Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.
A note on notation
This page has no equation renderer, so integrals are written in a compact form: ∫ (a to b) v(t) dt means the definite integral of v(t) from t = a to t = b. On paper, write a at the bottom of the integral sign and b at the top. Throughout, an object moves along a straight line, x(t) is its position, v(t) its velocity and a(t) its acceleration at time t.
Going backwards: from rates to amounts
In Topic 4.2 you differentiated: velocity is the derivative of position, v(t) = x′(t), and acceleration is the derivative of velocity, a(t) = v′(t). Now you go the other way. Integration undoes differentiation, but it only tells you how much something changed, not where it started.
Displacement: the integral of velocity
Since x is an antiderivative of v, the Fundamental Theorem gives
∫ (a to b) v(t) dt = x(b) − x(a)
The right side is the displacement: the net change in position, final minus initial. It is a signed quantity. Time spent moving in the negative direction (v < 0) cancels some of the time spent moving in the positive direction.
Rearranging gives the formula you will use most:
x(b) = x(a) + ∫ (a to b) v(t) dt v(b) = v(a) + ∫ (a to b) a(t) dt
The second line is the same idea one step down the chain: v is an antiderivative of a, so the integral of acceleration is the change in velocity.
Units. If v is in metres per second and t in seconds, ∫ v(t) dt is in (m/s) × s = metres. If a is in m/s², ∫ a(t) dt is in m/s.
Total distance: the integral of speed
Displacement can hide motion. An object that goes 5 m forward and 5 m back has displacement 0 but has travelled 10 m. To count every metre, integrate speed, |v(t)|, which is never negative:
Total distance travelled from t = a to t = b = ∫ (a to b) |v(t)| dt
By hand. You usually cannot find an antiderivative of |v| directly. Instead:
- Find where v(t) = 0 and check where v changes sign. These are the times the object changes direction.
- Split [a, b] at those times.
- Integrate v on each piece and take the absolute value of each result.
- Add.
With a calculator. Enter ∫ (a to b) |v(t)| dt directly, using the absolute value function.
Average velocity and average speed. By Topic 8.1, the average value of v on [a, b] is (1/(b − a)) ∫ (a to b) v(t) dt, which is displacement ÷ time. That is the average velocity. Similarly, the average speed is total distance ÷ time.
Read the question word by word. The wording tells you which integral to write:
| The question asks for… | Write |
|---|---|
| the position at time b | x(a) + ∫ (a to b) v(t) dt |
| the displacement, or the change in position | ∫ (a to b) v(t) dt |
| the total distance travelled | ∫ (a to b) |v(t)| dt |
| the velocity at time b, given a(t) | v(a) + ∫ (a to b) a(t) dt |
Worked example 1: from acceleration to position, by hand
Question. A particle moves along the x-axis. Its acceleration is a(t) = 6t − 12, in cm/s², for 0 ≤ t ≤ 4 seconds. At t = 0 its velocity is 9 cm/s and its position is x = 2 cm. (a) Find v(t), and check v(4) using an integral of a. (b) Find the position at t = 4. (c) Find the total distance travelled over 0 ≤ t ≤ 4.
(a) v(t) = v(0) + ∫ (0 to t) a(s) ds = 9 + [3s² − 12s] from 0 to t = 3t² − 12t + 9, in cm/s. (The letter s is used inside because t is the upper limit.)
Check: v(4) = v(0) + ∫ (0 to 4) (6t − 12) dt = 9 + (48 − 48) = 9. And 3(16) − 48 + 9 = 9. They agree.
(b) x(4) = x(0) + ∫ (0 to 4) (3t² − 12t + 9) dt. An antiderivative is t³ − 6t² + 9t, which equals 64 − 96 + 36 = 4 at t = 4 and 0 at t = 0. So the displacement is 4 cm and x(4) = 2 + 4 = 6 cm.
(c) Factor: v(t) = 3(t − 1)(t − 3). It is positive on [0, 1), negative on (1, 3) and positive on (3, 4]. The particle changes direction at t = 1 and t = 3. Integrate each piece:
| Time interval | ∫ v(t) dt | Distance on that piece |
|---|---|---|
| 0 to 1 | 4 | 4 |
| 1 to 3 | −4 | 4 |
| 3 to 4 | 4 | 4 |
Total distance = 4 + 4 + 4 = 12 cm.
Check with positions. x(1) = 2 + 4 = 6, x(3) = 6 − 4 = 2 and x(4) = 2 + 4 = 6. The particle goes from 2 to 6, back to 2, then out to 6 again: 4 + 4 + 4 = 12 cm. The displacement is only 4 cm.
Averages. Average velocity = 4/4 = 1 cm/s. Average speed = 12/4 = 3 cm/s.
Worked example 2: reading a velocity graph
Question. A robot moves along a straight track. Its velocity v(t), in metres per second, is shown in Figure 2 for 0 ≤ t ≤ 8 seconds. At t = 0 the robot is at x = −5 m. (a) Find the robot’s position at t = 4 and at t = 8. (b) Find its displacement and its total distance travelled over 0 ≤ t ≤ 8. (c) At what time is the robot farthest in the positive direction? Justify. (d) Find its average velocity and average speed over 0 ≤ t ≤ 8.
(a) Use signed areas.
- x(4) = −5 + ∫ (0 to 4) v(t) dt = −5 + (8 + 4) = 7 m.
- x(8) = x(4) + ∫ (4 to 8) v(t) dt = 7 + (−2 − 4) = 1 m.
(b) Displacement = 8 + 4 − 2 − 4 = 6 m (also x(8) − x(0) = 1 − (−5)). Total distance = 8 + 4 + 2 + 4 = 18 m.
(c) v > 0 on [0, 4) and v < 0 on (4, 8]. So x increases until t = 4 and decreases after. The robot changes direction at t = 4, and that is when it is farthest in the positive direction, at x = 7 m.
(d) Average velocity = 6/8 = 0.75 m/s. Average speed = 18/8 = 2.25 m/s.
Interpretation. The robot drove 12 m forward, then 6 m back. It ended 6 m from where it started, after covering 18 m of track.
Worked example 3: with a graphing calculator
Question. A particle moves along a line with velocity v(t) = t²e^(−0.5t) − 1, for 0 ≤ t ≤ 6. At t = 0 it is at x = 3. A graphing calculator is allowed. (a) Find the position at t = 6. (b) Find the total distance travelled over 0 ≤ t ≤ 6. (c) Find the particle’s leftmost position on 0 ≤ t ≤ 6.
(a) x(6) = 3 + ∫ (0 to 6) v(t) dt. The calculator gives ∫ (0 to 6) v(t) dt ≈ 3.22896. So x(6) ≈ 6.229.
(b) Total distance = ∫ (0 to 6) |v(t)| dt ≈ 4.937 (entered directly with the absolute value).
Check by splitting. v(0) = −1, and solving v(t) = 0 on the calculator gives t ≈ 1.430, where v changes from negative to positive. ∫ (0 to 1.430) v(t) dt ≈ −0.854 and ∫ (1.430 to 6) v(t) dt ≈ 4.083. Then 0.854 + 4.083 = 4.937, which agrees.
(c) The particle moves left until t ≈ 1.430, then right. So its leftmost position is at t ≈ 1.430: x ≈ 3 − 0.854 ≈ 2.146. (Compare the end points: x(0) = 3 and x(6) ≈ 6.229 are both larger.)
Good habits. Write each integral before giving its value. Store intermediate values in the calculator instead of retyping rounded decimals, and round only at the end.
Common misconceptions
- “∫ v(t) dt is the distance.” It is the displacement. Distance needs |v|.
- Taking |∫ v(t) dt| as the distance. The absolute value must go inside the integral. |∫ v dt| = |displacement|.
- Forgetting the starting value. ∫ (a to b) v(t) dt is a change in position. Add x(a) to get x(b).
- “x(b) = ∫ (0 to b) v(t) dt.” Only if x(0) = 0.
- Not splitting at direction changes when finding distance by hand.
- Splitting at the wrong times, such as where a(t) = 0. The object changes direction where v changes sign, not where a does.
- Mixing up average velocity and average speed. Average velocity uses displacement; average speed uses distance.
- Wrong units. ∫ v dt is in units of length; ∫ a dt is in units of velocity.
Where this leads
Topic 8.2 is the motion case of a general rule: integrating a rate gives the net change in the amount. In Topic 8.3, Using Accumulation Functions and Definite Integrals in Applied Contexts, the same reasoning handles water in a tank, people in a queue and many other rates. The averages here use Topic 8.1, Finding the Average Value of a Function on an Interval, and the derivative relationships come from Topic 4.2. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.
Try the practice questions now, then use the revision notes and the checklist to consolidate.
Quick questions
Is this page for Calculus AB or Calculus BC?
Both. Topic 8.2 is common content, so the same page serves AB and BC students. BC students later extend the same ideas to motion in a plane.
Why do I need a starting position?
Velocity tells you how position changes, not where the object is. The integral gives the change in position; you add it to a known position to get a new one.
When are displacement and distance equal?
Distance equals the absolute value of displacement when the velocity never changes sign on the interval, so the object moves in one direction only. If that direction is positive, the two are equal; if it is negative, the displacement is the negative of the distance.