Definite Integrals in Context

Lots of real situations give you a rate (litres per minute, people per hour, cars per day) but ask about an amount. Integrating a rate over a time interval gives the total change in the amount over that time. This is one of the most common question types on the AP exam, usually calculator active, and it builds directly on accumulation of change.

Key ideas

Integrating a rate gives a net change

If $R(t)$ is the rate of change of some quantity $Q(t)$, then $R = Q'$ and the Fundamental Theorem of Calculus says

$$\int_a^b R(t)\, dt = Q(b) - Q(a)$$

The integral of a rate is the net change in the amount from $t = a$ to $t = b$.

Amount = starting amount + net change

To find the amount at a particular time, start with a known amount and add the change:

$$Q(t) = Q(0) + \int_0^t R(u)\, du$$

Rates in and rates out

Many problems have something flowing in and something flowing out at the same time. With an in-rate $E(t)$ and an out-rate $L(t)$:

$$\text{amount at time } t = \text{starting amount} + \int_0^t E(u)\, du - \int_0^t L(u)\, du$$

and the amount is changing at the rate

$$Q'(t) = E(t) - L(t)$$

• If $E(t) \gt L(t)$, more is coming in than going out, so the amount is increasing.
• If $E(t) \lt L(t)$, the amount is decreasing.

When is the amount largest?

A maximum or minimum of $Q$ happens at an endpoint or where $Q'(t) = E(t) - L(t) = 0$, that is, where in-rate = out-rate. Use the candidates test: evaluate $Q$ at the endpoints and at those critical points, and compare.

Units

The integral’s units are (rate units) × (time units). Litres per minute × minutes = litres. People per hour × hours = people. Always include units in a context answer, and use the units to check your setup.

Estimating from a table

If the rate is only given as a table, estimate the integral with a Riemann sum or a trapezoidal sum, using the intervals in the table (they may be unequal widths).

Worked examples

Example 1: A single rate

Oil leaks from a tank at a rate of $r(t) = 2 + 0.6t$ litres per hour, where $t$ is in hours. How much oil leaks out during the first $5$ hours?

Solution.

$$\int_0^5 (2 + 0.6t)\, dt = \Big[ 2t + 0.3t^2 \Big]_0^5 = 10 + 7.5 = 17.5$$

$17.5$ litres leak out. (Units: litres per hour × hours = litres.)

Example 2: In and out, with a maximum

A tank holds $120$ L of water at $t = 0$. Water flows in at a constant $12$ L/min and drains out at $3\sqrt{t}$ L/min, for $0 \le t \le 25$ minutes.

• (a) How much water is in the tank at $t = 9$?
• (b) At what time is the amount of water greatest? How much is in the tank then? Justify.

Solution. Let $W(t)$ be the amount of water in litres. Then

$$W(t) = 120 + \int_0^t \big( 12 - 3\sqrt{u} \big)\, du = 120 + 12t - 2t^{3/2}$$

(a) $W(9) = 120 + 108 - 2(27) = 174$ L.

(b) $W'(t) = 12 - 3\sqrt{t}$, which is $0$ when $\sqrt{t} = 4$, so $t = 16$. Candidates:

$t$	$0$	$16$	$25$
$W(t)$	$120$	$120 + 192 - 128 = 184$	$120 + 300 - 250 = 170$

The greatest amount is $184$ L, at $t = 16$ minutes. (Also, $W'$ changes from positive to negative at $t = 16$.)

The tank fills while the in-rate is above the out-rate, so the amount peaks where the two rates cross, at $t = 16$.

Example 3: Estimating from a table

Visitors enter a museum at the rates below, where $t$ is hours after 9 a.m.

$t$ (hours)	$0$	$1$	$3$	$4$	$6$	$8$
$R(t)$ (people per hour)	$40$	$120$	$210$	$180$	$150$	$60$

• (a) Use a trapezoidal sum with the five intervals in the table to estimate $\displaystyle\int_0^8 R(t)\, dt$. Explain what it means.
• (b) Estimate the average rate at which visitors entered from 9 a.m. to 5 p.m.

Solution. (a) Each trapezoid is (width) × (average of the two heights). The widths are $1, 2, 1, 2, 2$:

$$\begin{aligned} \int_0^8 R(t)\, dt &\approx 1\cdot\frac{40 + 120}{2} + 2\cdot\frac{120 + 210}{2} + 1\cdot\frac{210 + 180}{2} + 2\cdot\frac{180 + 150}{2} + 2\cdot\frac{150 + 60}{2} \\ &= 80 + 330 + 195 + 330 + 210 = 1145 \end{aligned}$$

About $1145$ people entered the museum between 9 a.m. and 5 p.m.

(b) The average value of $R$ on $[0, 8]$ is $\dfrac{1}{8}\displaystyle\int_0^8 R(t)\, dt \approx \dfrac{1145}{8} \approx 143.125$ people per hour.

Example 4: Calculator active

A festival’s gates open at $t = 0$ with nobody inside. For $0 \le t \le 10$ hours, people arrive at a rate of $E(t) = 1200\sin\!\left(\dfrac{\pi t}{10}\right)$ people per hour and leave at a rate of $L(t) = 80t$ people per hour. (Radian mode.)

• (a) How many people are at the festival at $t = 10$?
• (b) Is the number of people increasing or decreasing at $t = 4$? Explain.
• (c) At what time is the crowd largest, and about how many people are there then?

Solution. Let $N(t) = \displaystyle\int_0^t \big( E(u) - L(u) \big)\, du$ be the number of people at time $t$.

(a)

$$N(10) = \int_0^{10} \big( E(t) - L(t) \big)\, dt \approx 3639.437$$

About $3639$ people.

(b) $N'(4) = E(4) - L(4) \approx 1141.268 - 320 = 821.268 \gt 0$, so the number of people is increasing at $t = 4$.

(c) $N'(t) = 0$ when $E(t) = L(t)$. Solving with a calculator gives $t \approx 8.167$ in $(0, 10)$. Candidates: $N(0) = 0$, $N(8.167) \approx 4355.427$, $N(10) \approx 3639.437$. The crowd is largest at about $t = 8.167$ hours, with about $4355$ people.

Common mistakes

Forgetting the starting amount. The integral gives the change, not the amount. In Example 2, the answer to (a) is $174$ L, not $54$ L.

Subtracting the wrong way. The amount changes at rate (in) − (out). If you write (out) − (in), every sign flips and “increasing” becomes “decreasing.”

Confusing the rate with the amount. “How much water is in the tank?” wants $W(t)$, an integral. “How fast is the amount changing?” wants $W'(t) = E(t) - L(t)$, no integral.

Only checking the critical point. To justify a maximum on a closed interval, compare the endpoints too (or use a sign change of $Q'$ with a clear argument).

Missing or wrong units. The integral of people per hour over hours is people. AP graders often require the units in context answers.

Rounding too early. Store full calculator values and round only the final answer (to 3 decimal places, or to whole people when it makes sense).

Practice

1. (Warm-up) Water flows into a pool at $r(t) = 0.5t + 4$ cubic metres per hour. How much water flows in during the first $6$ hours?

Solution

$$\int_0^6 (0.5t + 4)\, dt = \Big[ 0.25t^2 + 4t \Big]_0^6 = 9 + 24 = 33 \text{ m}^3$$

2. (Warm-up) $P(t)$ is the population of a town (people) $t$ years after 2020. Explain the meaning of $\displaystyle\int_0^{10} P'(t)\, dt = 350$, with units.

Solution

$\displaystyle\int_0^{10} P'(t)\, dt = P(10) - P(0)$. The town’s population increased by $350$ people from 2020 to 2030.

3. (Warm-up) A tank holds $40$ L at $t = 0$, and $\displaystyle\int_0^3 r(t)\, dt = 15$, where $r(t)$ is the rate (L/min) at which the amount of water changes. How much water is in the tank at $t = 3$?

Solution

$40 + 15 = 55$ L.

4. (Core) A lake has $5000$ fish at $t = 0$. The fish population changes at a rate of $r(t) = 200 - 24t$ fish per year.

• (a) How many fish are in the lake at $t = 10$?
• (b) Is the population increasing or decreasing at $t = 10$?

Solution

(a)

$$5000 + \int_0^{10} (200 - 24t)\, dt = 5000 + \Big[ 200t - 12t^2 \Big]_0^{10} = 5000 + 800 = 5800 \text{ fish}$$

(b) $r(10) = 200 - 240 = -40 \lt 0$, so the population is decreasing at $t = 10$.

5. (Core) (Calculator active.) People arrive at a stadium at a rate of $A(t) = 1500\,t\,e^{-0.5t}$ people per hour, for $0 \le t \le 4$ hours after the gates open.

• (a) How many people arrive during these $4$ hours?
• (b) At what time has the $3000$th person arrived?

Solution

(a)

$$\int_0^4 1500\,t\,e^{-0.5t}\, dt \approx 3563.965$$

About $3564$ people.

(b) Solve $\displaystyle\int_0^T 1500\,t\,e^{-0.5t}\, dt = 3000$ with a calculator: $T \approx 3.357$ hours.

6. (Core) Water flows from a river into a reservoir. The rate is measured on some days:

$t$ (days)	$0$	$2$	$4$	$7$	$10$
$F(t)$ (thousand m³ per day)	$14$	$18$	$21$	$16$	$12$

Use a trapezoidal sum with the four intervals to estimate the total water that flows in from $t = 0$ to $t = 10$. Include units.

Solution

$$2\cdot\frac{14 + 18}{2} + 2\cdot\frac{18 + 21}{2} + 3\cdot\frac{21 + 16}{2} + 3\cdot\frac{16 + 12}{2} = 32 + 39 + 55.5 + 42 = 168.5$$

About $168.5$ thousand cubic metres ($168\,500$ m³).

7. (Core) A rain barrel holds $W(t)$ litres of water at time $t$ hours. Rain flows in at $r(t)$ L/h, and water leaks out at $d(t)$ L/h.

• (a) Explain the meaning of $W(0) + \displaystyle\int_0^5 \big( r(t) - d(t) \big)\, dt$.
• (b) What does $r(3) \gt d(3)$ tell you about the water in the barrel?

Solution

(a) It is $W(5)$: the number of litres of water in the barrel at $t = 5$ hours.

(b) At $t = 3$ hours, water is flowing in faster than it leaks out, so the amount of water in the barrel is increasing at that moment.

8. (Challenge) (Calculator active.) A water tower holds $8000$ L at $t = 0$. For $0 \le t \le 12$ hours, water is pumped in at $P(t) = 600 + 200\sin\!\left(\dfrac{t}{3}\right)$ L/h, and the town uses water at $U(t) = 500 + 30t$ L/h.

• (a) How much water is in the tower at $t = 12$?
• (b) Is the amount increasing or decreasing at $t = 8$? Explain.
• (c) Find the maximum amount of water in the tower on $[0, 12]$. Justify.

Solution

Let $A(t) = 8000 + \displaystyle\int_0^t \big( P(u) - U(u) \big)\, du$.

(a) $A(12) = 8000 + \displaystyle\int_0^{12} \big( P(t) - U(t) \big)\, dt \approx 8032.186$ L.

(b) $A'(8) = P(8) - U(8) \approx -48.545 \lt 0$, so the amount is decreasing at $t = 8$.

(c) $A'(t) = P(t) - U(t) = 0$ at $t \approx 7.436$ (the only solution in $[0, 12]$). Candidates:

$t$	$0$	$7.436$	$12$
$A(t)$	$8000$	$\approx 8987.105$	$\approx 8032.186$

The maximum is about $8987.105$ L, at $t \approx 7.436$ hours.

9. (Challenge) Suppose the tank in Example 2 keeps running past $t = 25$ with the same in-rate and out-rate. At what time does it get back to $120$ L?

Solution

$W(t) = 120 + 12t - 2t^{3/2}$. Set $W(t) = 120$:

$$12t - 2t^{3/2} = 0 \quad\Rightarrow\quad 2t\big( 6 - \sqrt{t} \big) = 0$$

So $t = 0$ (the start) or $\sqrt{t} = 6$, that is, $t = 36$ minutes.

This means the water gained from $t = 0$ to $16$ exactly matches the water lost from $t = 16$ to $36$.