Choose a rate and accumulation interval
Rain in a barrel
The curve shows how fast rain runs into a barrel, in liters per hour, and the input \(t\) counts the hours after 10 a.m.
Noon is 2 hours after 10 a.m., and 3 p.m. is 5 hours after. The rain from noon to 3 p.m. then falls between \(t = 2\) and \(t = 5\), where the button sets the start and the end.
The teal region lies between the curve and the horizontal axis, from the start to the end. Each thin strip of it has a height in liters per hour and a width in hours, so its area is in liters.
From noon to 3 p.m., the area of the whole region is the rain collected, 54 liters.
Any rate
The curve now shows any rate against any input \(t\). Move the start \(a\) and the end \(b\), or slide the whole interval. The teal region is the amount added between them.
The question sets where the region starts and ends, and the curve sets how tall it is.
Notation
The rate \(R(t)\) tells how fast an amount changes with the input \(t\), and its units are amount units per unit of \(t\).
\[\cy{R(t)}\]The question gives the interval from \(t = a\) to \(t = b\). If it gives clock times, convert them to values of \(t\) first.
\[a \le t \le b\]The amount added equals the area between the rate curve and the horizontal axis, from \(a\) to \(b\).
\[\cb{\int_a^b \cy{R(t)}\,dt}\]The rate must be continuous on the interval.
The units of the integral are the units of the rate times the units of the input.
\[\text{liters per hour} \times \text{hours} = \text{liters}\]
When it fails
The rate is in liters per hour, so \(t\) counts hours. If you write the interval in minutes, noon to 3 p.m. becomes 120 to 300.
The region then runs to the right edge of the picture, and 300 lies far beyond the 7 hours the curve describes. The ends of the interval must use the unit of the input, so convert them before you integrate.
Example 1: the extra cost of more chairs
A factory has the marginal cost \(C'(x) = 50 + 0.2x\) dollars per chair, where \(x\) is the number of chairs made. Find the extra cost of raising production from 100 chairs to 200 chairs.
The rate is the marginal cost, in dollars per chair. The input is the number of chairs, not time.
\[C'(x) = 50 + 0.2x\]The question names 100 chairs and 200 chairs. These are the ends of the interval, so the extra cost is the area between them.
\[\cb{\int_{100}^{200}} (50 + 0.2x)\,dx\]An antiderivative of 50 is \(50x\), and an antiderivative of \(0.2x\) is \(0.1x^2\). Substitute the two ends.
\[\begin{aligned} &\Big[\, 50x + 0.1x^2 \,\Big]_{100}^{200} \\ &= 14000 - 6000 \end{aligned}\]The units are dollars per chair times chairs, so the extra cost is 8000 dollars.
\[\cb{8000 \text{ dollars}}\]
Example 2: an oil leak
Oil leaks from a pipe at \(L(t) = 90 - t\) gallons per minute, where \(t\) is the number of minutes since the pipe broke. How much oil leaks between minute 20 and minute 50?
"Between minute 20 and minute 50" means \(t = 20\) to \(t = 50\).
\[20 \le t \le 50\]A tempting wrong move: Start the interval at 0, because the leak began at \(t = 0\).
\[\int_0^{50} (90 - t)\,dt = 3250\]This is the oil from minute 0 to minute 50. The picture now also shades the first 20 minutes, and the question leaves out their area, the red 1600 gallons.
Start the interval where the question starts it.
\[\cb{\int_{20}^{50}} (90 - t)\,dt = \Big[\, 90t - \tfrac{t^2}{2} \,\Big]_{20}^{50}\]The antiderivative is 3250 at \(t = 50\) and 1600 at \(t = 20\), so the leak is the difference.
\[\cb{3250 - 1600 = 1650 \text{ gallons}}\]
Practice
Cars enter a toll plaza at \(c(t) = 240 - 20t\) cars per hour, where \(t\) is the number of hours after 6 a.m. How many cars enter between 8 a.m. and 10 a.m.?
Move the start and the end to the times in the question. 8 a.m. is how many hours after 6 a.m.?
Solution
8 a.m. is 2 hours after 6 a.m., and 10 a.m. is 4 hours after, so the interval is \(t = 2\) to \(t = 4\).
\[\int_2^4 (240 - 20t)\,dt\]Find an antiderivative and substitute the two ends.
\[\Big[\, 240t - 10t^2 \,\Big]_2^4 = 800 - 440\]The units are cars per hour times hours, so 360 cars enter the plaza.
\[\cb{360 \text{ cars}}\]
Questions
The graph shows how fast a pump fills a pool, in gallons per minute, during the first 12 minutes.
Which integral gives the water pumped during the last 4 minutes?
- \(\int_8^{12} r(t)\,dt\)
- \(\int_0^{4} r(t)\,dt\)
- \(r(12) - r(8)\)
- \(4\,r(12)\)
Answer
\(\int_8^{12} r(t)\,dt\)
What are the units of \(\int_2^6 r(t)\,dt\)?
- gallons
- gallons per minute
- minutes
- gallons per minute squared
Answer
gallons
Gravel pours onto a pile at \(g(t) = 3t\) kilograms per minute, where \(t\) is in minutes. How much gravel pours in the first half hour?
- 1350 kg
- 90 kg
- 0.375 kg
- 2700 kg
Answer
1350 kg
Summary
You write the ends of the interval in the unit of the input, and you integrate the rate between them.
This works because height times a small width is a small amount, and the area adds these amounts.
You use it when a problem gives how fast a quantity changes and asks how much it changes over an interval.
This is one of five skills in topic 8.3 of Which & Why: AP Calculus.