Topic 8.1 Notes – Finding the Average Value of a Function on an Interval

What the Average Value of a Function Is

If $f$ is continuous on $[a,b]$, the average value of $f$ on that interval is

$$\text{Average Value} = \frac{1}{b-a}\int_a^b f(x)\,dx$$

Think about what each piece means:

• $\int_a^b f(x)\,dx$
  → total signed area (net accumulation) between $a$ and $b$
• $b-a$
  → length of the interval

So you’re doing:

$$\text{average} = \frac{\text{total accumulation}}{\text{interval length}}$$

It’s the continuous version of

$$\text{average} = \frac{\text{sum}}{\text{number of values}}$$

Except instead of adding up individual y-values, the integral adds up infinitely many values smoothly.

Also important: the average value has the same units as $f(x)$. If $f$ is in meters per second, the average value is meters per second.

The Formula and What Each Part Means

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

Here’s how to interpret it correctly:

• If most of the graph is above the x-axis → average value is positive.
• If most is below → average value is negative.
• If positive and negative areas cancel → the average could be small or even zero.

This is about signed area, not total area. That difference matters a lot on tests.

There’s also a helpful geometric interpretation. The average value is the height $k$ of a rectangle that has:

• Base $b-a$
• The same area as the region under $f(x)$

In equation form:

$$k(b-a) = \int_a^b f(x)\,dx$$

Picture a rectangle drawn from $x=a$ to $x=b$ with a constant height $k$ that matches the area under the curve on that interval.

Average value as an equivalent-area rectangle

The rectangle’s height is exactly the average value.

How to Find the Average Value

When you’re given a formula, the process is mechanical:

1. Write
   $$\frac{1}{b-a}\int_a^b f(x)\,dx$$
2. Find an antiderivative $F(x)$.
3. Evaluate $F(b) - F(a)$.
4. Divide by $b-a$.

Example (quick one):

Find the average value of $f(x) = 3x^2 - 4$ on $[0,2]$.

$$\frac{1}{2-0} \int_0^2 (3x^2 - 4)\,dx$$

Antiderivative:
$$x^3 - 4x$$

Evaluate:
$$[x^3 - 4x]_0^2 = (8 - 8) - 0 = 0$$

Divide by 2:

$$f_{\text{avg}} = 0$$

Even though the function isn’t zero everywhere, the positive and negative accumulation cancel over that interval.

From a Graph

If they give you a graph:

• Break the region into geometric shapes.
• Add signed areas (below the x-axis counts negative).
• Divide by $b-a$.

This shows up often in no-calculator multiple choice. If you forget to treat below-axis area as negative, you’ll pick the trap answer.

Common Confusions

Average value vs. average rate of change

• Average value uses an integral:
  $$\frac{1}{b-a}\int_a^b f(x)\,dx$$
• Average rate of change uses endpoints:
  $$\frac{f(b)-f(a)}{b-a}$$

These are completely different ideas. One averages outputs over an interval. The other measures slope between endpoints.

Forgetting the denominator

A lot of students stop after computing the definite integral. That’s only the total accumulation, not the average.

Using total area instead of signed area

Definite integrals include negatives automatically. Don’t convert everything to positive unless the question explicitly says “total area.”