Interpret the result in context
A hiker on a trail
The line shows how fast a hiker's elevation changes, in meters per minute, from a start at 200 meters above sea level.
Move the End slider to the right. The teal region is where the hiker climbs, and the red region below the axis is where the hiker descends, so it counts as negative. The lower graph shows the net change so far.
At t = 5 the teal area is 27 and the red area is 12. The net change in elevation, the climb minus the descent, is 27 − 12 = 15 meters, so the hiker is at 200 + 15 = 215 meters.
Press the button. The red region flips above the axis and counts as positive. The lower graph now shows the total: the hiker moved 27 + 12 = 39 meters up and down together.
Three numbers from one rate
The curve is any rate. The net change up to the orange edge is the signed area: the teal area minus the red area.
Press the button. The red part flips above the axis, and the lower graph rises to the total, which counts both directions as positive.
Whatever the curve does, the amount at the edge is the amount at the start plus the net change. The total does not give the amount.
Notation
The net change from a to b is the integral of the rate f. It equals \(A(b) - A(a)\), the amount at the end minus the amount at the start.
\[\cb{\int_a^b} f(t)\,dt = A(b) - A(a)\]The total is the integral of the absolute value of the rate, so the red part flips up and counts as positive.
\[\cb{\int_a^b} |f(t)|\,dt\]The amount at the end is the amount at the start plus the net change.
\[A(b) = A(a) + \int_a^b f(t)\,dt\]A statement about a result names which number it is, the net change, the total or the amount, with its interval and units.
When it fails
Until \(t = 3\) the rate is positive, so the net change and the total are both 27 meters. Show the total next to the net change.
After t = 3 the hiker descends. The net change falls to 15 meters while the total rises to 39, so the net change alone hides the descent.
Example 1: a town's population
A town's population changes at \(p(t) = 30 - 20t\) people per year, where \(t\) is the number of years after 2020. The integral of \(p\) from 0 to 5 is \(-100\). What does \(-100\) mean?
The integral of a rate is a change in the amount, not the amount itself. The teal area minus the red area is \(22.5 - 122.5 = -100\).
\[\int_0^5 p(t)\,dt = -100\]The interval \(t = 0\) to \(t = 5\) is the years 2020 to 2025. The integral is negative, so the population fell.
The population in 2025 is 100 people lower than in 2020. This does not give the population in 2025, because the starting population is unknown.
\[\cr{-100 \text{ people}}\]
Example 2: a robot in a hallway
A robot moves along a hallway with velocity \(v(t) = 12 - 4t\) meters per minute for \(0 \le t \le 6\). It starts 5 meters from the door. How far does the robot travel, and how far from the door is it at \(t = 6\)?
The velocity is positive before \(t = 3\) and negative after, so the robot moves forward and then backward.
\[\begin{aligned} &\cb{\int_0^3 (12 - 4t)\,dt = 18} \\ &\cr{\int_3^6 (12 - 4t)\,dt = -18} \end{aligned}\]A tempting wrong move: The distance traveled is the integral of the velocity from 0 to 6.
\[\int_0^6 (12 - 4t)\,dt = 0\]The integral is 0 because the red area cancels the teal area. But the robot moved 18 meters forward and 18 back, so the distance is not 0.
The distance traveled is the total. The red part flips up and the areas add.
\[\cb{18 + 18 = 36 \text{ meters}}\]The net change is 0, so the robot ends 5 meters from the door, where it began.
\[5 + 0 = 5\]
Practice
A sand pile has a mass of 40 tons at \(t = 0\) hours. Its mass changes at \(m'(t) = 8 - 2t\) tons per hour for \(0 \le t \le 6\). Find the net change, the total sand added and removed, and the mass at \(t = 6\).
Press the flip button. Does the flipped picture show the net change or the total?
Solution
The rate is positive before \(t = 4\) and negative after, so sand is added and then removed.
\[\begin{aligned} &\cb{\int_0^4 (8 - 2t)\,dt = 16} \\ &\cr{\int_4^6 (8 - 2t)\,dt = -4} \end{aligned}\]The net change subtracts the red area, and the total adds it.
\[\begin{aligned} 16 - 4 &= 12 \text{ tons} \\ 16 + 4 &= 20 \text{ tons} \end{aligned}\]The mass at t = 6 is the starting mass plus the net change.
\[40 + 12 = 52 \text{ tons}\]
Questions
The graph shows the net rate \(r(t)\) at which water enters a reservoir, inflow minus outflow, in millions of gallons per day. The reservoir holds 50 million gallons at day 0.
What does \(\int_0^8 r(t)\,dt = 0\) tell you?
- The reservoir holds the same amount at day 8 as at day 0.
- No water entered or left the reservoir.
- The reservoir is empty at day 8.
Answer
The reservoir holds the same amount at day 8 as at day 0.
Which statement does the graph justify?
- The reservoir holds 74 million gallons at day 4.
- The reservoir gained 74 million gallons during the first 4 days.
- The reservoir holds 24 million gallons at day 4.
Answer
The reservoir holds 74 million gallons at day 4.
A particle moves on a line with velocity \(v(t) = 3t^2 - 12\) meters per second. What is the total distance it travels from \(t = 0\) to \(t = 3\)?
- 23 meters
- −9 meters
- 9 meters
- 7 meters
Answer
23 meters
Summary
You say whether a number is the net change, the total or the amount, with its interval and units.
This works because the integral of the rate subtracts the red area, while the total counts every area as positive.
You use it when a rate changes sign and a problem asks what an integral or a distance means.
This is one of five skills in topic 8.3 of Which & Why: AP Calculus.