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

1. What the Average Value of a Function Is

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

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

Break that formula apart:

• $\int_a^b f(x)\,dx$ → total accumulated change (signed area).
• $b-a$ → length of the interval.
• So you’re doing
  (total accumulation) ÷ (interval length).

This mirrors the discrete average:

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

Integration replaces the “sum of values” with an infinite sum over the interval.

Geometric meaning

Now think visually. The definite integral gives the area under the curve on $[a,b]$. The average value is the height of a horizontal line so that a rectangle with base $b-a$ has the same area as that region.

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

Geometric interpretation of average value on $[a,b]$

In the middle panel, the rectangle’s height is the average value. In the right panel, the horizontal line balances the areas above and below it so the total signed area matches the integral.

So the average value is literally a “balance height” for the function on that interval.

2. How to Find the Average Value

When you’re asked to find it, the process is mechanical.

1. Write the formula

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

2. Evaluate the definite integral

   • Find an antiderivative $F(x)$.
   • Compute $F(b) - F(a)$.

3. Divide by $b-a$.

Quick example

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

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

Antiderivative:

$$\int (3x^2-2x)\,dx = x^3 - x^2$$

Evaluate:

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

Divide:

$$\frac{1}{2} \cdot 4 = 2$$

Average value = 2.

That’s a typical no-calculator FRQ part. Clean setup and algebra matter.

If You’re Given a Graph

You may not have a formula. Then:

• Compute signed area.
  • Above x-axis → positive
  • Below x-axis → negative
• Add all pieces.
• Divide by $b-a$.

If a region dips below the axis, that lowers the average. Students often forget the sign and accidentally compute total area instead.

If the Function Is Defined by an Integral

Suppose

$$g(x)=\int_1^x f(t)\,dt.$$

To find the average value of $g$ on $[2,5]$, you must compute:

$$\frac{1}{5-2}\int_2^5 g(x)\,dx.$$

You are averaging $g$, not $f$. That means another integral. On FRQs, this is where FTC Part 1 or Part 2 shows up. Slow down and track which function is being averaged.

3. Why This Shows Up on Tests

You’ll see three main versions.

Straight computation

Very common as an early FRQ part. They want:

• Correct setup
• Proper antiderivative
• Final simplified number

Missing the $\frac{1}{b-a}$ costs easy points.

Conceptual reasoning

Questions like:

• Is the average value positive or negative?
• Is it greater than 1?
• Between which two values does it lie?

If most of the graph lies above the x-axis, expect a positive average. Large negative regions pull it down.

Connection to the Mean Value Theorem for Integrals

If $f$ is continuous on $[a,b]$, then there exists some $c \in (a,b)$ such that

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

So the function actually hits its average value somewhere.

Average value and the Mean Value Theorem for Integrals

In the middle panel, the rectangle with width $b-a$ has the same area as the shaded region under $f$. That rectangle’s height is the average value. In the right panel, the horizontal line at that height intersects the curve, showing the point $c$ where $f(c)$ equals the average.

You usually don’t solve for $c$, but you should understand what it represents.

4. Average Value vs. Average Rate of Change

Students mix these up constantly.

One measures height. The other measures slope.