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Average Value and Accumulated Quantities · Tutorial 1290 of 1495

Average-Value Problems with Unknown Parameters

Turn a specified average into an equation for an unknown parameter, then verify that each solution is allowed by the model.

Intermediate 10 min read

What You'll Learn

  • Write the average-value equation for a parameterized function on a fixed interval.
  • Distinguish a specified average from a specified integral total.
  • Solve linear and nonlinear parameter equations obtained from integrals.
  • Check every candidate against the stated parameter domain.
  • Recognize when an average condition has more than one admissible solution.

Introduction

A specified average can reveal an unknown parameter in a function model. For example, if a model depends on a coefficient that has not yet been measured, knowing the function’s average over an interval may provide an equation for that coefficient.

The key is to translate the average condition into an integral equation with the correct interval length. Solving that equation gives candidate parameter values; the model’s domain then determines which candidates are permitted. An average condition does not always determine exactly one value.

Setting Up the Parameter Equation

Suppose a model is described by a family of functions \(f_p\), where \(p\) is an unknown real parameter. On an interval \([u,v]\) with \(u<v\), the average value of an integrable function \(f_p\) is its integral divided by the interval length.

Definition: The average value of \(f_p\) on \([u,v]\) is \[ f_{\mathrm{avg}}(p)=\frac{1}{v-u}\int_u^v f_p(x)\,dx. \] If the specified average is \(M\), the parameter must satisfy \[ \frac{1}{v-u}\int_u^v f_p(x)\,dx=M, \] or, equivalently, \[ \int_u^v f_p(x)\,dx=M(v-u). \]

The interval must have positive length, so \(u<v\). Continuity of \(f_p\) on \([u,v]\) is a standard sufficient condition for the integral to exist. The model may also restrict the parameter, for example by requiring \(p\geq 0\). Any solution to the integral equation must be checked against such restrictions.

Why the Average Equation Works

The equation comes directly from the definition of average value. The integral measures the signed accumulation over the interval, while dividing by \(v-u\) expresses that accumulation per unit of interval length. Thus, the specified average \(M\) requires a total integral equal to \(M\) times the interval length.

$$ \frac{1}{v-u}\int_u^v f_p(x)\,dx=M \quad\Longleftrightarrow\quad \int_u^v f_p(x)\,dx=M(v-u). $$

Because \(v-u\) is a fixed positive number, multiplying both sides by it preserves the equality. This equivalence is useful when evaluating the integral is simpler than working with the average formula throughout. It also highlights a common distinction: a specified average \(M\) corresponds to a specified integral \(M(v-u)\), not generally to an integral of \(M\).

Worked Examples

Worked Example: Finding a Linear Coefficient

Suppose \(f_p(x)=px+2\) on \([0,4]\), and the average value is \(7\). The model requires \(p\geq 0\). Find the permitted value of \(p\).

The interval length is \(4-0=4\). Set the average equal to \(7\):

$$ \frac{1}{4}\int_0^4 (px+2)\,dx=7. $$

Integrate each term and evaluate at the endpoints:

$$ \int_0^4 (px+2)\,dx = \left[\frac{p x^2}{2}+2x\right]_0^4 = \left(\frac{16p}{2}+8\right)-0 = 8p+8. $$

Therefore,

$$ \frac{8p+8}{4}=7 \quad\Longrightarrow\quad 2p+2=7 \quad\Longrightarrow\quad 2p=5 \quad\Longrightarrow\quad p=\frac{5}{2}. $$

This candidate meets the restriction \(p\geq 0\). To check it, substitute \(p=\frac{5}{2}\) into the average:

$$ \frac{1}{4}\int_0^4 \left(\frac{5}{2}x+2\right)\,dx = \frac{1}{4}\left(8\cdot\frac{5}{2}+8\right) = \frac{1}{4}(20+8) = 7. $$

The average is \(7\), as specified, so \(p=\frac{5}{2}\) is the permitted solution.

Worked Example: Two Parameters Give the Same Average

Let \(f_p(x)=(x-p)^2\) on \([0,1]\), with parameter domain \(0\leq p\leq 1\). Find all parameter values for which the average is \(\frac{1}{3}\).

Expand the function, then set its average equal to the target:

$$ \frac{1}{1-0}\int_0^1 (x-p)^2\,dx = \int_0^1 (x^2-2px+p^2)\,dx = \frac{1}{3}. $$

Since \(p\) is constant with respect to \(x\), evaluate the integral term by term:

$$ \left[\frac{x^3}{3}-p x^2+p^2x\right]_0^1 = \left(\frac{1}{3}-p+p^2\right)-0 = \frac{1}{3}. $$

Subtract \(\frac{1}{3}\) from both sides and factor:

$$ \frac{1}{3}-p+p^2=\frac{1}{3} \quad\Longrightarrow\quad p^2-p=0 \quad\Longrightarrow\quad p(p-1)=0. $$

Thus, the candidates are \(p=0\) and \(p=1\). Both are in the required domain \([0,1]\). Check each one in the average:

$$ \frac{1}{1-0}\int_0^1 (x-0)^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}, $$ $$ \frac{1}{1-0}\int_0^1 (x-1)^2\,dx = \left[\frac{x^3}{3}-x^2+x\right]_0^1 = \left(\frac{1}{3}-1+1\right)-0 = \frac{1}{3}. $$

Each value produces the required average. In this model, the average condition does not identify a unique parameter.

Worked Example: Applying a Parameter Restriction

Consider \(g_c(t)=c^2t\) on \([0,2]\), where the model requires \(c\geq 0\). Find the parameter value for which the average is \(2\).

The interval length is \(2\). The average condition gives:

$$ \frac{1}{2}\int_0^2 c^2t\,dt=2. $$

Since \(c^2\) is constant with respect to \(t\),

$$ \frac{1}{2}\left[c^2\frac{t^2}{2}\right]_0^2 = \frac{1}{2}\left(c^2\frac{4}{2}-0\right) = \frac{1}{2}(2c^2) = c^2. $$

Therefore \(c^2=2\), which gives the algebraic candidates \(c=\sqrt{2}\) and \(c=-\sqrt{2}\). The model requires \(c\geq 0\), so only \(c=\sqrt{2}\) is admissible. Substitution checks the average:

$$ \frac{1}{2}\int_0^2 (\sqrt{2})^2t\,dt = \frac{1}{2}\int_0^2 2t\,dt = \frac{1}{2}\left[t^2\right]_0^2 = \frac{1}{2}(4) = 2. $$

Thus \(c=\sqrt{2}\) satisfies both the average condition and the parameter restriction. The negative root solves the squared equation, but it is not a value allowed by this model.

A Reliable Procedure

For a parameterized model, it helps to separate the calculation from the model check. The equation may produce candidates that are inadmissible, or several candidates that all fit the stated average.

1
Record the interval and the target.
Identify the endpoints \(u\) and \(v\), the interval length \(v-u\), and the specified average \(M\).
2
Write the average equation.
Use \(\frac{1}{v-u}\int_u^v f_p(x)\,dx=M\), or its equivalent total-integral form.
3
Evaluate the integral and solve.
Keep the parameter in the expression and solve the resulting equation for all candidates.
4
Apply the model’s domain.
Discard any candidate that violates the stated parameter restrictions or makes the model undefined on the interval.
5
Verify the remaining candidates.
Substitute each one into the original average condition and confirm that it gives the specified value.

Common Pitfall: Confusing an Average with a Total

A frequent error is to set the integral equal to the specified average without accounting for the interval length. In the first example, the incorrect equation would be

$$ \int_0^4 (px+2)\,dx=7. $$

Using the evaluated integral gives \(8p+8=7\), so this incorrect setup yields \(p=-\frac{1}{8}\). But that value does not produce an average of \(7\): its integral is \(7\), and its average over an interval of length \(4\) is \(\frac{7}{4}\).

The correct total integral must be the target average times the interval length. Here, it must be \(7\cdot 4=28\). Accordingly, the correct equation is \(8p+8=28\), which gives \(p=\frac{5}{2}\), and the check confirms the average is \(7\).

Why the Domain Check Matters

The integral equation describes which parameter values produce the specified average mathematically. The model’s domain describes which parameter values are allowed. These are separate requirements: a value can solve the equation and still be disallowed. The condition may also have no solutions or multiple permitted solutions, so it is important to report what the calculation actually establishes rather than assuming uniqueness.

When the model represents a quantity with restrictions, such as a coefficient required to be nonnegative, those restrictions are part of the problem. Even without a physical interpretation, stating the admissible parameter set makes the conclusion precise. A complete answer gives the candidate values, identifies which ones meet the domain condition, and checks the retained values in the original average equation.

Check Your Understanding

For each question, identify both the equation to solve and the check needed to validate the result.

  1. On \([0,2]\), a function is \(h_p(x)=px+1\), and its average is \(5\). Find \(p\), assuming \(p\) is unrestricted.
  2. Why does a target average \(M\) on \([u,v]\) require an integral of \(M(v-u)\), rather than usually an integral of \(M\)?
  3. A parameter equation gives two real roots, but the model requires \(p\geq 0\). What must be done before giving the final answer?
  4. For \(f_p(x)=(x-p)^2\) on \([0,1]\), the average is \(\frac{1}{3}\) when \(p=0\) and when \(p=1\). What does this show about uniqueness?
  5. What should be checked after solving the average equation, even if every algebraic step is correct?

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