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Average Value and Accumulated Quantities · Tutorial 1283 of 1495

How Shifts and Scaling Change an Average

See why adding a constant shifts an average by that constant, while multiplying a function by a constant scales its average by the same factor.

Intermediate 9 min read

What You'll Learn

  • Define the average of an integrable function on a nonzero interval.
  • Derive how vertical shifts change function averages.
  • Derive how constant multiplication changes function averages.
  • Combine the rules for a general affine transformation.
  • Apply the rules to a Celsius-to-Fahrenheit temperature model.
  • Avoid confusing a vertical transformation with a change to the input variable.

Introduction

A function average summarizes the values of a function across an interval. Once that average is known, a simple change to every function value has a predictable effect: adding the same amount to every value shifts the average by that amount, and multiplying every value by the same constant scales the average by that constant.

These rules are useful when a model is expressed in a new scale or has a constant baseline added to it. They also give a quick way to find an average after a transformation, without recomputing the transformed function’s integral from scratch.

Definition and Conditions

Let \(f\) be integrable on an interval \([a,b]\), where \(a<b\). Its average value on that interval is the definite integral divided by the interval’s length.

Definition: The average value of \(f\) on \([a,b]\) is \[ f_{\mathrm{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx. \]

The condition \(a<b\) ensures the interval has positive length, so the denominator is not zero. Integrability ensures the definite integral exists. For example, every continuous function on a closed interval is integrable there.

In this tutorial, a shift means adding a constant to the function’s output: \(f(x)+d\). A scaling means multiplying the output by a constant: \(cf(x)\). Neither operation changes the input \(x\) or the interval \([a,b]\).

The Two Transformation Rules

Suppose \(f\) is integrable on \([a,b]\), and \(c\) and \(d\) are real constants. The shift and scaling rules are:

Theorem: On the same interval \([a,b]\), \[ \bigl(f+d\bigr)_{\mathrm{avg}}=f_{\mathrm{avg}}+d \qquad\text{and}\qquad \bigl(cf\bigr)_{\mathrm{avg}}=c f_{\mathrm{avg}}. \] Therefore, for the combined transformation \(g(x)=cf(x)+d\), \[ g_{\mathrm{avg}}=c f_{\mathrm{avg}}+d. \]

The transformed function is integrable as well: adding a constant to an integrable function, or multiplying it by a constant, preserves integrability. The rules apply to positive, zero, and negative constants. If \(c=0\), the transformed function is identically zero and its average is zero.

Why the Rules Work

The reason is the linearity of the definite integral and the fact that the integral of a constant over an interval is that constant times the interval’s length. Here is the full derivation for \(g(x)=cf(x)+d\).

1
Write the average of the transformed function.
Since \(g(x)=cf(x)+d\) on the same interval, its average is \(\frac{1}{b-a}\int_a^b (cf(x)+d)\,dx\).
2
Use linearity of the integral.
For integrable \(f\) and constants \(c,d\), \(\int_a^b (cf(x)+d)\,dx=c\int_a^b f(x)\,dx+\int_a^b d\,dx\).
3
Integrate the constant and divide by the interval length.
Because \(\int_a^b d\,dx=d(b-a)\), dividing by \(b-a\) gives \(c\frac{1}{b-a}\int_a^b f(x)\,dx+d\).
4
Recognize the original average.
The expression is \(c f_{\mathrm{avg}}+d\), which proves the combined rule and, by setting one constant to zero or one as appropriate, the shift and scaling rules.

In equation form, the derivation is:

$$ \begin{aligned} g_{\mathrm{avg}} &=\frac{1}{b-a}\int_a^b \bigl(cf(x)+d\bigr)\,dx\\ &=\frac{1}{b-a}\left(c\int_a^b f(x)\,dx+\int_a^b d\,dx\right)\\ &=\frac{c}{b-a}\int_a^b f(x)\,dx+\frac{d(b-a)}{b-a}\\ &=c f_{\mathrm{avg}}+d. \end{aligned} $$

This proof also explains why the length of the interval matters in the definition, but not as an extra adjustment in the final transformation rule: the constant contribution \(d(b-a)\) is divided by that same interval length.

Worked Examples

Worked Example: Adding a Constant to a Function

Let \(f(t)=3t+2\) on \([0,4]\), and define \(g(t)=f(t)+5\). Find the average of \(g\), using the original average and the shift rule. Then check by direct integration.

First find the original average. The interval length is \(4-0=4\), and an antiderivative of \(3t+2\) is \(\frac{3}{2}t^2+2t\). Thus

$$ \begin{aligned} f_{\mathrm{avg}} &=\frac{1}{4}\int_0^4 (3t+2)\,dt\\ &=\frac{1}{4}\left[\frac{3}{2}t^2+2t\right]_0^4\\ &=\frac{1}{4}\left(\left(\frac{3}{2}\cdot16+8\right)-0\right)\\ &=\frac{1}{4}(24+8)=8. \end{aligned} $$

Since \(g=f+5\), the shift rule gives \(g_{\mathrm{avg}}=8+5=13\). To check, write \(g(t)=3t+7\) and calculate its average directly:

$$ \begin{aligned} \frac{1}{4}\int_0^4 (3t+7)\,dt &=\frac{1}{4}\left[\frac{3}{2}t^2+7t\right]_0^4\\ &=\frac{1}{4}\left((24+28)-0\right)\\ &=\frac{52}{4}=13. \end{aligned} $$

The direct computation agrees with the shift rule.

Worked Example: Scaling by a Negative Constant

Let \(f(x)=x+2\) on \([0,2]\), and define \(h(x)=-2f(x)\). Find the average of \(h\) by scaling the original average, then verify it directly.

The interval length is \(2-0=2\). The original average is

$$ \begin{aligned} f_{\mathrm{avg}} &=\frac{1}{2}\int_0^2 (x+2)\,dx\\ &=\frac{1}{2}\left[\frac{x^2}{2}+2x\right]_0^2\\ &=\frac{1}{2}\left((2+4)-0\right)\\ &=3. \end{aligned} $$

Since \(h=-2f\), the scaling rule gives \(h_{\mathrm{avg}}=-2(3)=-6\). For a direct check, \(h(x)=-2x-4\), so

$$ \begin{aligned} \frac{1}{2}\int_0^2 (-2x-4)\,dx &=\frac{1}{2}\left[-x^2-4x\right]_0^2\\ &=\frac{1}{2}\left((-4-8)-0\right)\\ &=\frac{-12}{2}=-6. \end{aligned} $$

The negative scaling factor reverses the sign of the average, as the result shows. The direct integral confirms the calculation.

Worked Example: Applying an Affine Temperature Conversion

Suppose a temperature model in degrees Celsius is \(C(t)=4+t\) for \(0\leq t\leq6\), where \(t\) is measured in hours. Convert it to Fahrenheit using \(F(t)=\frac{9}{5}C(t)+32\), and find the average Fahrenheit temperature over the interval.

First calculate the average Celsius temperature. The interval length is \(6-0=6\), so

$$ \begin{aligned} C_{\mathrm{avg}} &=\frac{1}{6}\int_0^6 (4+t)\,dt\\ &=\frac{1}{6}\left[4t+\frac{t^2}{2}\right]_0^6\\ &=\frac{1}{6}\left((24+18)-0\right)\\ &=\frac{42}{6}=7\text{ degrees Celsius}. \end{aligned} $$

The conversion has the form \(F=\frac{9}{5}C+32\). Therefore the combined rule gives

$$ F_{\mathrm{avg}} =\frac{9}{5}(7)+32 =\frac{63}{5}+\frac{160}{5} =\frac{223}{5} =44.6\text{ degrees Fahrenheit}. $$

To check, substitute \(C(t)=4+t\) into the conversion formula: \(F(t)=\frac{9}{5}(4+t)+32=\frac{196}{5}+\frac{9}{5}t\). Direct integration gives

$$ \begin{aligned} \frac{1}{6}\int_0^6 \left(\frac{196}{5}+\frac{9}{5}t\right)\,dt &=\frac{1}{6}\left[\frac{196}{5}t+\frac{9}{10}t^2\right]_0^6\\ &=\frac{1}{6}\left(\frac{1176}{5}+\frac{324}{10}\right)\\ &=\frac{1}{6}\left(\frac{1176}{5}+\frac{162}{5}\right)\\ &=\frac{1}{6}\cdot\frac{1338}{5} =\frac{223}{5} =44.6\text{ degrees Fahrenheit}. \end{aligned} $$

Both methods give the same average. The factor \(\frac{9}{5}\) scales the Celsius values, while \(32\) shifts them. The conversion is an affine transformation of the output values.

Interpreting and Using the Rules

The combined formula is especially useful when a model is transformed in one step. If \(g(x)=cf(x)+d\), first find or identify \(f_{\mathrm{avg}}\); then calculate \(c f_{\mathrm{avg}}+d\). A separate integral of \(g\) is not needed, although it can be useful as a check.

The constant \(d\) shifts every function value by the same amount. The average shifts by precisely the same amount. A factor \(c\) changes every function value in the same proportion, so the average changes by that proportion too. If \(c<0\), the sign of the average is reversed and its magnitude is multiplied by \(|c|\). If \(c=0\), every transformed value is zero.

Units should remain consistent. If \(f\) represents a quantity with a particular unit, then an added constant \(d\) must have that same unit to be added to \(f(x)\). The factor \(c\) can be dimensionless, as in a numerical rescaling, or can carry units that convert the output into another unit. In a conversion such as Celsius to Fahrenheit, both the scaling and offset are part of the conversion of output values.

These rules concern vertical transformations of the function: changing \(f(x)\) to \(cf(x)+d\). They do not describe changing the input, such as replacing \(f(x)\) by \(f(x+2)\), or changing the interval of averaging. Those operations alter the function or interval in a different way and are not covered by these formulas.

A Common Mistake

A frequent error is to apply a scale factor to the shift as well as to the original average. For example, if \(f_{\mathrm{avg}}=4\) and \(g(x)=2f(x)+3\), an incorrect calculation is

$$ g_{\mathrm{avg}}\ \text{(incorrect)}=2(4+3)=14. $$

This calculation treats the function as \(2(f(x)+3)\), which would be \(2f(x)+6\), not \(2f(x)+3\). The actual transformation scales \(f\) first and adds \(3\) afterward. Thus

$$ g_{\mathrm{avg}}=2f_{\mathrm{avg}}+3=2(4)+3=11. $$

The order and grouping in the expression matter. For \(cf+d\), multiply the original average by \(c\), then add \(d\).

Summary

For an integrable function on an interval \([a,b]\) with \(a<b\), the average is its integral divided by the interval length. On that same interval, adding a constant adds it to the average; multiplying by a constant multiplies the average by that constant. In combined form, if \(g(x)=cf(x)+d\), then \(g_{\mathrm{avg}}=c f_{\mathrm{avg}}+d\). These statements follow directly from linearity of the definite integral.

Check Your Understanding

Use the shift and scaling rules, and state the interval whenever an average is involved.

  1. If \(f_{\mathrm{avg}}=6\) and \(g(x)=f(x)-4\), what is \(g_{\mathrm{avg}}\)?
  2. If \(f_{\mathrm{avg}}=-3\) and \(h(x)=5f(x)\), what is \(h_{\mathrm{avg}}\)?
  3. If \(f_{\mathrm{avg}}=2\) and \(q(x)=-3f(x)+1\), what is \(q_{\mathrm{avg}}\)?
  4. What condition on the interval ensures the average-value formula has a nonzero denominator?
  5. In one sentence, explain why adding \(d\) to every function value adds \(d\) to the average.

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