8.1 - Finding the Average Value of a Function on an Interval

Related Ideas Worth Untangling First

These four things sound alike and get mixed up constantly.

Intermediate Value Theorem (Unit 1)

f(x)

is continuous on

[a, b]

, then

f

takes on every value between

f(a)

and

f(b)

somewhere in the interval.

8.1 - Finding the Average Value of a Function on an Interval

Average Rate of Change (Unit 2)

\text{Average Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}

That’s the slope of the secant line connecting

(a, f(a))

and

(b, f(b))

Mean Value Theorem (Unit 5)

f

is continuous on

[a, b]

and differentiable on

(a, b)

, there’s at least one point

c

(a, b)

where

f'(c) = \frac{f(b) - f(a)}{b - a}

In other words, the instantaneous rate of change equals the average rate of change somewhere.

Average Value of a Function

For a continuous function

f(x)

[a, b]

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

In words: integrate the function over the interval, then divide by the length of the interval.

Intuition: the integral adds up all the y-values (weighted by width), and dividing by

(b - a)

turns that total into an average.

Example 1: Finding Average Value

Find the average value of

f(x) = 4 - x^2

[0, 2]

Step 1: Set up.

f_{\text{avg}} = \frac{1}{2} \int_0^2 (4 - x^2) \, dx

Step 2: Evaluate.

= \frac{1}{2} \left[4x - \frac{x^3}{3}\right]_0^2 = \frac{1}{2}\left[8 - \frac{8}{3}\right] = \frac{1}{2} \cdot \frac{16}{3} = \frac{8}{3}

Answer:

\frac{8}{3} \approx 2.67

Example 2: Where Does the Function Equal Its Average?

Continuing from Example 1, at what

x

does

f(x) = 4 - x^2

equal its average value of

\frac{8}{3}

4 - x^2 = \frac{8}{3} \quad \Rightarrow \quad x^2 = \frac{4}{3} \quad \Rightarrow \quad x = \pm\frac{2\sqrt{3}}{3}

We only care about

[0, 2]

, so take the positive value:

x = \frac{2\sqrt{3}}{3} \approx 1.155

Example 3: Average Height

A ball’s height in feet is

h(t) = -16t^2 + 48t + 5

, with

t

in seconds. Find the average height from

t = 0

t = 2

Key insight: “average height” means average value of

h(t)

h_{\text{avg}} = \frac{1}{2} \int_0^2 (-16t^2 + 48t + 5) \, dt

= \frac{1}{2} \left[-\frac{16t^3}{3} + 24t^2 + 5t\right]_0^2 = \frac{1}{2}\left[-\frac{128}{3} + 96 + 10\right]

= \frac{1}{2} \cdot \frac{190}{3} = \frac{95}{3} \approx 31.67 \text{ feet}

Critical Distinction: Average Value vs. Average Rate of Change

This is where people lose points.

Scenario: Given a Position Function

Say

s(t) = t^3 - 6t^2 + 9t

is position in meters,

t

in seconds.

Question A: What is the average velocity from $t = 0$ to $t = 3$ ?

Velocity is the rate of change of position, so use the slope formula:

v_{\text{avg}} = \frac{s(3) - s(0)}{3 - 0}

Question B: What is the average of the velocity from $t = 0$ to $t = 3$ ?

First find

v(t) = s'(t) = 3t^2 - 12t + 9

, then take its average value:

v_{\text{avg}} = \frac{1}{3} \int_0^3 (3t^2 - 12t + 9) \, dt

Here’s the thing: both methods give the same answer. By the Fundamental Theorem of Calculus,

\frac{1}{b-a}\int_a^b v(t) \, dt = \frac{1}{b-a}\int_a^b s'(t) \, dt = \frac{s(b) - s(a)}{b-a}

Example 4: Rate Function Context

Water flows into a tank at a rate of

R(t) = 5 + 2\sin(t) \text{ gallons per minute}

with

t

in minutes.

Part A: Average Rate

Question: what is the average rate of flow from

t = 0

t = \pi

R(t)

is already a rate, and we want the average value of that rate:

R_{\text{avg}} = \frac{1}{\pi} \int_0^\pi (5 + 2\sin t) \, dt = \frac{1}{\pi}\left[5t - 2\cos t\right]_0^\pi

= \frac{1}{\pi}\left[(5\pi + 2) - (-2)\right] = \frac{5\pi + 4}{\pi} = 5 + \frac{4}{\pi} \approx 6.27 \text{ gal/min}

Part B: Average Change in the Rate

Question: what is the average change of the rate from

t = 0

t = \pi

Now we want the rate of change of the rate, so use the slope formula:

\frac{R(\pi) - R(0)}{\pi - 0} = \frac{[5 + 2\sin\pi] - [5 + 2\sin 0]}{\pi} = \frac{5 - 5}{\pi} = 0

Units here are gallons per minute per minute.

Decision Tree: Which Formula?

First ask: is the given function already a rate?

If NO (it’s a position, area, population, etc.):

Question asks for	Use
Average value of the function	$\frac{1}{b-a}\int_a^b f(x)\,dx$
Average rate of change	$\frac{f(b) - f(a)}{b - a}$

If YES (it’s a velocity, growth rate, flow rate, etc.):

Question asks for	Use
Average rate	$\frac{1}{b-a}\int_a^b R(x)\,dx$ (average value)
Average change of the rate	$\frac{R(b) - R(a)}{b - a}$ (slope)

Example 5: Complete Problem

Let

f(x) = x^2 - 4x + 5

[1, 4]

Part A: Average value of $f$ .

f_{\text{avg}} = \frac{1}{3} \int_1^4 (x^2 - 4x + 5) \, dx = \frac{1}{3}\left[\frac{x^3}{3} - 2x^2 + 5x\right]_1^4

x = 4

\frac{64}{3} - 32 + 20 = \frac{64}{3} - 12

x = 1

\frac{1}{3} - 2 + 5 = \frac{1}{3} + 3

= \frac{1}{3}\left[\frac{63}{3} - 15\right] = \frac{1}{3}\left[21 - 15\right] = \frac{6}{3} = 2

Average value: $2$

Part B: Average rate of change on $[1, 4]$ .

\frac{f(4) - f(1)}{4 - 1} = \frac{5 - 2}{3} = 1

Part C: Find all $c$ in $[1, 4]$ where $f(c) = f_{\text{avg}}$ .

x^2 - 4x + 5 = 2 \quad \Rightarrow \quad x^2 - 4x + 3 = 0 \quad \Rightarrow \quad (x-1)(x-3) = 0

c = 1 \text{ and } c = 3

Both are in

[1, 4]

Calculator Tips

Common mistake: computing only

\int_a^b f(x)\,dx

and forgetting to divide by

(b - a)

Correct approach:


(1/(b-a)) * fnInt(f(x), x, a, b)

Reusing functions: store the function in Y₁, then reference it as Y₁ in your calculations. Y₁(3) evaluates it at

x = 3

Common Mistakes

Mistake 1: confusing average value with average rate. Given a position function

s(t)

, average velocity is

\frac{s(b) - s(a)}{b - a}

\frac{1}{b-a}\int_a^b v(t)\,dt

, NOT

\frac{1}{b-a}\int_a^b s(t)\,dt

Mistake 2: forgetting the $\frac{1}{b-a}$ factor. The integral alone gives total accumulation, not an average.

Mistake 3: misreading the function type. Always ask whether you’ve been handed an amount or a rate.

Mistake 4: wrong units. Average value carries the same units as the function’s outputs. The average of a height in feet is in feet, and the average of a flow rate in gallons per minute is in gallons per minute (like Example 4, Part A). Dividing by

(b - a)

cancels the time unit the integral picked up, so you never tack on an extra “per time.”

Key Takeaways

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

Don’t forget the $\frac{1}{b-a}$ on the calculator

Average rate of change uses the slope formula: $\frac{f(b) - f(a)}{b - a}$

The Mean Value Theorem guarantees $f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $c$

Read carefully: is the given function a rate or an amount? That decides the formula

The average of a rate uses the average value formula (an integral)

Average velocity can be computed either way, they agree

Average value has the same units as the function’s output

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