \(1\). The estimated signed areas are:
The total estimated integral is \(3+6+1=10\). Dividing by the interval width \(4-0=4\) gives
Check: The piecewise-linear graph used for the estimate has total area \(10\). A rectangle of width \(4\) and height \(2.5\) also has area \(4(2.5)=10\). This checks the arithmetic under the stated linear-interpolation assumption; the original samples alone do not guarantee that the actual average is \(2.5\).
Why Point Samples Do Not Fix an Exact Average
Knowing a function’s values at a finite list of inputs restricts its graph at those points. It does not, in general, determine how the graph behaves between them. The following example gives two continuous functions with identical values at three sample inputs but different average values.
Worked Example: Same Samples, Different Averages
On \([0,2]\), compare \(f_0(x)=0\) with \(f_{15}(x)=15x(x-1)(x-2)^2\). At \(x=0\), \(x=1\), and \(x=2\), both functions have value \(0\), because the second function has a zero at each of those inputs.
The zero function has integral \(0\), so its average value is \(0/2=0\). For the other function, first expand the polynomial:
Then integrate term by term over \([0,2]\):
Multiplying by \(15\) gives \(\int_0^2 f_{15}(x)\,dx=-4\). Its average value is therefore \((-4)/(2-0)=-2\).
Check: The two functions match at every listed sample input, yet their averages are \(0\) and \(-2\). Thus those point samples cannot determine a unique exact average.
A Common Mistake
A tempting shortcut is to average the listed function values and call the result the function’s average value. That is generally incorrect. The function’s average value is defined using an integral over an interval, not the arithmetic mean of a few samples.
For example, in the trapezoidal-estimate example, the arithmetic mean of the four listed values is \((2+4+2+0)/4=2\). It is not the trapezoidal estimate \(2.5\). The spacing is unequal: the interval \([1,3]\) has width \(2\), while the other two subintervals have width \(1\). The trapezoidal method accounts for widths by weighting each pair of endpoint values by its subinterval width. Even with equally spaced samples, the arithmetic mean is not automatically the exact function average; the samples still do not specify all the values between them.
Check Your Understanding
For each question, decide whether the requested result is exact or estimated, and show how the interval width is used.
- A piecewise-linear graph on \([0,4]\) has signed areas \(3\) on \([0,1]\), \(-2\) on \([1,3]\), and \(1\) on \([3,4]\). Find its average value.
- A table gives \(\int_2^5 f(x)\,dx=9\) and \(\int_5^7 f(x)\,dx=1\). Find the average value on \([2,7]\).
- Why do values of \(f\) at five selected inputs not, in general, determine the exact average over the interval containing them?
- When is the trapezoidal calculation from neighboring graph points exact, and when should it be described as an estimate?
- A student averages all the ordinates in an unequally spaced table. Explain why this need not give the function’s average value.
