Motion with Integrals — AP® Calculus AB Notes

In Unit 4 you started from a position function $x(t)$ and differentiated: velocity $v = x'$ , acceleration $a = v'$ . Now run the film backward. Given the velocity, integrals recover how far the particle moved and where it is. Two different questions hide here, and the exam asks both: how far did it end up from its start, and how far did it actually travel?

Displacement: the integral of velocity

Because $v$ is the derivative of $x$ , the Fundamental Theorem gives

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

This net change in position is the displacement on $[a, b]$ . On a velocity graph it is the signed area: area above the $t$ -axis (moving right, or up, or forward) counts as positive, area below (moving left) counts as negative. So a trip right followed by an equal trip left has displacement 0.

Total distance: the integral of speed

A car's odometer does not subtract when you drive back home. Speed is $|v(t)|$ , and the total distance traveled is

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

On the graph, that is the sum of the areas, every one counted as positive.

The velocity v(t) = t² − 4t + 3 m/s for 0 ≤ t ≤ 4 s. It is positive on (0, 1) and (3, 4), where the region between the graph and the t-axis is shaded crimson with area 4/3 each, and negative on (1, 3), where the region below the axis is shaded blue with area 4/3. Displacement adds signed areas: 4/3 m. Total distance adds the areas: 4 m. — **Figure 1.** $v(t) = t^2 - 4t + 3$ : the particle moves right on $(0, 1)$ , left on $(1, 3)$ , right on $(3, 4)$ . Each shaded region has area $\frac{4}{3}$ . LevlPrep original diagram.

Worked example 1 (no calculator). A particle moves with $v(t) = t^2 - 4t + 3$ meters per second for $0 \le t \le 4$ . Find its displacement and its total distance.

Antiderivative: $F(t) = \frac{t^3}{3} - 2t^2 + 3t$ .

Displacement: $F(4) - F(0)$ $= \frac{64}{3} - 32 + 12$ $= \frac{4}{3}$ meters.

Distance: $v = (t - 1)(t - 3)$ changes sign at $t = 1$ and $t = 3$ , so split there. $F(1) - F(0) = \frac{4}{3}$ ; $F(3) - F(1)$ $= 0 - \frac{4}{3}$ $= -\frac{4}{3}$ ; $F(4) - F(3) = \frac{4}{3}$ .

Add the sizes: $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ meters. The particle traveled 4 m but ended only $\frac{4}{3}$ m to the right of where it began.

Without a calculator you cannot integrate $|v|$ directly. Find where $v$ changes sign, integrate $v$ on each piece, and add the absolute values. Check every zero, not just whole-number times.

Position from a starting point

Displacement tells you the change. To know the position itself, you need one known position, then add the change:

$x(t) = x(t_0) + \int_{t_0}^{t} v(s)\,ds.$

Worked example 2 (graph). A particle's velocity is made of segments through $(0, 0)$ , $(2, 4)$ , $(5, 4)$ , $(8, -2)$ and $(10, -2)$ , in m/s, and $x(0) = -3$ . Find $x(10)$ , the total distance, and the time it is farthest right.

Areas: 4 on $[0, 2]$ , 12 on $[2, 5]$ , 4 on $[5, 7]$ (where $v$ reaches 0), then 1 and 4 below the axis on $[7, 8]$ and $[8, 10]$ .

$x(10)$ $= -3$ $+ (4 + 12 + 4 - 1 - 4)$ $= -3$ $+ 15$ $= 12$ m. Total distance: $4 + 12 + 4 + 1 + 4 = 25$ m.

The particle moves right while $v > 0$ and left after $v$ changes sign at $t = 7$ , so it is farthest right at $t = 7$ : $x(7) = -3 + 20 = 17$ m.

The same idea gives velocity from acceleration: $v(t)$ $= v(t_0)$ $+ \int_{t_0}^{t} a(s)\,ds$ . With $a(t) = 6t$ and $v(0) = -3$ , $v(2) = -3 + 12 = 9$ m/s.

With a calculator

When the velocity has no easy antiderivative, the calculator does the integral, and it can integrate $|v|$ directly. Write the setup every time, then the value to three decimal places:

Displacement: $\int_0^6 (4 - t + 2\sin t)\,dt$ $\approx 6.080$ feet.
Total distance: $\int_0^6 |4 - t$ $+ 2\sin t|\,dt$ $\approx 17.430$ feet.
Position, if $x(0) = 5$ : $x(6)$ $= 5$ $+ \int_0^6 v(t)\,dt$ $\approx 11.080$ feet.

Radian mode, and keep full precision until the final answer.

Displacement or distance?

Question asks for	Integral	Can it be negative?
displacement, change in position, net distance	$\int_a^b v(t)\,dt$	yes
total distance traveled	$\int_a^b \|v(t)\|\,dt$	no
position at time $b$	$x(a) + \int_a^b v(t)\,dt$	yes

Common slips

Giving the displacement when the question asks for total distance, or the reverse.
Forgetting the initial position, or starting the integral at 0 when the known position is at another time.
Putting the absolute value outside: $\left|\int v\,dt\right|$ is the size of the displacement, not the distance.
Units: the integral of feet per second over seconds is feet.