Position, velocity, acceleration via integration — AP Calculus BC
1. Core Definitions and Relationships ★☆☆☆☆ ⏱ 2 min
This topic applies integration to reverse the derivative relationships between motion quantities learned in differential calculus, for rectilinear (straight-line) motion, the primary context tested on the AP exam.
When given acceleration or velocity and asked to find velocity or position, integration is the required tool. A key skill you will develop is distinguishing between net displacement (net change in position) and total distance traveled (total path length) over an interval. According to the AP Calculus CED, Applications of Integration make up 10-15% of the BC exam, with this subtopic accounting for roughly 2-4% of total exam points, appearing in both multiple-choice and free-response sections.
2. Finding Velocity and Position with Initial Conditions ★★☆☆☆ ⏱ 4 min
Since acceleration is the rate of change of velocity, and velocity is the rate of change of position, we reverse these relationships via integration:
v(t) = \int a(t) \, dt + C \\ s(t) = \int v(t) \, dt + C
We can also write this in definite integral form to avoid carrying an unknown constant, where $\tau$ is a dummy variable of integration:
v(t) = v(0) + \int_0^t a(\tau) d\tau \\ s(t) = s(0) + \int_0^t v(\tau) d\tau
The constant of integration $C$ is solved using an initial condition: a known value of velocity or position at a given starting time. Intuition: Integration accumulates all small changes in acceleration over time to get the total change in velocity, then adding the starting velocity gives the current velocity, with the same logic applying to position from velocity.
Exam tip: Always solve for the constant of integration immediately after integrating each quantity, don’t wait until the end to solve for multiple constants. This avoids arithmetic errors from carrying unknown values through multiple steps.
3. Calculating Net Displacement Over a Time Interval ★★☆☆☆ ⏱ 3 min
Net displacement is the net change in a particle’s position between two times $t=a$ and $t=b$, or $\Delta s = s(b) - s(a)$. By the Fundamental Theorem of Calculus Part 2, since $s(t)$ is the antiderivative of $v(t)$, we get:
\Delta s = s(b) - s(a) = \int_a^b v(t) \, dt
Displacement is a net quantity: positive velocity (movement in the positive direction) adds to displacement, while negative velocity (movement in the negative direction) subtracts from displacement. Displacement can be positive, negative, or zero, depending on where the particle ends up relative to its starting position. You do not need an initial condition to calculate displacement, because the constant of integration cancels out when evaluating the definite integral.
Exam tip: If a question only asks for displacement, you do not need to solve for the position function or find the constant of integration — just evaluate the definite integral of velocity directly.
4. Calculating Total Distance Traveled Over a Time Interval ★★★☆☆ ⏱ 5 min
Total distance traveled is the total length of the path a particle takes over $[a,b]$, regardless of direction of movement. Unlike displacement, it is always non-negative. Any movement (forward or backward) adds to total distance, so we have to account for negative velocity by making it positive before integrating. The formula for total distance is:
D = \int_a^b |v(t)| \, dt
- Find all times $t$ in $(a,b)$ where $v(t) = 0$ (these are turning points, where the particle changes direction).
- Split the original interval into subintervals where $v(t)$ is entirely positive or entirely negative.
- On subintervals where $v(t) < 0$, replace $|v(t)|$ with $-v(t)$ to make it positive; leave $v(t)$ as-is on intervals where $v(t) > 0$.
- Integrate each subinterval and add the results.
Exam tip: If velocity is never zero on the interval, the particle never changes direction, so total distance equals displacement. Always confirm whether turning points exist inside the interval before splitting.
Common Pitfalls
Why: Students mix up the definitions of displacement and total distance, especially on rushed FRQs.
Why: You remember to split the interval but rush and leave $v(t)$ as-is, leading to a total distance that is too low.
Why: You are used to finding position functions, so you add an unnecessary step that introduces arithmetic errors.
Why: The question gives you acceleration and you forget the chain of relationships, skipping a step.
Why: Rushing through multiple integration steps mixes up which initial condition goes with which integration.