Accumulation of Change

So far in calculus you have started with an amount and found its rate of change. This unit goes the other way: if you know the rate, how much has the amount changed? The key idea is surprisingly simple. The area between a rate graph and the horizontal axis is the accumulated change in the amount. This idea leads straight to the definite integral.

Key ideas

Rate times time

If a quantity changes at a constant rate, the change is

$$\text{change} = \text{rate} \times \text{time}.$$

A tap that fills a tub at $8$ L/min for $5$ min adds $8 \times 5 = 40$ L. On a graph of rate against time, that’s the area of a rectangle with height $8$ and width $5$.

Area under a rate graph

When the rate is not constant, the same idea still works: the accumulated change is the area between the rate graph and the horizontal axis. You can think of it as many thin rectangles, each with “rate times a tiny bit of time”, all added up. (That’s exactly what Riemann sums do.)

If the rate graph is made of straight lines, you can find the area with rectangles, triangles, and trapezoids.

Units

The units of the area are (units of the rate) × (units of the input). Some examples:

Rate	Input	Area means
velocity in m/s	time in s	displacement in m
water flow in L/min	time in min	litres of water added
rainfall in mm/h	time in h	millimetres of rain
power in kW	time in h	energy in kWh

Always check the units. If you multiply “people per hour” by “hours”, you get “people”.

Positive and negative contributions

When the rate is positive, the amount is increasing, and the area counts as positive change. When the rate is negative (below the axis), the amount is decreasing, and that area counts as negative change.

• Net change = (area above the axis) − (area below the axis).
• Total (for example, total distance travelled) = (area above) + (area below).

For motion: the net change in position is the displacement, and the total of all the areas is the distance travelled.

Final amount

$$\text{final amount} = \text{initial amount} + \text{net change}$$

The amount reaches a maximum where the rate changes from positive to negative, and a minimum where it changes from negative to positive (check the endpoints too).

Worked examples

Example 1: A constant velocity

A cyclist rides at a constant $6$ m/s for $40$ s. How far does she travel?

Solution. The velocity graph is a horizontal line at height $6$. The area under it from $t = 0$ to $t = 40$ is a rectangle:

$$6 \ \tfrac{\text{m}}{\text{s}} \times 40 \ \text{s} = 240 \ \text{m}$$

Notice how the seconds cancel and leave metres.

Example 2: Water in a tank

Water flows into and out of a tank at the rate $r(t)$ litres per minute shown below, for $0 \le t \le 10$ minutes. When $r(t)$ is negative, water is draining out.

The blue area is water flowing in; the orange area is water flowing out.

(a) How much water flows in from $t = 0$ to $t = 6$?

(b) What is the net change in the amount of water from $t = 0$ to $t = 10$?

(c) The tank holds $50$ L at $t = 0$. How much does it hold at $t = 10$? When is the amount of water greatest?

Solution.

(a) Split the region above the axis into simple shapes:

$$\begin{aligned} 0 \le t \le 2: &\quad \text{trapezoid } \tfrac{2 + 4}{2} \times 2 = 6 \\ 2 \le t \le 4: &\quad \text{rectangle } 4 \times 2 = 8 \\ 4 \le t \le 6: &\quad \text{triangle } \tfrac{1}{2} \times 2 \times 4 = 4 \end{aligned}$$

That’s $6 + 8 + 4 = 18$ L flowing in.

(b) Below the axis, from $t = 6$ to $t = 10$, there is one triangle with base $4$ and height $2$, so its area is $\tfrac{1}{2}(4)(2) = 4$. This counts as $-4$ L. The net change is $18 - 4 = 14$ L.

(c) At $t = 10$ the tank holds $50 + 14 = 64$ L.

The amount increases while $r(t) \gt 0$ and decreases while $r(t) \lt 0$. Since $r$ changes from positive to negative at $t = 6$, the amount is greatest at $t = 6$, when the tank holds $50 + 18 = 68$ L.

Example 3: Displacement versus distance

A toy car moves along a straight track. Its velocity is $3$ m/s for $0 \le t \le 4$, then decreases at a steady rate to $-3$ m/s at $t = 6$ (passing $0$ at $t = 5$), then stays at $-3$ m/s until $t = 8$. Find the car’s displacement and the total distance it travels over $0 \le t \le 8$.

Solution. Sketch the velocity graph and find each area:

• $0 \le t \le 4$: rectangle, $3 \times 4 = 12$ (above the axis).
• $4 \le t \le 5$: triangle, $\tfrac{1}{2}(1)(3) = 1.5$ (above).
• $5 \le t \le 6$: triangle, $\tfrac{1}{2}(1)(3) = 1.5$ (below).
• $6 \le t \le 8$: rectangle, $3 \times 2 = 6$ (below).

$$\begin{aligned} \text{displacement} &= 12 + 1.5 - 1.5 - 6 = 6 \text{ m} \\ \text{distance} &= 12 + 1.5 + 1.5 + 6 = 21 \text{ m} \end{aligned}$$

The car ends $6$ m forward of where it started, but it travelled $21$ m in all, because it reversed for part of the time.

Example 4: Interpreting an area

The rate at which visitors enter a museum is $V(t)$ people per hour, where $t$ is hours after 9 a.m. What does the area under the graph of $V$ from $t = 0$ to $t = 3$ represent?

Solution. The units are $\dfrac{\text{people}}{\text{hour}} \times \text{hours} = \text{people}$. The area is the number of people who entered the museum between 9 a.m. and noon.

On the AP exam, an interpretation like this should mention the quantity, its units, and the time interval.

Common mistakes

Adding areas below the axis as positive when you want net change. Area below the axis is a negative contribution. Subtract it for net change (displacement, net amount of water). Only add it when the question asks for a total like distance travelled.

Forgetting the starting amount. The area gives the change, not the amount. In Example 2 the answer to “how much is in the tank” is $50 + 14 = 64$ L, not $14$ L.

Getting the units wrong. Multiply the units of the two axes. Velocity (m/s) times time (s) gives metres, not m/s.

Thinking the amount is largest where the rate is largest. In Example 2 the rate is highest from $t = 2$ to $t = 4$, but water keeps coming in until $t = 6$. The amount is greatest where the rate changes from positive to negative.

Reading the rate graph as the amount graph. A rate graph going down does not mean the amount is going down. The amount only decreases when the rate is below the axis.

Practice

1. (Warm-up) A pump moves water into a pool at a constant $12$ L/min for $15$ minutes. How much water does it add?

Solution

$$12 \ \tfrac{\text{L}}{\text{min}} \times 15 \ \text{min} = 180 \ \text{L}$$

2. (Warm-up) During a storm, rain falls at a rate of $R(t)$ millimetres per hour, where $t$ is in hours. What are the units of the area under the graph of $R$ from $t = 0$ to $t = 6$, and what does the area mean?

Solution

$\dfrac{\text{mm}}{\text{h}} \times \text{h} = \text{mm}$. The area is the total depth of rain, in millimetres, that fell during the first $6$ hours of the storm.

3. (Warm-up) A particle moves with a constant velocity of $-2$ m/s for $6$ seconds. Find its displacement and the distance it travels.

Solution

The rectangle is below the axis, with area $2 \times 6 = 12$. The displacement is $-12$ m (it moves $12$ m in the negative direction), and the distance travelled is $12$ m.

4. (Core) Use the rate graph in Example 2. How much water drains out between $t = 6$ and $t = 8$? How much water is in the tank at $t = 8$?

Solution

From $t = 6$ to $t = 8$ the graph is a triangle below the axis with base $2$ and height $2$: area $\tfrac{1}{2}(2)(2) = 2$. So $2$ L drains out.

At $t = 8$:

$$50 + 18 - 2 = 66 \text{ L}$$

5. (Core) A cyclist starts from rest. Her velocity is $v(t) = 2t$ m/s for $0 \le t \le 5$, and then she rides at a constant $10$ m/s until $t = 20$. How far does she travel in the first $20$ seconds?

Solution

From $0$ to $5$ the graph is a triangle with base $5$ and height $v(5) = 10$: area $\tfrac{1}{2}(5)(10) = 25$.

From $5$ to $20$ it’s a rectangle: $10 \times 15 = 150$.

She travels $25 + 150 = 175$ m.

6. (Core) A robot moves along a straight line. Its velocity is $4$ m/s for $0 \le t \le 3$ and $-2$ m/s for $3 \lt t \le 7$. It starts at position $x = -5$ m.

• (a) Find its displacement over $0 \le t \le 7$.
• (b) Find the total distance it travels.
• (c) Find its position at $t = 7$.

Solution

(a) Above the axis: $4 \times 3 = 12$. Below: $2 \times 4 = 8$. Displacement $= 12 - 8 = 4$ m.

(b) Distance $= 12 + 8 = 20$ m.

(c) Position $= -5 + 4 = -1$ m.

7. (Core) Water flows into a barrel at the rate $r(t) = 6 - 2t$ litres per minute for $0 \le t \le 5$ (negative values mean water is flowing out).

• (a) Find the net change in the amount of water from $t = 0$ to $t = 5$.
• (b) At what time is the amount of water in the barrel greatest? Justify your answer.

Solution

(a) The graph is a line from $(0, 6)$ to $(5, -4)$, crossing the axis at $t = 3$.

Above: triangle with base $3$ and height $6$, area $9$. Below: triangle with base $2$ and height $4$, area $4$.

Net change $= 9 - 4 = 5$ L.

(b) $r(t)$ changes from positive to negative at $t = 3$, so the amount of water increases before $t = 3$ and decreases after. The amount is greatest at $t = 3$ minutes.

8. (Challenge) A tank holds $30$ L at $t = 0$. For $0 \le t \le 8$ minutes, water flows in at rate $r(t)$ L/min, where $r(t) = -4$ for $0 \le t \le 2$, then $r$ increases in a straight line from $-4$ at $t = 2$ to $4$ at $t = 6$, then $r(t) = 4$ for $6 \le t \le 8$. Find the least and greatest amounts of water in the tank during $0 \le t \le 8$, and when they happen.

Solution

The line from $(2, -4)$ to $(6, 4)$ crosses the axis at $t = 4$. Areas:

• $0 \le t \le 2$: rectangle below, $-8$.
• $2 \le t \le 4$: triangle below, $-\tfrac{1}{2}(2)(4) = -4$.
• $4 \le t \le 6$: triangle above, $+4$.
• $6 \le t \le 8$: rectangle above, $+8$.

The amount decreases until $t = 4$ (where $r$ changes from negative to positive), then increases. Amounts: $A(0) = 30$, $A(4) = 30 - 8 - 4 = 18$, $A(8) = 18 + 4 + 8 = 30$.

The least amount is $18$ L at $t = 4$. The greatest is $30$ L, at both $t = 0$ and $t = 8$.

9. (Challenge) Snow builds up on a flat roof at $2$ cm/h for the first $3$ hours of a storm. Then the sun comes out and the snow melts at a constant $1.5$ cm/h. The snow was $5$ cm deep at the start. When is it back to $5$ cm deep?

Solution

After $3$ hours the depth is $5 + 2(3) = 11$ cm.

To lose the extra $6$ cm at $1.5$ cm/h takes $\dfrac{6}{1.5} = 4$ hours.

So the snow is back to $5$ cm deep at $t = 3 + 4 = 7$ hours after the storm started.