Distinguish net change from total accumulation
A cart that turns around
A cart rolls out along a straight track and then back. The graph shows its velocity \(v\) in meters per second, positive while the cart moves forward and negative while it moves back.
Drag the orange edge to the right. The lower graph shows the position of the cart, which rises to 12 meters and then falls back to 0.
The teal area is 12 meters forward and the red area is 12 meters back. The red area counts as negative, so the net change in position is 0.
Flip the red part up. The red area now counts as positive, so nothing cancels, and the lower graph rises to the 24 meters the cart traveled.
Subtract or add the red area
Without numbers, the teal region is where the rate is positive and the red region is where it is negative.
The net change is the teal area minus the red area. The quantity grows while the rate is positive and shrinks while it is negative, so part of the change cancels.
The total accumulation is the teal area plus the red area. Flip the red part up. The lower graph now only rises, because every change counts as positive.
Notation
The rate is \(f(t)\), from \(a\) to \(\cy{b}\). Measure the teal area \(\cb{P}\) and the red area \(\cr{N}\) as positive numbers.
The net change is the integral of the rate, which counts the red area as negative. For a velocity, it is the displacement, the change in position.
\[\int_a^b f(t)\,dt = \cb{P} - \cr{N}\]The total accumulation counts every region as positive. It is the integral of \(|f(t)|\), the rate without its sign, which flips the red part of the curve up. For a velocity, it is the distance traveled.
\[\int_a^b |f(t)|\,dt = \cb{P} + \cr{N}\]To find the total, split the interval at the point \(c\) where \(f\) changes sign. Integrate over each piece. Add the absolute values of the results.
\[\text{total} = \left|\int_a^c f\,dt\right| + \left|\int_c^b f\,dt\right|\]
When it fails
The net change and the total are equal only when the rate has one sign on the whole interval. Raise the curve until it stays above the axis.
There is then no red area, so nothing cancels and the net change equals the total. With a red part, a net change of 0 can hide a long trip, which only the total shows.
Example 1: A particle on a line
A particle has velocity \(v(t) = 2t - 8\) meters per second for \(0 \le t \le 6\). Find its displacement and the distance it travels.
The velocity is negative before \(t = 4\) and positive after it, so the particle moves backward and then forward.
\[2t - 8 = 0 \Rightarrow t = 4\]The red triangle has base 4 and height 8, so its area is 16. The teal triangle has base 2 and height 4, so its area is 4. The displacement is the integral, which counts the red area as negative.
\[\int_0^6 (2t - 8)\,dt = -\cr{16} + \cb{4} = -12\]For the distance, flip the red part up, so both areas count as positive.
\[\int_0^6 |2t - 8|\,dt = 16 + 4 = 20\]The particle ends 12 meters behind its start, and it travels 20 meters.
Example 2: A particle with two turns
A particle moves along a line. The graph shows its velocity \(v(t)\) in meters per second for \(0 \le t \le 5\). Find the distance it travels.
The graph crosses the axis at \(t = 2\) and at \(t = 4\), so the particle moves forward, then backward, then forward.
A tempting wrong move: The distance is the integral of the velocity from 0 to 5, the teal areas minus the red area.
\[\int_0^5 v\,dt = 4 - 2 + 1 = 3\]This integral is the displacement, the final position, marked by the red dashed line. From \(t = 2\) to \(t = 4\) the particle moves backward, so the integral subtracts a distance that it still travels.
Find the area of each triangle. The bases are 2, 2 and 1, and the heights are 4, 2 and 2.
\[\begin{aligned} \tfrac12\cdot2\cdot4 &= 4 \\ \tfrac12\cdot2\cdot2 &= 2 \\ \tfrac12\cdot1\cdot2 &= 1 \end{aligned}\]Flip the red part up and add the three areas. The particle travels 7 meters.
\[4 + 2 + 1 = 7\]
Practice
The graph shows the velocity \(v(t)\) of a particle, in meters per second, for \(0 \le t \le 4\). Find its displacement and the distance it travels.
The graph crosses the axis once, so there are two triangles. The displacement subtracts the red one, and the distance adds it.
Solution
The graph is below the axis before \(t = 3\) and above it after.
The red triangle has base 3 and height 6. The teal triangle has base 1 and height 4.
\[\tfrac12\cdot3\cdot6 = \cr{9},\qquad \tfrac12\cdot1\cdot4 = \cb{2}\]The displacement subtracts the red area, and the distance adds it.
\[\begin{aligned} -9 + 2 &= -7 \\ 9 + 2 &= 11 \end{aligned}\]
Questions
A particle starts at position 0. The graph shows its velocity in meters per second.
What distance does the particle travel from \(t = 0\) to \(t = 12\)?
- 18 meters
- 6 meters
- 14 meters
Answer
18 meters
What is the particle's position at \(t = 8\)?
- 2 meters
- 14 meters
- 8 meters
- \(-6\) meters
Answer
2 meters
A particle has velocity \(v(t) = t - 1\) meters per second for \(0 \le t \le 3\). What distance does it travel?
- \(\tfrac52\) meters
- \(\tfrac32\) meters
- 2 meters
Answer
\(\tfrac52\) meters
Summary
The integral of the rate gives the net change, and the integral of its absolute value, with the red part flipped up, gives the total.
Red regions subtract from the net change, but the quantity still changes there, so they add to the total.
You use the net change for a final difference such as displacement, and the total for how much changed in all, such as distance traveled.
This is one of four skills in topic 6.1 of Which & Why: AP Calculus.