Represent change as a definite integral
Water filling a tank
Water flows into an empty tank. The curve shows the rate of flow \(r\), in liters per minute, at minute \(t\).
Drag the orange edge to the right. The teal region under the curve grows, because each new minute adds water. The lower graph shows the total added so far.
The green strip beside the edge is one minute wide and as tall as the rate. Liters per minute times minutes gives liters, so its area is the water that enters during that minute.
At \(t = 4\) the teal area is 12, so 12 liters entered during the first 4 minutes.
The area is the amount added
The curve shows any rate. The teal region runs from a start time, the dashed line, to an end time, the orange edge.
Move either slider. The region and its area change, and the lower graph shows the area from the start to the edge.
The area is the amount that enters between the two times, because each thin strip is a rate times a short time.
Notation
The rate is the amount added per minute, and its graph is the curve.
\[r(t)\]The limits are the start time \(a\) and the end time \(\cy{b}\) of the question, the two ends of the shaded region.
\[a\ \text{to}\ \cy{b}\]A strip \(r(t)\) tall and \(\cg{dt}\) wide has area \(\cg{r(t)\,dt}\), the small amount added in a short time. The strips from \(a\) to \(b\) add up to the shaded area, so the integral is the change in the amount.
\[\cb{\int_a^b r(t)\,dt}\]This needs \(r\) to be continuous on the interval from \(a\) to \(b\).
The units of the integral are the units of the rate times the units of the time.
\[\frac{\text{liters}}{\text{minute}}\times\text{minutes}=\text{liters}\]
When it fails
The integral multiplies the rate by the time, so both must use the same time unit. Here \(r\) is in liters per minute, so the limits must be in minutes. Write the time in hours.
Liters per minute times hours is not a number of liters. If a problem gives the limits 0 and 4 hours, convert them to 0 and 240 minutes. Otherwise each strip is 60 times too narrow, so the integral is 60 times too small.
Example 1: A solar panel
A solar panel produces power \(P(t) = 3t\) kilowatts, where \(t\) is the number of hours after 8 AM. Find the energy it produces from 9 AM to 12 PM.
Power is energy produced per hour, so it is the rate.
\[P(t) = 3t\]The variable \(t\) counts hours after 8 AM, so 9 AM is \(t = 1\) and 12 PM is \(t = 4\). These are the limits.
\[a = 1,\quad b = 4\]The energy is the integral of the power, the area of the shaded trapezoid. Its parallel sides are \(P(1) = \cy{3}\) and \(P(4) = \cy{12}\), and it is 3 hours wide.
\[\cb{\int_1^4 3t\,dt} = \frac{3 + 12}{2} \times 3 = \tfrac{45}{2}\]Kilowatts times hours gives kilowatt-hours, so the panel produces \(\tfrac{45}{2}\) kilowatt-hours.
Example 2: An oil leak
Oil leaks from a tank at the rate \(L(t) = 2 + 2t\) gallons per hour, with \(t\) in hours. Find the amount that leaks from \(t = 1\) to \(t = 4\).
The amount is the integral of the rate from 1 to 4.
\[\int_1^4 (2 + 2t)\,dt\]A tempting wrong move: The rate at \(t = 4\) is 10 gallons per hour, and the leak lasts 3 hours, so multiply 10 by 3.
\[10 \times 3 = 30\]The rate is 10 only at \(t = 4\). Earlier it is smaller, so the dashed rectangle reaches above the line. The red part above the line is 9 gallons that did not leak.
Use the whole line. The shaded trapezoid has parallel sides \(L(1) = \cy{4}\) and \(L(4) = \cy{10}\) and is 3 hours wide.
\[\frac{4 + 10}{2} \times 3 = 21\]The tank loses 21 gallons, not the 30 that counted the red part.
Practice
A bakery makes loaves at the rate \(20 + 10t\) loaves per hour, where \(t\) is the number of hours after 6 AM. How many loaves does it make from 8 AM to 10 AM?
Convert the clock times to values of t. The shaded area between them is the number of loaves.
Solution
Because \(t\) counts hours after 6 AM, 8 AM is \(t = 2\) and 10 AM is \(t = 4\). Integrate the rate between them.
\[\cb{\int_2^4 (20 + 10t)\,dt}\]The shaded trapezoid has parallel sides \(\cy{40}\) and \(\cy{60}\) and is 2 hours wide.
\[\frac{40 + 60}{2} \times 2 = 100\]The bakery makes 100 loaves.
Questions
The graph shows the rate \(e(t) = 20 - 4t\) people per minute at a stadium gate, \(t\) minutes after it opens.
What does the area under the graph from \(t = 1\) to \(t = 3\) measure?
- The number of people who enter from minute 1 to minute 3.
- The rate at which people enter at minute 3.
- The number of people inside the stadium at minute 3.
- The average rate between minute 1 and minute 3.
Answer
The number of people who enter from minute 1 to minute 3.
Compare the area under the graph from \(t = 0\) to \(t = 1\) with the area from \(t = 3\) to \(t = 4\). Which statement is correct?
- The first area is larger, so more people enter in the first minute.
- The second area is larger, because more time has passed.
- The areas are equal, because both intervals are one minute wide.
Answer
The first area is larger, so more people enter in the first minute.
Cars enter a garage at the rate \(40 + 6t\) cars per hour, where \(t\) is the number of hours after 6 AM. How many cars enter from 8 AM to 11 AM?
- 183 cars
- 147 cars
- 210 cars
- 291 cars
Answer
183 cars
Summary
To represent a change, you name the rate, write the limits in the rate's time unit, and integrate between them.
This works because a thin strip under the rate graph is a rate times a short time, which is a small amount.
You use it when a problem gives a rate and asks how much of a quantity builds up between two times.
This is one of four skills in topic 6.1 of Which & Why: AP Calculus.