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How the Interval Affects an Average Value

Learn to compare interval averages by weighting each portion according to its length and comparing the added portion’s average with the original average.

Intermediate 10 min read

What You'll Learn

  • Express an interval’s average value as its integral divided by its length.
  • Derive the weighted-average formula for adjacent intervals.
  • Predict how extending an interval changes its average.
  • Determine when the average rises, falls, or stays unchanged.
  • Identify when an unweighted mean of two averages is valid.

Why the Interval Matters

The average value of a function depends not only on the function but also on the interval over which it is averaged. If an interval is extended, the new average reflects both the original interval and the added portion. The key is to compare the average on each portion while accounting for its length.

For example, a short portion with a high average may affect the overall average less than a much longer portion with a moderately high average. This is why simply taking the arithmetic mean of two interval averages is not always correct.

Definition: If \(f\) is integrable on an interval \([a,b]\) with \(a<b\), its average value on that interval is \[ f_{\mathrm{avg},[a,b]}=\frac{1}{b-a}\int_a^b f(x)\,dx. \] The denominator \(b-a\) is the length of the interval.

Continuity on \([a,b]\) is a standard sufficient condition for the integral and average value to exist. In the comparisons below, all intervals have positive length, so each average is defined.

How an Added Portion Changes an Average

Suppose an interval \([a,b]\) is extended to \([a,c]\), where \(a<b<c\). The added portion is \([b,c]\). Let \(A_1\) be the average of \(f\) on the original interval, and let \(A_2\) be its average on the added portion. Their lengths are \(L_1=b-a\) and \(L_2=c-b\).

The average on the extended interval is a weighted average of \(A_1\) and \(A_2\). The weights are the lengths of the portions, not merely the number of portions.

$$ f_{\mathrm{avg},[a,c]} = \frac{L_1A_1+L_2A_2}{L_1+L_2}. $$

This formula gives a quick comparison rule: the extended average moves toward the average on the added portion. If that portion’s average is greater than the original average, the extended average rises. If it is less, the extended average falls. If the two averages are equal, extending the interval leaves the average unchanged.

Why the Weighted-Average Formula Works

The integral over the extended interval is the sum of the integrals over its two pieces. By the definitions of \(A_1\) and \(A_2\), the integral over each piece equals its length multiplied by its average. These facts yield the formula directly.

1
Write each integral using its average.
\(\int_a^b f(x)\,dx=L_1A_1\) and \(\int_b^c f(x)\,dx=L_2A_2\), because each average is the integral divided by its interval length.
2
Split the integral over the full interval.
Additivity of the definite integral gives \(\int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx=L_1A_1+L_2A_2\).
3
Divide by the full length.
The length of \([a,c]\) is \(L_1+L_2\), so its average is \((L_1A_1+L_2A_2)/(L_1+L_2)\), as claimed.

The formula also shows precisely how much the average changes. Subtracting \(A_1\) from both sides and simplifying gives

$$ f_{\mathrm{avg},[a,c]}-A_1 = \frac{L_2}{L_1+L_2}(A_2-A_1). $$

Since \(L_2/(L_1+L_2)\) is positive, the sign of the change is determined by \(A_2-A_1\). This proves the comparison rule: the full-interval average increases, decreases, or stays the same exactly as the added portion’s average is greater than, less than, or equal to the original average.

Worked Example: Extending an Interval for \(f(x)=x^2\)

Worked Example: Extending an Interval for \(f(x)=x^2\)

Find the average value of \(f(x)=x^2\) on \([0,5]\) by comparing the original interval \([0,2]\) with the added portion \([2,5]\). Then explain how the average changes.

First find the average on the original interval. Its length is \(2-0=2\), and

$$ \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}-0 = \frac{8}{3}. $$

Thus the original average is \((8/3)/2=4/3\).

The added interval has length \(5-2=3\). Its integral and average are

$$ \int_2^5 x^2\,dx = \left[\frac{x^3}{3}\right]_2^5 = \frac{125}{3}-\frac{8}{3} = \frac{117}{3} = 39, \qquad A_2=\frac{39}{3}=13. $$

The added portion’s average, \(13\), is greater than \(4/3\), so the extended average must increase. Applying the weighted formula,

$$ f_{\mathrm{avg},[0,5]} = \frac{2\left(\frac{4}{3}\right)+3(13)}{2+3} = \frac{\frac{8}{3}+39}{5} = \frac{\frac{8}{3}+\frac{117}{3}}{5} = \frac{\frac{125}{3}}{5} = \frac{25}{3}. $$

Check directly on the full interval:

$$ \frac{1}{5-0}\int_0^5 x^2\,dx = \frac{1}{5}\left[\frac{x^3}{3}\right]_0^5 = \frac{1}{5}\cdot\frac{125}{3} = \frac{25}{3}. $$

The direct calculation agrees with the weighted calculation, and \(25/3\) is greater than the original average \(4/3\), as predicted.

Adjacent Intervals Use the Same Reasoning

The same formula applies when two adjacent intervals are combined. For \([a,b]\) and \([b,c]\), let their lengths be \(L_1\) and \(L_2\), and their average values be \(A_1\) and \(A_2\). The average over their union \([a,c]\) is

$$ f_{\mathrm{avg},[a,c]} = \frac{L_1A_1+L_2A_2}{L_1+L_2}. $$

This is not generally the unweighted mean \((A_1+A_2)/2\). The unweighted mean is generally valid when the two intervals have equal lengths. It also gives the correct result when their averages happen to be equal, even if their lengths differ.

Worked Example: Combining Adjacent Intervals

Worked Example: Combining Adjacent Intervals

Let \(f(x)=2x+1\). Find the average on \([-1,1]\), the average on \([1,4]\), and the average on their union \([-1,4]\).

The first interval has length \(1-(-1)=2\). Its integral is

$$ \int_{-1}^{1}(2x+1)\,dx = \left[x^2+x\right]_{-1}^{1} = (1+1)-(1-1) = 2. $$

Its average is \(2/2=1\). The second interval has length \(4-1=3\), and

$$ \int_1^4(2x+1)\,dx = \left[x^2+x\right]_1^4 = (16+4)-(1+1) = 20-2 = 18. $$

Its average is \(18/3=6\). The weighted average over the union is therefore

$$ f_{\mathrm{avg},[-1,4]} = \frac{2(1)+3(6)}{2+3} = \frac{2+18}{5} = 4. $$

Check using the full interval integral:

$$ \frac{1}{4-(-1)}\int_{-1}^{4}(2x+1)\,dx = \frac{1}{5}\left[x^2+x\right]_{-1}^{4} = \frac{(16+4)-(1-1)}{5} = \frac{20}{5} = 4. $$

The second interval has the greater average, and its length is greater as well. The union’s average, \(4\), lies between \(1\) and \(6\), as a weighted average should.

Worked Example: Adding a Portion with a Lower Average

Worked Example: Adding a Portion with a Lower Average

For \(f(x)=4-x\), compare its average on \([0,2]\) with its average on \([0,5]\). Use the added interval to explain the change.

On \([0,2]\), the length is \(2\), and

$$ \int_0^2(4-x)\,dx = \left[4x-\frac{x^2}{2}\right]_0^2 = (8-2)-0 = 6. $$

The original average is \(6/2=3\). The added interval \([2,5]\) has length \(3\). Its integral is

$$ \int_2^5(4-x)\,dx = \left[4x-\frac{x^2}{2}\right]_2^5 = \left(20-\frac{25}{2}\right)-(8-2) = \frac{15}{2}-6 = \frac{3}{2}. $$

Its average is \((3/2)/3=1/2\), which is below the original average \(3\). The extended average is

$$ f_{\mathrm{avg},[0,5]} = \frac{2(3)+3\left(\frac{1}{2}\right)}{2+3} = \frac{6+\frac{3}{2}}{5} = \frac{\frac{12}{2}+\frac{3}{2}}{5} = \frac{15}{2}\cdot\frac{1}{5} = \frac{3}{2}. $$

Checking directly,

$$ \frac{1}{5}\int_0^5(4-x)\,dx = \frac{1}{5}\left[4x-\frac{x^2}{2}\right]_0^5 = \frac{1}{5}\left(20-\frac{25}{2}\right) = \frac{1}{5}\cdot\frac{15}{2} = \frac{3}{2}. $$

The extended average is \(3/2\), below the original average \(3\) and above the added portion’s average \(1/2\).

A Common Mistake: Averaging the Averages Equally

Suppose \(f(x)=x\) on the adjacent intervals \([0,1]\) and \([1,4]\). Their lengths are \(1\) and \(3\). The first average is

$$ \frac{1}{1-0}\int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}. $$

The second average is

$$ \frac{1}{4-1}\int_1^4 x\,dx = \frac{1}{3}\left[\frac{x^2}{2}\right]_1^4 = \frac{1}{3}\left(8-\frac{1}{2}\right) = \frac{1}{3}\cdot\frac{15}{2} = \frac{5}{2}. $$

A tempting but incorrect calculation for the union’s average is the unweighted mean:

$$ \frac{\frac{1}{2}+\frac{5}{2}}{2} = \frac{3}{2}. $$

This treats a length-\(1\) interval and a length-\(3\) interval as if they contributed equally. The correct weighted calculation is

$$ \frac{1\left(\frac{1}{2}\right)+3\left(\frac{5}{2}\right)}{1+3} = \frac{\frac{1}{2}+\frac{15}{2}}{4} = \frac{8}{4} = 2. $$

Check directly:

$$ \frac{1}{4-0}\int_0^4 x\,dx = \frac{1}{4}\left[\frac{x^2}{2}\right]_0^4 = \frac{1}{4}\cdot 8 = 2. $$

Thus \(3/2\) is the wrong result, while \(2\) is correct. The unweighted mean can still be correct for unequal lengths if the two component averages happen to be equal: in that case, both the weighted and unweighted calculations return that shared average.

A Reliable Comparison Strategy

When an interval is extended or two adjacent intervals are combined, use the following process:

  1. Find the length of each interval.
  2. Find each portion’s average by dividing its integral by its own length.
  3. Combine the averages using length as the weight.
  4. Compare the added portion’s average with the original average to predict the direction of change.
  5. Check the result using the integral over the full interval when practical.

The added portion is important as a whole. Its average, rather than a single function value at its endpoint, determines whether extending the interval raises or lowers the original average. A high value at one point does not by itself determine the average over an entire added interval.

Key takeaway: An interval average is an integral divided by that interval’s length. When portions are combined, their averages must be weighted by their lengths; extending an interval moves its average toward the average on the added portion.

Check Your Understanding

Use interval lengths and component averages to answer each question. Show the weighted calculation where appropriate.

  1. An interval of length \(2\) has average \(5\). An adjacent interval of length \(4\) has average \(8\). What is the average on their union?
  2. An interval has average \(6\). An added portion has average \(2\). Does extending the interval raise, lower, or preserve the average?
  3. Two adjacent intervals have lengths \(3\) and \(3\), with averages \(4\) and \(10\). Find the union’s average. Explain why the unweighted mean works here.
  4. Two intervals have unequal lengths but the same average \(7\). What is the average over their union, and why?
  5. An interval of length \(1\) has average \(0\), and an adjacent interval of length \(5\) has average \(3\). Explain why the union’s average is not generally the unweighted mean of the two averages, and find its value.

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