AP Calculus AB Unit 8 Notes: Average Value, Net Change, and Motion via Integration

Finding the Average Value of a Function on an Interval

What “average value” means (and why it’s an integral idea)

When you hear “average,” you probably think “add up the values and divide by how many values there are.” That works for a finite list of numbers. But a function on an interval has infinitely many values—one for each input. Average value of a function is the continuous-version of an average: you are averaging the function’s outputs over every input in an interval.

The key idea is that an integral naturally represents “total accumulated amount.” If you want an average, you take:

• total amount over the interval
• divided by length of the interval

This mirrors “sum divided by count,” but “count” becomes “interval length,” because in a continuous setting your inputs are spread over distance (or time) rather than being a list.

A very helpful picture is area. If f(x) is above the x-axis on [a,b], then \int_a^b f(x)\,dx is the area under the curve. The **average value** is the height of a rectangle with the same base b-a that has the same area as the region under the curve. In other words: average value is the “constant height” that would give the same total accumulation.

The average value formula

If f is integrable on [a,b], the **average value** of f on [a,b] is

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

Interpretation of each part:

• \int_a^b f(x)\,dx is the total accumulation of f across the interval.
• b-a is the interval length.
• Dividing gives “accumulation per unit of x.”

Why this matters in AP Calculus (and beyond)

Average value shows up because it connects three big themes:

1. Riemann sums to integrals: average value is essentially a limit of averages of sample values.
2. Modeling: average temperature, average density, average rate, average velocity.
3. Mean Value Theorem for Integrals (often tested conceptually): for a continuous function, the average value is actually achieved somewhere on the interval.

How average value connects to Riemann sums

Suppose you split [a,b] into n equal subintervals of width \Delta x = \frac{b-a}{n}. Pick sample points x_i^*. The average of the sampled function values is

\frac{1}{n}\sum_{i=1}^n f(x_i^*)

Now multiply and divide by b-a in a strategic way:

\frac{1}{n}\sum_{i=1}^n f(x_i^*) = \frac{1}{b-a}\sum_{i=1}^n f(x_i^*)\Delta x

As n \to \infty, the sum \sum f(x_i^*)\Delta x becomes \int_a^b f(x)\,dx. That limit leads directly to the formula

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

So average value isn’t a random formula—you’re literally averaging infinitely many values via a limiting process.

Mean Value Theorem for Integrals (conceptual anchor)

If f is continuous on [a,b], then there exists at least one number c in [a,b] such that

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

This is sometimes called the Mean Value Theorem for Integrals. It says the function hits its average height somewhere.

Common pitfall: students sometimes think you’re supposed to find c uniquely. On AP problems, you might be asked to **set up an equation** to find c, but there can be multiple solutions (or the problem may only ask you to justify existence).

Worked Example 1: Average value of a function

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

Step 1: Write the average value formula.

f_{avg} = \frac{1}{3-0}\int_0^3 x^2\,dx

Step 2: Compute the integral.

\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3}-0 = 9

Step 3: Divide by the interval length.

f_{avg} = \frac{1}{3}\cdot 9 = 3

Interpretation: On average, x^2 takes the value 3 over [0,3].

Worked Example 2: Average value and “equal area rectangle” intuition

Suppose a tank’s inflow rate (in gallons per minute) is r(t) for 0 \le t \le 10. The total gallons added is \int_0^{10} r(t)\,dt. The average inflow rate over the 10 minutes is

r_{avg} = \frac{1}{10}\int_0^{10} r(t)\,dt

This is exactly “total gallons added divided by total minutes.” That unit check is a great way to avoid mistakes:

• r(t) has units gallons per minute.
• \int r(t)\,dt has units gallons.
• Divide by minutes gives gallons per minute again.

Common misconceptions (and how to avoid them)

1. Mixing up average value with average rate of change.

   • Average rate of change of f on [a,b] is \frac{f(b)-f(a)}{b-a} (a slope).
   • Average value of f on [a,b] is \frac{1}{b-a}\int_a^b f(x)\,dx (an average height).
     They are different ideas with different units.

2. Forgetting that integrals can be negative.
   If f(x) dips below the axis, the integral counts signed area. The “average value” is then a signed average. That’s still valid mathematically, but you need to interpret it correctly (for example, an average velocity can be negative).

3. Using the wrong interval length.
   The denominator is always b-a, not a-b, and not something like \int_a^b 1\,dx unless you recognize those are the same value.

Exam Focus

• Typical question patterns:
  • “Find the average value of f(x) on [a,b]” (straight computation).
  • “A quantity has rate r(t); find the average rate on a time interval” (interpretation + units).
  • “Use the Mean Value Theorem for Integrals to justify there exists c such that f(c)=f_{avg}” (conceptual justification and/or equation setup).
• Common mistakes:
  • Using \frac{f(a)+f(b)}{2} (that’s only the average of endpoints, not average value in general).
  • Confusing average value with average rate of change (integral average vs secant slope).
  • Dropping the parentheses in \frac{1}{b-a} and effectively computing \int_a^b f(x)\,dx - (b-a) instead of dividing.

Connecting Position, Velocity, and Acceleration Using Integrals

The big picture: derivatives describe motion, integrals rebuild it

In motion problems, you usually work with three related functions:

• Position s(t) (where you are)
• Velocity v(t) (how fast and in what direction you’re moving)
• Acceleration a(t) (how velocity changes)

The derivative relationships are:

v(t)=s'(t)

a(t)=v'(t)=s''(t)

AP Calculus often flips the direction: if derivatives break down motion into rates, then integrals rebuild the original quantities from their rates. This is the core idea of accumulation and the Fundamental Theorem of Calculus.

Accumulation and net change (the unifying idea)

A central principle for Unit 8 is the Net Change Theorem:

If F'(x)=f(x) on [a,b], then

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

This says: integrating a rate of change over an interval gives the total change in the original quantity.

For motion, that becomes:

• Integrate velocity to get change in position (displacement).
• Integrate acceleration to get change in velocity.

Displacement vs total distance (a crucial distinction)

If you know velocity v(t) on [a,b], then

Displacement (net change in position) is

s(b)-s(a)=\int_a^b v(t)\,dt

Displacement is signed: moving left (negative velocity) subtracts from your position.

Total distance traveled is

\int_a^b |v(t)|\,dt

This counts all movement as positive. On AP questions, this difference is frequently the entire point.

A quick reality check:

• If you walk 3 miles east and then 3 miles west, your displacement is 0 but your distance is 6.

From acceleration to velocity: using initial conditions

Since a(t)=v'(t), integrating acceleration gives the change in velocity:

v(b)-v(a)=\int_a^b a(t)\,dt

If you are given an initial velocity like v(0)=5, then you can reconstruct velocity:

v(t)=v(0)+\int_0^t a(u)\,du

Notice the dummy variable u. Using a different letter inside the integral helps you avoid a common confusion: the variable of integration is a placeholder.

From velocity to position: again, initial conditions matter

Similarly, because v(t)=s'(t),

s(b)-s(a)=\int_a^b v(t)\,dt

And with an initial position s(0)=2,

s(t)=s(0)+\int_0^t v(u)\,du

This is the accumulation-function viewpoint: position is “starting position plus accumulated velocity.”

Notation reference (helps with reading AP questions)

You’ll see multiple notations for the same ideas:

How to think about signs (so you don’t get trapped)

• If v(t) > 0, position is increasing (moving right).
• If v(t) < 0, position is decreasing (moving left).
• If a(t) > 0, velocity is increasing (could mean speeding up if v(t) > 0, or slowing down if v(t) < 0).

A very common mistake is to equate “positive acceleration” with “speeding up.” Speed is |v(t)|, not v(t). You speed up when v(t) and a(t) have the same sign.

Worked Example 1: Displacement and total distance from a velocity function

Let v(t)=t^2-4t+3 on [0,4].

A. Find displacement on [0,4].

\text{Displacement}=\int_0^4 (t^2-4t+3)\,dt

Compute:

\int_0^4 (t^2-4t+3)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_0^4

Evaluate:

\left(\frac{64}{3}-32+12\right)-0 = \frac{64}{3}-20 = \frac{4}{3}

So s(4)-s(0)=\frac{4}{3}. Net movement is slightly to the right.

B. Find total distance traveled on [0,4].

Total distance is

\int_0^4 |t^2-4t+3|\,dt

You must find where velocity changes sign:

t^2-4t+3=(t-1)(t-3)

Zeros at t=1 and t=3. Test intervals:

• On [0,1], pick t=0: v(0)=3>0.
• On [1,3], pick t=2: v(2)=4-8+3=-1

So

\int_0^4 |v(t)|\,dt = \int_0^1 v(t)\,dt - \int_1^3 v(t)\,dt + \int_3^4 v(t)\,dt

Use the antiderivative from part A:

\int_0^1 v(t)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_0^1 = \frac{1}{3}-2+3 = \frac{4}{3}

\int_1^3 v(t)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_1^3

Compute values:

\left(\frac{27}{3}-18+9\right)-\left(\frac{1}{3}-2+3\right) = (9-18+9)-\frac{4}{3} = 0-\frac{4}{3} = -\frac{4}{3}

So -\int_1^3 v(t)\,dt = \frac{4}{3}.

\int_3^4 v(t)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_3^4 = \left(\frac{64}{3}-20\right)-0 = \frac{4}{3}

Total distance:

\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4

Notice what happened: the displacement was \frac{4}{3}, but the total distance is 4 because you reversed direction.

Worked Example 2: Reconstructing velocity and position from acceleration

A particle has acceleration a(t)=6t-4 for 0 \le t \le 3. Given v(0)=2 and s(0)=5, find v(t) and s(t).

Step 1: Build velocity from acceleration.

Use

v(t)=v(0)+\int_0^t a(u)\,du

Compute the integral:

\int_0^t (6u-4)\,du = \left[3u^2-4u\right]_0^t = 3t^2-4t

So

v(t)=2+3t^2-4t

Step 2: Build position from velocity.

Use

s(t)=s(0)+\int_0^t v(u)\,du

Substitute v(u)=2+3u^2-4u:

\int_0^t (2+3u^2-4u)\,du = \left[2u+u^3-2u^2\right]_0^t = 2t+t^3-2t^2

Therefore

s(t)=5+2t+t^3-2t^2

Why initial conditions matter: without v(0), you’d only know velocity up to a constant. Without s(0), you’d only know position up to a constant. Integrals introduce constants; initial conditions pin them down.

Average value shows up in motion, too

Because velocity is a function, you can talk about its average value on a time interval. The average velocity on [a,b] is

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

But notice:

\int_a^b v(t)\,dt = s(b)-s(a)

So

v_{avg} = \frac{s(b)-s(a)}{b-a}

This is a great connection: in motion, “average value of velocity” matches the familiar physics idea “displacement divided by time.”

Be careful: this is average velocity, not average speed. Average speed would be

\text{average speed} = \frac{1}{b-a}\int_a^b |v(t)|\,dt

Accumulation functions in context (how AP likes to phrase it)

Sometimes you’re given a rate function and asked to define a new function using an integral, for example:

A(t)=\int_0^t v(x)\,dx

Conceptually, A(t) represents the accumulated displacement from time 0 to time t. By the Fundamental Theorem of Calculus,

A'(t)=v(t)

AP questions often test whether you can interpret:

• what A(t) means (units and real-world meaning)
• what A'(t) means (it recovers the original rate)
• how to use an initial value to shift from displacement to actual position

For instance, if position is s(0)=10, then

s(t)=10+A(t)

Common misconceptions (and how to avoid them)

1. Confusing displacement with distance.
   If the question says “total distance,” you must use absolute value and split at sign changes of v(t).

2. Forgetting to add initial conditions.
   Students will compute an antiderivative and stop, effectively assuming the constant is 0. Always look for given values like v(0) or s(0).

3. Thinking “positive acceleration” always means speeding up.
   Speed increases only when v(t) and a(t) share a sign.

4. Not checking units.
   Units are a built-in error detector. If v is meters per second and you integrate with respect to seconds, the result must be meters.

Exam Focus

• Typical question patterns:
  • Given v(t), compute displacement and then compute total distance (requires finding where v(t)=0).
  • Given a(t) and v(0), find v(t); then use s(0) to find s(t) (accumulation with initial conditions).
  • Interpret an integral-defined function like \int_0^t r(x)\,dx in context (meaning + units) and use FTC to relate derivatives.
• Common mistakes:
  • Using \int_a^b v(t)\,dt for “distance traveled” without absolute value.
  • Solving v(t)=0 incorrectly (missing sign-change points leads to wrong distance).
  • Dropping the constant of integration or ignoring a provided initial value.