Finding an Input That Attains the Average Value

From an Average Value to an Input

The average value of a function is a height. A natural follow-up question is: at what input does the function actually reach that height? For a continuous function on a closed interval, the Mean Value Theorem for Integrals guarantees at least one such input in the interval’s interior.

To find it, first calculate the average value, then set the function equal to that number and solve. The final step matters: a solution to the equation is an answer only if it lies in the required interval.

Definition: If \(f\) is continuous on \([a,b]\), where \(a<b\), its average value on that interval is

f_{\mathrm{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

An input \(c\) that attains the average value satisfies \(f(c)=f_{\mathrm{avg}}\).

The Mean Value Theorem for Integrals

The theorem supplies the existence guarantee. Its hypotheses are that \(f\) is continuous on the entire closed interval \([a,b]\) and that \(a<b\). Under those conditions, there is at least one \(c\) in the open interval \((a,b)\) such that \(f(c)\) equals the average value.

Theorem: If \(f\) is continuous on \([a,b]\), where \(a<b\), then there exists at least one \(c\in(a,b)\) such that

f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx.

The theorem guarantees an input, but it does not directly tell you what that input is. To find one, you must solve the displayed equation for \(c\). An algebraic solution outside \((a,b)\) does not satisfy the theorem’s conclusion.

Why the Theorem Guarantees Such an Input

Here is a justification of the existence claim. Because \(f\) is continuous on the closed interval, it has a minimum value \(m\) and a maximum value \(M\) there. Thus \(m\leq f(x)\leq M\) for every \(x\in[a,b]\). Integrating these inequalities and dividing by the positive interval length \(b-a\) gives

m\leq \frac{1}{b-a}\int_a^b f(x)\,dx\leq M.

Write the middle quantity as \(A\). If \(m<A<M\), choose points \(p\) and \(q\) where \(f\) attains \(m\) and \(M\), respectively. Since \(f(p)=m<A<M=f(q)\), the Intermediate Value Theorem gives a point between \(p\) and \(q\) where \(f\) equals \(A\). Because \(A\) differs from both endpoint values \(m\) and \(M\), this point lies strictly between \(p\) and \(q\), and therefore in \((a,b)\).

It remains to consider the boundary cases. If \(A=m\), then

\int_a^b (f(x)-m)\,dx = \int_a^b f(x)\,dx-m(b-a) = (b-a)(A-m) = 0.

The integrand \(f(x)-m\) is continuous and nonnegative. If it were positive at any point, continuity would make it positive on a small interval, giving a positive integral. Therefore it is zero everywhere, so \(f(x)=m=A\) throughout \([a,b]\). Any interior point works. The case \(A=M\) follows in the same way by applying the argument to \(M-f(x)\). This proves that at least one required input exists.

A Reliable Solving Process

In a particular problem, the existence proof is usually not the way to calculate \(c\). Use the average-value equation directly:

Compute the average.
Evaluate the definite integral and divide by the interval length \(b-a\).

Set the function equal to the average.
Write \(f(c)=f_{\mathrm{avg}}\), using the same function and interval as in the average calculation.

Solve for the input.
Use algebra or an appropriate inverse function to obtain candidate values.

Check and report.
Keep a candidate only if it lies in \((a,b)\), then substitute it into \(f\) to verify that it gives the average.

Worked Examples

Worked Example: A Quadratic on an Interval Starting at Zero

Find an input \(c\in(0,3)\) where \(f(x)=x^2\) equals its average value on \([0,3]\).

First calculate the average. The interval length is \(3-0=3\), and

\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3}-0 = 9.

Therefore,

f_{\mathrm{avg}}=\frac{9}{3}=3.

Set \(f(c)\) equal to \(3\) and solve:

c^2=3 \qquad\Longrightarrow\qquad c=\pm\sqrt{3}.

The interval is \((0,3)\), so \(-\sqrt{3}\) is outside it and must be discarded. Since \(0<\sqrt{3}<3\), the required input is \(c=\sqrt{3}\). Check by substitution:

f(\sqrt{3})=(\sqrt{3})^2=3=f_{\mathrm{avg}}.

Thus \(c=\sqrt{3}\) is an input inside the interval where the function attains its average value.

Worked Example: A Linear Function

Find an input \(c\in(1,4)\) where \(f(x)=2x+1\) equals its average value on \([1,4]\).

The interval length is \(4-1=3\). An antiderivative of \(2x+1\) is \(x^2+x\), so

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

The average value is \(18/3=6\). Now solve \(f(c)=6\):

2c+1=6 \qquad\Longrightarrow\qquad 2c=5 \qquad\Longrightarrow\qquad c=\frac{5}{2}.

The value \(5/2\) is between \(1\) and \(4\). Substitution confirms the answer:

f\left(\frac{5}{2}\right) = 2\left(\frac{5}{2}\right)+1 = 5+1 = 6 = f_{\mathrm{avg}}.

So \(c=5/2\) is an input in the required interval that attains the average.

Worked Example: An Exponential Function

Find an input \(c\in(0,1)\) where \(f(x)=e^x\) equals its average value on \([0,1]\).

The interval length is \(1-0=1\), and

\int_0^1 e^x\,dx = [e^x]_0^1 = e-1.

Dividing by the interval length leaves the average value \(f_{\mathrm{avg}}=e-1\). Solve the equation:

e^c=e-1 \qquad\Longrightarrow\qquad c=\ln(e-1).

To check that this input lies in the interval, use \(e>2\), which implies \(e-1>1\). Also \(e-1<e\). Since the natural logarithm is increasing,

0=\ln(1)<\ln(e-1)<\ln(e)=1.

Thus \(c=\ln(e-1)\) is in \((0,1)\). Substituting verifies the equation:

f(c)=e^{\ln(e-1)}=e-1=f_{\mathrm{avg}}.

The exact input is \(c=\ln(e-1)\).

Common Pitfall: Using the Integral Instead of the Average

The definite integral gives the total signed accumulation, not the average height by itself. The average is the integral divided by the interval length. Forgetting this division changes the equation you need to solve.

For example, on \([0,3]\), the integral of \(x^2\) is \(9\), but the average value is \(9/3=3\). A mistaken calculation might set \(c^2=9\), giving \(c=3\) as a candidate. That value is not even in the open interval \((0,3)\), and substitution gives \(f(3)=9\), not the correct average \(3\). The corrected equation is \(c^2=3\), which gives \(c=\sqrt{3}\) in the interval.

Key check: Before solving for \(c\), confirm that the right-hand side is \(\frac{1}{b-a}\int_a^b f(x)\,dx\), not just \(\int_a^b f(x)\,dx\). After solving, check both the interval and the equation \(f(c)=f_{\mathrm{avg}}\).

What the Solution Tells You

The average-value equation converts an integral into a target function value. Solving it identifies an input where the function reaches that target. In a contextual problem, the input may represent time, distance, or another independent variable; the interval check ensures the answer belongs to the time period or range under consideration.

Keep the roles of the quantities distinct: the integral accumulates values across the interval, the average value is a height, and \(c\) is an input. Continuity guarantees that a suitable input exists, while the equation and interval check identify a valid one for the specific function.

Check Your Understanding

For each question, show the average calculation, solve the equation, and check the proposed input against the interval.

For \(f(x)=x+2\) on \([0,4]\), find an input where \(f\) equals its average value.
For \(f(x)=x^2\) on \([1,3]\), write and solve the average-value equation. Which candidate lies in \((1,3)\)?
For \(f(x)=3x\) on \([0,2]\), calculate the average value and find the required input.
What conditions on \(f\) and \([a,b]\) allow the Mean Value Theorem for Integrals to guarantee an input in \((a,b)\)?
Why is dividing by \(b-a\) essential before solving for \(c\)?