Accumulation and Motion

Integrate a velocity and you get how far the position changed. Integrate its absolute value and you get how far the object actually went. The two differ whenever it turns back. The page shades the forward and backward parts, draws the position, and marks the moments the velocity equals its own average.

Distance against displacement	$\int_{a}^{b} v d t = s (b) - s (a), d = \int_{a}^{b} ∣ v ∣ d t$
Average value	$\overset{ˉ}{f} = \frac{1}{b - a} \int_{a}^{b} f (t) d t$

Worked example — Accumulation and Motion

Displacement	1.3333
Distance travelled, split at the turns	4
Average velocity	0.3333
Average speed	1
Times when v equals its average	0.8453, 3.1547

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Accumulation and Motion

For a velocity v(t) on [a, b], the displacement is ∫ₐᵇ v dt and the distance travelled is ∫ₐᵇ |v| dt. The average value of a function f on [a, b] is (1/(b − a))∫ₐᵇ f dt: the height of the rectangle on [a, b] with the same area as f.

What each symbol means

v̄: the average velocity, displacement divided by elapsed time

Where the formula comes from

Distance: split at the turns. Where v changes sign the object turns back, and |v| has a different formula on each side. Integrate between neighbouring turns and add the absolute values.
$\int_{a}^{b} ∣ v ∣ d t = i \sum \int_{t_{i}}^{t_{i + 1}} v d t, v (t_{i}) = 0$
The average is reached. A continuous f on [a, b] takes every value between its minimum m and maximum M, and its average lies between them, so f equals its average at some c.
$m \leq \frac{1}{b - a} \int_{a}^{b} f d t \leq M \Rightarrow f (c) = \overset{ˉ}{f}$

Problems solved in full

Average value, and where it is reached 2 steps
Find the average value of f(t) = t² on [0, 3], and the time where f takes it.

Open this problem in the tool
1. Divide the integral by the length of the interval:
  
  $\overset{ˉ}{f} = \frac{1}{3} \int_{0}^{3} t^{2} d t = \frac{1}{3} \cdot 9 = 3$
2. Solve f(c) = 3 inside the interval:
  
  $c^{2} = 3 \Rightarrow c = 3$
Answer

Average 3, reached at c = √3 ≈ 1.7321. Not at the midpoint 1.5, where t² is only 2.25.

$\overset{ˉ}{f} = 3, c = 3 \approx 1.7321$
Displacement and distance 2 steps
A particle moves with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 4. Find its displacement and the distance it travels.

Open this problem in the tool
Show solution
1. Displacement is the plain integral:
  
  $\int_{0}^{4} (t^{2} - 4 t + 3) d t = [\frac{t ^{3}}{3} - 2 t^{2} + 3 t]_{0}^{4} = \frac{4}{3}$
2. It turns where v = 0, at t = 1 and t = 3, so split the distance there:
  
  $\frac{4}{3} + - \frac{4}{3} + \frac{4}{3} = 4$
Answer

Displacement 4/3, distance 4. Each of the three stretches covers 4/3; the middle one goes backward.

$Δ s = \frac{4}{3}, d = 4$
Position from velocity and a starting point 2 steps
v(t) = 3t² − 6t and s(0) = 2. Find s(t), and the position at t = 3.
Show solution
1. Position is the starting point plus the accumulated velocity:
  
  $s (t) = 2 + \int_{0}^{t} (3 τ^{2} - 6 τ) d τ = t^{3} - 3 t^{2} + 2$
2. At t = 3:
  
  $s (3) = 27 - 27 + 2 = 2$
Answer

s(t) = t³ − 3t² + 2, and s(3) = 2. Back where it started: it moved backward until t = 2, where s = −2, then forward again.

$s (t) = t^{3} - 3 t^{2} + 2, s (3) = 2$
The average of a sine 1 steps
Find the average value of sin t on [0, π].

Open this problem in the tool
Show solution
1. The integral over the arch, divided by its width:
  
  $\overset{ˉ}{f} = \frac{1}{π} \int_{0}^{π} sin t d t = \frac{1}{π} [- cos t]_{0}^{π} = \frac{2}{π}$
Answer

2/π ≈ 0.6366. The page marks the two times it is reached, at 0.6901 and π − 0.6901 = 2.4515.

$\overset{ˉ}{f} = \frac{2}{π} \approx 0.6366$
Accumulation from a rate 1 steps
Water flows into a tank at r(t) = 20 − 4t litres per minute for 0 ≤ t ≤ 5. How much water enters?
Show solution
1. The amount is the integral of the rate:
  
  $\int_{0}^{5} (20 - 4 t) d t = [20 t - 2 t^{2}]_{0}^{5} = 100 - 50$
Answer

50 litres. The rate falls to 0 at t = 5, so the amount is the area of a triangle with base 5 and height 20, a check that needs no calculus.

$50 L$
A round trip 2 steps
v(t) = sin t on [0, 2π]. Find the average velocity and the average speed.

Open this problem in the tool
Show solution
1. The displacement over the full turn is 0:
  
  $\int_{0}^{2 π} sin t d t = 0$
2. The distance splits at π:
  
  $\frac{1}{2 π} (\int_{0}^{π} sin t d t - \int_{π}^{2 π} sin t d t) = \frac{4}{2 π} = \frac{2}{π}$
Answer

Average velocity 0, average speed 2/π ≈ 0.6366. A round trip always has average velocity 0, however fast it went.

$\overset{v}{ˉ} = 0, \frac{2}{π} \approx 0.6366$

Displacement says where it ended up; only distance says how far it went 🖖

Press Three times as far. The velocity t² − 4t + 3 is positive until t = 1, negative from 1 to 3, and positive again after. Integrate it over [0, 4] and the backward stretch cancels part of the forward ones: the object ends 1.3333 from its start. But it travelled 4, three times that, because each of the three stretches covers 4/3 and only the absolute value counts them all as distance. A velocity integral answers where; for how far, split at every turn and add the pieces as positive numbers.

The average is reached somewhere, and rarely at the middle 🖖

Press t²: not at the middle. The average of t² over [0, 3] is 9/3 = 3, and the dashed line meets the curve once, at t = √3 ≈ 1.7321. The midpoint of the interval is 1.5, where t² is only 2.25. For a continuous function the average is always reached somewhere, by the intermediate value theorem, since the average lies between the smallest and largest values. Where it is reached depends on the shape: a function that grows faster at the end, like t², reaches its average late.

A full swing has average velocity zero and an average speed that is not 🖖

Press A full swing. sin t over a whole turn goes forward and comes back to where it began: displacement 0, so the average velocity is 0, though between start and finish it stopped only at the turn. The average speed, distance over time, is 4/(2π) = 2/π ≈ 0.6366. Now press Half a swing: the average of sin t over [0, π] is also 2/π. Not by chance: the second half of the swing covers the same distance as the first in the same time, so the speed averaged over the full turn equals the velocity averaged over the half.

Learning path

Applications of integration

References (2)

Displacement against distance as the page computes them: the Net Change Theorem (Theorem 5.6, F(b) = F(a) + ∫ₐᵇ F′(x) dx) and Examples 5.24 and 5.25, net displacement from ∫v and total distance from ∫|v| split where v changes sign: G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 5.4, “Integration Formulas and the Net Change Theorem”.
The average value reached somewhere: the Mean Value Theorem for Integrals (Theorem 5.3): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 5.3, “The Fundamental Theorem of Calculus”.

By the 42Math team · Reviewed against published sources. Corrections welcome.