Why Bound an Average Value?
An average value summarizes how a function behaves across an entire interval. Sometimes an exact calculation is useful; sometimes a reliable range is enough. In either case, knowing that a function stays between a minimum and a maximum immediately limits how large or small its average can be.
This gives a practical check on a computed average. If a calculation produces a number smaller than every function value, or larger than every function value, something has gone wrong. Bounds can also provide an estimate when the exact integral is not needed.
Definition and Bounds
For a continuous function \(f\) on an interval \([a,b]\), where \(a<b\), its average value is the accumulated signed value divided by the interval’s length:
Because \(f\) is continuous on the closed interval \([a,b]\), it attains both a minimum and a maximum there. Write
Thus, for every \(x\) in the interval, \(m\le f(x)\le M\). The average-value bound follows from these pointwise inequalities.
The theorem does not say that the average must equal an endpoint value or that it is halfway between the minimum and maximum. It says only that the average cannot fall outside the range of values the function actually takes.
Why the Bounds Hold
The key idea is that integrating preserves inequalities over an interval. Since \(f(x)-m\ge 0\) throughout \([a,b]\), its definite integral is nonnegative. Likewise, \(M-f(x)\ge 0\), so that integral is nonnegative too.
Using the integral of a constant, the first inequality gives
so \(\int_a^b f(x)\,dx\ge m(b-a)\). The second gives
so \(\int_a^b f(x)\,dx\le M(b-a)\). Combining these results,
Because \(b-a>0\), dividing all three parts by \(b-a\) preserves the inequality and yields the average-value bound. This argument also works when \(f\) takes negative values: the interval length is still positive, and the inequalities account for the function’s signed values.
Use the Bounds to Check a Computed Average
The bound is most useful as a check when the minimum and maximum are easy to identify. First find or establish the function’s range on the whole interval. Then calculate the average if needed and compare it with that range.
Worked Example: Checking an Exact Average
Find the average value of \(f(x)=x^2+1\) on \([0,2]\), and check the result using minimum and maximum bounds.
On \([0,2]\), \(x^2\) increases from \(0\) to \(4\). Therefore \(f(x)=x^2+1\) has minimum \(m=1\) and maximum \(M=5\). The theorem says the average must lie between \(1\) and \(5\).
Now compute the average. An antiderivative of \(x^2+1\) is \(\frac{x^3}{3}+x\), so
Since \(1\le \frac{7}{3}\le 5\), the computed value passes the bound check. The average is not required to be halfway between \(1\) and \(5\); here it is \(\frac{7}{3}\).
As a direct check of the integration, differentiating the antiderivative gives \(\frac{d}{dx}\left(\frac{x^3}{3}+x\right)=x^2+1\), the original function.
Use Bounds for an Estimate in Context
An average value has the same units as the function. If \(r(t)\) is a rate, its average over a time interval is an average rate. The integral of the rate gives accumulated change; the bound on the average rate therefore also provides a bound on that accumulation after multiplication by the interval length.
Worked Example: Bounding an Average Rate
A flow rate is modeled by \(r(t)=12+0.2t\) liters per minute for \(0\le t\le30\), where \(t\) is measured in minutes. Bound its average rate using the minimum and maximum, then calculate the average rate and total volume to check the bounds.
Since the rate increases with \(t\), its minimum occurs at \(t=0\) and its maximum at \(t=30\):
Thus \(12\le r(t)\le18\) liters per minute throughout the interval. Its average rate must also lie between \(12\) and \(18\) liters per minute.
To compute the average rate exactly, integrate and divide by the interval length:
Indeed, \(12\le15\le18\). The accumulated volume is \(450\) liters, which is also between the amounts obtained by holding the minimum or maximum rate constant for \(30\) minutes:
The bounds give a useful range even before the integral is evaluated. Here the exact average and accumulation provide a further check.
A Tighter Range Gives a Tighter Estimate
The minimum and maximum should describe the function over the entire interval. If a broad range is all that is known, the resulting average bound may be broad too. If the function’s values can be confined to a narrower range, the average bound narrows with them.
Worked Example: Estimating an Average on a Short Interval
Bound the average value of \(f(x)=\sqrt{x}\) on \([1,1.21]\), then calculate it to see how informative the bounds are.
The square-root function increases on this interval, so its minimum and maximum are
Therefore the average value is between \(1\) and \(1.1\). To calculate it, note that \(1.21=\frac{121}{100}\), so the interval length is \(\frac{21}{100}\). An antiderivative of \(\sqrt{x}\) is \(\frac{2}{3}x^{3/2}\):
The result satisfies \(1\le\frac{331}{315}\le\frac{11}{10}\). In decimal form, the average is approximately \(1.05079\), which is consistent with the bound from \(1\) to \(1.1\). The exact fraction is the computed value; the decimal is rounded to five places.
This example shows the role of the interval: because all function values are confined to a narrow range, the average is confined to that same narrow range.
A Common Mistake: Checking Only the Endpoints
The values at the endpoints do not always give the minimum and maximum on the interval. Using them as bounds without checking the function’s behavior can lead to an incorrect average bound.
Worked Example: Correcting an Endpoint-Only Bound
Consider \(f(x)=x^2\) on \([-1,1]\). Both endpoint values equal \(1\). A mistaken argument might declare the minimum and maximum to be \(1\), and then conclude that the average must be \(1\).
That conclusion fails because the function takes smaller values between the endpoints: \(f(0)=0\). In fact, \(x^2\ge0\) throughout the interval and reaches its maximum value \(1\) at both endpoints, so the actual bounds are \(m=0\) and \(M=1\).
The correct average is
Now the bound check is valid: \(0\le\frac13\le1\). Differentiating the antiderivative \(\frac{x^3}{3}\) gives \(x^2\), confirming the integral calculation. The error was not in the average-value formula; it was in mistaking endpoint values for the function’s range.
Using Bounds Carefully
Before applying the theorem, confirm that the interval has positive length and that the function is continuous on the closed interval. Then identify bounds that hold for every point in that interval. Continuity guarantees that a minimum and maximum exist, but finding them still requires attention to the function’s behavior across the full interval, not just at its endpoints.
The resulting bounds are inclusive: the average can equal a minimum or maximum in cases where the function is constant. If the function is not constant, its average is still between its minimum and maximum, but this basic bound does not usually tell you the exact value. It is a check or estimate, not a replacement for an integral when an exact average is required.
Check Your Understanding
For each question, identify the interval-wide bounds before drawing a conclusion about the average.
- A continuous function satisfies \(3\le f(x)\le8\) on an interval \([a,b]\), where \(a<b\). What bounds must its average value satisfy?
- For \(g(x)=2x+1\) on \([0,3]\), find its minimum and maximum, then state the resulting bounds for its average value.
- If a continuous rate \(r(t)\) is between \(5\) and \(9\) meters per second for \(12\) seconds, what bounds follow for its average rate and its accumulated distance?
- Why is it not enough to check only the endpoint values when bounding an average?
- A computed average is larger than the maximum value of the function on the interval. What does that tell you about the calculation or the proposed maximum?
