From an Integral Average to a Point
An integral can describe the total accumulation of a varying quantity across an interval. Dividing that total by the interval’s length gives its average value. The Integral Mean Value Theorem makes a stronger connection: under a continuity condition, the average is not merely a numerical summary. The function actually takes that value somewhere in the interval.
This result concerns existence. It guarantees at least one point with the required function value, but it does not generally say that the point is unique or provide a formula for finding it.
The Integral Average
Let \(f\) be integrable on an interval \([a,b]\) with \(a<b\). The length of the interval is \(b-a\), which is positive. The integral average of \(f\) on \([a,b]\) is defined by dividing the definite integral by that length.
The numerator records the signed accumulation, and the denominator accounts for how long the interval is. Thus, if \(f\) represents a rate over time, its integral average is an average rate over that time interval. This is the average-value formula developed in the earlier material on average values.
The theorem below does not merely define this number. It states that for a continuous function, the number is attained as an ordinary function value at some point strictly between the endpoints.
The Integral Mean Value Theorem
The conditions have specific roles. The interval must have positive length, so that the average is defined. Continuity on the entire closed interval ensures the hypotheses needed for the standard integral and derivative results used in the proof. The theorem gives \(c\) in the open interval \((a,b)\), not merely somewhere in the closed interval.
It is important to distinguish the conclusion from a method for locating \(c\). The theorem guarantees that at least one such point exists. If an exact point is requested in a particular problem, it may take additional calculation to identify one.
Why the Theorem Is True
The proof packages the accumulated integral into a new function. Its rate of change is the original function. The Mean Value Theorem then identifies a point where this rate of change equals the average rate of change of the accumulated integral across the whole interval.
Set \(F(x)=\int_a^x f(t)\,dt\) for \(x\in[a,b]\). Because \(f\) is continuous, \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\).
There is some \(c\in(a,b)\) such that \(F'(c)=\frac{F(b)-F(a)}{b-a}\).
The Fundamental Theorem gives \(F'(c)=f(c)\), while \(F(b)-F(a)=\int_a^b f(t)\,dt\), since \(F(a)=0\).
Substitution yields \(f(c)=\frac{1}{b-a}\int_a^b f(t)\,dt\), which is exactly the claimed equality.
This proves the theorem: the accumulated integral changes from \(0\) to the total \(\int_a^b f(t)\,dt\) across a distance \(b-a\). Its average rate of change is the integral average of \(f\), and the Mean Value Theorem says that this average rate is the derivative \(F'(c)=f(c)\) at some interior point.
Worked Applications
Worked Example: A Quadratic on a Closed Interval
Consider \(f(x)=x^2\) on \([0,3]\). This polynomial is continuous on the closed interval, and its length is \(3-0=3\). First compute the integral:
Therefore the integral average is \(9/3=3\). To check a point where the function equals this average, solve \(c^2=3\). The solution in \((0,3)\) is \(c=\sqrt{3}\). Indeed, \(0<\sqrt{3}<3\), and \(f(\sqrt{3})=(\sqrt{3})^2=3\), exactly the average. The theorem guarantees such a point; this calculation verifies the specific point.
Worked Example: A Linear Function
Let \(f(x)=2x+1\) on \([1,4]\). The function is continuous there, and the interval length is \(4-1=3\). Integrate and evaluate at both endpoints:
The integral average is \(18/3=6\). To verify an interior point, solve \(2c+1=6\): subtracting \(1\) gives \(2c=5\), so \(c=5/2\). Since \(1<5/2<4\), it lies in the required open interval. Substitution gives \(f(5/2)=2(5/2)+1=5+1=6\), the integral average.
Worked Example: Two Points Can Realize the Average
Consider \(f(x)=x^2-2x+2\) on \([0,2]\). It is continuous, and the interval length is \(2\). Its integral is
Thus the integral average is \(\frac{8}{3}/2=\frac{4}{3}\). Solve for points where \(f(c)=4/3\):
Hence \(c=1\pm\frac{1}{\sqrt{3}}\). Both values are in \((0,2)\), because \(0<1-\frac{1}{\sqrt{3}}\) and \(1+\frac{1}{\sqrt{3}}<2\). For either value, \((c-1)^2=1/3\), and rewriting \(f(c)=(c-1)^2+1\) gives \(f(c)=1/3+1=4/3\). This example verifies two points, illustrating that the theorem guarantees at least one, not exactly one.
Hypotheses and a Common Pitfall
A frequent error is to confuse the integral with the integral average. The integral is the total signed accumulation; the average also divides by the interval length. For \(f(x)=2x+1\) on \([1,4]\), the integral is \(18\), but the average is \(18/(4-1)=6\). If one mistakenly claims the average is \(18\), solving \(2c+1=18\) gives \(c=17/2\), which is not in \((1,4)\). The problem is not a failure of the theorem: the calculation used the wrong value. With the correct average, \(c=5/2\) works.
Continuity is the theorem’s stated hypothesis, and it cannot simply be dropped from the general claim. For example, define \(f(x)=0\) for \(0\le x<1/2\) and \(f(x)=2\) for \(1/2\le x\le1\). This function is integrable on \([0,1]\), and its integral is \(0\cdot(1/2)+2\cdot(1/2)=1\), so its integral average is \(1\). But the function takes only the values \(0\) and \(2\), never \(1\). Thus there is no point \(c\) with \(f(c)\) equal to its integral average.
This example shows that continuity is an important sufficient condition for the theorem, not that continuity is necessary in every individual case. Some discontinuous functions may still take their integral average. The theorem promises the conclusion when its hypotheses hold; it does not assert that those hypotheses are the only possible route to that conclusion.
What the Result Establishes
The Integral Mean Value Theorem turns an accumulated total into a pointwise statement. When \(f\) is continuous on \([a,b]\), the total integral can be written as the value of \(f\) at some interior point multiplied by the interval length. In that sense, the function’s behavior across the entire interval has an exact representative value somewhere inside it.
The result depends on continuity over the closed interval, a nonzero interval length, the Fundamental Theorem of Calculus, and the Mean Value Theorem. Its conclusion is about existence, not uniqueness. These distinctions are useful when interpreting an average rate or accumulated quantity: a computed average corresponds to an actual function value under the stated conditions, but one should not infer more than the theorem guarantees.
Check Your Understanding
Answer each question using the hypotheses and conclusion of the theorem.
- State the Integral Mean Value Theorem, including the condition on the interval and the continuity requirement.
- Why must \(a<b\) in the definition of the integral average?
- In the proof, what function is defined from the integral, and what is its derivative on \((a,b)\)?
- For \(f(x)=x^2\) on \([0,3]\), what is the integral average, and which interior point from the worked example realizes it?
- Why does the discontinuous example on \([0,1]\) show that the continuity hypothesis cannot simply be omitted from the theorem?
