AP Calculus BC Unit 8 Notes: Average Value and One-Dimensional Motion with Integrals

Finding the Average Value of a Function on an Interval

What “average value” means (and why integrals show up)

The average value of a function is the calculus version of “the average height” of a graph over an interval. If you have a bunch of equally spaced function values, you might average them by adding them up and dividing by how many there are. Calculus takes that same idea but replaces “add up many sample values” with “add up infinitely many infinitesimal contributions,” which is exactly what an integral does.

Suppose a function stays near 10 for most of an interval and spikes to 50 for a short time. Your intuition says the “typical” value should be close to 10, not halfway between 10 and 50. The average value captures this intuition because it weights values by how long (or how wide in $x$) the function stays near them.

Geometrically, the definite integral $\int_a^b f(x)\,dx$ represents the **signed area** between the graph and the $x$-axis on $[a,b]$. If you want a constant function $y=c$ that has the _same area_ over $[a,b]$, then its area is a rectangle with height $c$ and width $b-a$. Matching areas gives the key formula.

The average value formula

The average value of $f$ on $[a,b]$ is

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

Here:

• $a$ and $b$ are the endpoints of the interval (with $a<b$).
• $\int_a^b f(x)\,dx$ is the net (signed) accumulation.
• Dividing by $b-a$ converts “total accumulated area” into “area per unit of width,” which is the average height.

Units check (a great self-test): If $f(x)$ has units of, say, meters, then $\int_a^b f(x)\,dx$ has units “meters times (units of $x$).” Dividing by $b-a$ (units of $x$) brings you back to meters. Average value has the same units as the function.

Why average value matters in applications

Average value is everywhere you want a single representative number from a varying quantity:

• Average temperature over a day (continuous time).
• Average rate of water flow through a pipe over an hour.
• Average density along a rod.
• Average velocity over a time interval (which links directly to motion).

In AP Calculus, average value problems often test whether you can (1) set up the correct definite integral, (2) divide by the correct interval length, and (3) interpret what the result means in context.

A helpful comparison: average value vs average rate of change

Students often confuse these because both include “average.” They are different ideas:

• Average value of a function (a typical output level):

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

• Average rate of change (slope of the secant line):

$\frac{f(b)-f(a)}{b-a}$

The first uses an integral (accumulation). The second uses endpoint values only. A quick way to remember: average value is about average height; average rate of change is about average slope.

Worked Example 1: Compute an average value

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

Step 1: Set up the formula.

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

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.

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

Interpretation: The “average height” of $y=x^2$ from $x=0$ to $x=3$ is 3.

A common misconception here is thinking the average value should be halfway between $f(0)=0$ and $f(3)=9$, which would give $4.5$. That would be an average of endpoints, not an average over the full interval.

Worked Example 2: When the function dips below the axis

Let $f(x)=x-2$ on $[0,4]$.

Compute the average value:

$f_{\text{avg}}=\frac{1}{4-0}\int_0^4 (x-2)\,dx$

$\int_0^4 (x-2)\,dx=\left[\frac{x^2}{2}-2x\right]_0^4=\left(8-8\right)-0=0$

$f_{\text{avg}}=\frac{1}{4}\cdot 0=0$

What this is really saying: The positive area above the axis cancels the negative area below the axis. Average value uses signed area, so it can be zero even when the function is not zero most of the time.

If a context requires an average of a quantity that cannot be negative (like “average speed”), you must be careful: you may need an average of $|f(x)|$ instead, depending on what the function represents.

The Average Value Theorem (how average value connects to “some point hits the average”)

Once you compute an average value, a natural question is: does the function ever actually equal that average? For continuous functions, yes.

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 result is often called the Mean Value Theorem for Integrals (or Average Value Theorem). It’s the continuous-function analogue of “in a list of numbers, at least one number is at or above the average and at least one is at or below.”

Why it matters: On AP problems, you may be asked to (1) find the average value and (2) find a $c$ where the function equals that average. The first part is calculus (integral). The second part is algebra (solve $f(x)=f_{\text{avg}}$).

Worked Example 3: Find $c$ such that $f(c)=f_{\text{avg}}$

Let $f(x)=\sin x$ on $[0,\pi]$.

Step 1: Compute the average value.

$f_{\text{avg}}=\frac{1}{\pi-0}\int_0^\pi \sin x\,dx$

$\int_0^\pi \sin x\,dx=\left[-\cos x\right]_0^\pi=-\cos(\pi)-(-\cos(0))=1-(-1)=2$

$f_{\text{avg}}=\frac{2}{\pi}$

Step 2: Solve $\sin c=\frac{2}{\pi}$ for $c$ in $[0,\pi]$.
There will be two solutions because $\sin x$ is symmetric on $[0,\pi]$ (one in the first quadrant and one in the second). You would typically leave these as

$c=\arcsin\left(\frac{2}{\pi}\right)$

and

$c=\pi-\arcsin\left(\frac{2}{\pi}\right)$

On a calculator-active setting, you could approximate these.

Common “setup” issues to watch for

1. Wrong denominator: The denominator is always the interval length $b-a$, not $b$, not $a+b$.
2. Forgetting signed area: If $f$ goes below the axis, the integral subtracts. That’s correct for average value of the function, but it may not match the average of a nonnegative physical quantity.
3. Mixing up with average rate of change: If you see an integral, you’re in average value territory. If you see $f(b)-f(a)$, you’re in average rate of change territory.

Exam Focus

• Typical question patterns
  • “Find the average value of $f$ on $[a,b]$” (often requiring a definite integral and sometimes a calculator).
  • “Given a graph/table of $f$, find/estimate the average value on $[a,b]$” (you may use geometric areas or numerical approximation).
  • “Find a value $c$ satisfying $f(c)=f_{\text{avg}}$” (compute average value, then solve an equation).
• Common mistakes
  • Using $\frac{f(a)+f(b)}{2}$ instead of $\frac{1}{b-a}\int_a^b f(x)\,dx$.
  • Dividing by the wrong interval length (especially when $a\neq 0$).
  • Ignoring negative parts of the graph when the problem truly asks for average value (signed), or failing to use absolute value when the context demands a nonnegative average.

Connecting Position, Velocity, and Acceleration Using Integrals

The big picture: derivatives describe change; integrals recover totals

In one-dimensional motion along a line, the core quantities are:

• Position: where an object is.
• Velocity: how fast position is changing.
• Acceleration: how fast velocity is changing.

Calculus links them through derivatives:

$v(t)=s'(t)$

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

Here $s(t)$ is position at time $t$, $v(t)$ is velocity, and $a(t)$ is acceleration.

Integrals run these relationships backward. If derivatives tell you “instantaneous change,” integrals tell you “total change accumulated over time.” This is why integrals are so natural in motion: they convert a rate (velocity or acceleration) into a net change (displacement or change in velocity).

Notation and meaning (a quick reference)

Different textbooks and problems use different symbols for position, but they mean the same idea.

On AP problems, pay close attention to whether the function given is position, velocity, or acceleration; the required integrals and interpretations change accordingly.

Displacement vs distance traveled (the most tested conceptual distinction)

If you know velocity $v(t)$, you can compute how much the position changes from $t=a$ to $t=b$. This net change in position is called displacement:

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

Displacement is a signed quantity. If the object moves forward and then backward, the backward motion subtracts from the forward motion.

Distance traveled measures total ground covered and is always nonnegative. To compute distance traveled from velocity, you must account for direction changes:

$\text{Distance traveled} = \int_a^b |v(t)|\,dt$

This difference explains many AP free-response questions: they often ask for both displacement and total distance specifically to check that you understand the absolute value and the idea of direction.

From acceleration to velocity (using initial conditions)

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

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

If you are given an initial velocity, such as $v(t_0)=v_0$, you can build a velocity function:

$v(t)=v_0+\int_{t_0}^t a(u)\,du$

(The variable inside the integral is often changed to avoid confusion, but any dummy variable works.)

A common mistake is forgetting to include the initial velocity. An integral gives you a change, not an absolute value, unless you anchor it with an initial condition.

From velocity to position (again, initial conditions matter)

Similarly, integrating velocity gives the change in position:

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

With an initial position $s(t_0)=s_0$, you can write

$s(t)=s_0+\int_{t_0}^t v(u)\,du$

This is one of the most useful “AP-ready” forms because it turns a velocity graph or table into a position model.

How the Fundamental Theorem of Calculus ties everything together

The Fundamental Theorem of Calculus is the reason all these integral relationships work. In this setting, it says (informally): if velocity is the derivative of position, then the integral of velocity accumulates to give position change.

For instance, if you define an accumulation function

$F(t)=\int_{t_0}^t v(u)\,du$

then

$F'(t)=v(t)$

So the rate at which displacement accumulates is exactly the instantaneous velocity.

This idea also appears when AP asks you to interpret an integral like $\int_0^5 v(t)\,dt$: it is not just a computation, it is a physical quantity (net change in position over that time interval).

Worked Example 1: Displacement and distance from a velocity function

A particle has velocity $v(t)=t^2-4t+3$ for $0\le t\le 4$, and position $s(0)=2$.

(a) Find the displacement on $[0,4]$

Displacement is

$\int_0^4 (t^2-4t+3)\,dt$

Compute:

$\int_0^4 (t^2-4t+3)\,dt=\left[\frac{t^3}{3}-2t^2+3t\right]_0^4=\left(\frac{64}{3}-32+12\right)-0=\frac{64}{3}-20$

$\frac{64}{3}-20=\frac{64}{3}-\frac{60}{3}=\frac{4}{3}$

So displacement is

$\frac{4}{3}$

(b) Find the position $s(4)$

Use $s(4)=s(0)+\text{displacement}$:

$s(4)=2+\frac{4}{3}=\frac{10}{3}$

(c) Find the total distance traveled on $[0,4]$

Distance uses $|v(t)|$, so you must find where $v(t)=0$ (direction changes).

Solve

$t^2-4t+3=0$

Factor:

$(t-1)(t-3)=0$

So the velocity changes sign at $t=1$ and $t=3$.

Check signs: for $t=0$, $v(0)=3>0$; for $t=2$, $v(2)=4-8+3=-1<0$; for $t=4$, $v(4)=16-16+3=3>0$. So the motion is forward, then backward, then forward.

Distance traveled is

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

You could compute these integrals using the same antiderivative as in part (a). Let

$A(t)=\frac{t^3}{3}-2t^2+3t$

Then:

$\int_0^1 v(t)\,dt=A(1)-A(0)=\left(\frac{1}{3}-2+3\right)-0=\frac{4}{3}$

$\int_1^3 v(t)\,dt=A(3)-A(1)=\left(9-18+9\right)-\left(\frac{1}{3}-2+3\right)=0-\frac{4}{3}=-\frac{4}{3}$

So

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

And

$\int_3^4 v(t)\,dt=A(4)-A(3)=\left(\frac{64}{3}-32+12\right)-0=\frac{4}{3}$

Total distance traveled:

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

Key takeaway: Displacement was $\frac{4}{3}$, but distance was $4$. The difference comes from reversing direction between $t=1$ and $t=3$.

Worked Example 2: Building velocity and position from acceleration

Suppose a particle’s acceleration is $a(t)=6t-4$ for $0\le t\le 3$. You are told $v(0)=5$ and $s(0)=1$.

(a) Find $v(t)$

Use the accumulation form:

$v(t)=v(0)+\int_0^t (6u-4)\,du$

Compute the integral:

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

So

$v(t)=5+3t^2-4t$

(b) Find $s(t)$

Now integrate velocity:

$s(t)=s(0)+\int_0^t \left(5+3u^2-4u\right)\,du$

Compute:

$\int_0^t \left(5+3u^2-4u\right)\,du=\left[5u+u^3-2u^2\right]_0^t=5t+t^3-2t^2$

So

$s(t)=1+5t+t^3-2t^2$

What often goes wrong: Students integrate acceleration correctly to get $3t^2-4t$ but forget the $+5$ from $v(0)=5$, which shifts the entire velocity function.

Interpreting signs: what velocity and acceleration tell you qualitatively

Beyond computation, AP questions frequently ask you to interpret motion.

• If $v(t)>0$, position is increasing (moving right/forward).
• If $v(t)<0$, position is decreasing (moving left/backward).
• If $a(t)>0$, velocity is increasing.
• If $a(t)<0$, velocity is decreasing.

A subtle but important point: “speeding up” depends on both velocity and acceleration.

• The object is speeding up when $v(t)$ and $a(t)$ have the same sign.
• The object is slowing down when $v(t)$ and $a(t)$ have opposite signs.

This is because speed is $|v(t)|$, and acceleration increases speed only when it pushes in the same direction as motion.

Motion problems from graphs and tables (how integrals appear without formulas)

On AP exams, you are often given a graph of $v(t)$ or $a(t)$ and asked for quantities like displacement. The integral becomes geometry:

• Area above the axis contributes positively.
• Area below the axis contributes negatively.

For total distance, you add magnitudes of areas, treating below-axis regions as positive by flipping them.

If velocity values are given in a table, you might approximate the integral with a Riemann sum or trapezoidal sum. The conceptual meaning stays the same: you’re approximating accumulated change.

A strong habit: always attach meaning to an integral

When you see something like $\int_2^7 v(t)\,dt$, train yourself to say: “This is the displacement from $t=2$ to $t=7$.”

Similarly:

• $\int_2^7 a(t)\,dt$ is the change in velocity on $[2,7]$.
• $\frac{1}{b-a}\int_a^b v(t)\,dt$ is the **average velocity** on $[a,b]$.

This last point connects directly back to average value: average velocity is the average value of the velocity function.

Worked Example 3: Average velocity vs average speed

Suppose $v(t)$ is a velocity function on $[a,b]$.

• Average velocity is

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

This equals

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

because displacement is the integral of velocity.

• Average speed is

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

These are equal only if the object never changes direction (so $v(t)$ does not change sign).

Common misconception: calling $\frac{1}{b-a}\int_a^b v(t)\,dt$ “average speed.” That is only true when velocity is never negative on the interval.

Exam Focus

• Typical question patterns
  • Given $v(t)$ (formula, graph, or table), compute displacement $\int_a^b v(t)\,dt$ and/or position $s(b)=s(a)+\int_a^b v(t)\,dt$.
  • Compute total distance traveled $\int_a^b |v(t)|\,dt$, requiring you to find where $v(t)=0$ or identify sign changes from a graph.
  • Given $a(t)$ with an initial condition, find $v(t)$ and possibly $s(t)$ by integrating and applying constants/initial values.
• Common mistakes
  • Confusing displacement with distance (forgetting the absolute value for distance, or incorrectly using it for displacement).
  • Integrating but forgetting initial conditions (missing the “+C” idea in applied form: use the given initial position/velocity).
  • Not splitting the integral at direction changes (using $\int_a^b |v(t)|\,dt$ without locating where $v(t)=0$).