6.1 Accumulation of change

Achievable AP Calculus AB

6. Integration

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Accumulation of change

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What you’ll learn

Accumulation: Interpret the area under a rate function as a total accumulated amount.
Net vs. total area: Use signed area to find net change versus total distance or quantity.
Unit analysis: Determine and apply the correct units for accumulated quantities.

In the first part of AP Calculus, derivatives were used to find rates of change from a function that describes an amount.

The second part of AP Calculus reverses this process. Given a rate of change, such as speed, a growth rate, or an inflow/outflow rate, you can determine the accumulated change of a quantity, or how much it has increased or decreased over a specific interval.

Accumulation under a constant rate

When a rate of change is constant, total accumulation is simply the rate $\times$ the time. For example,

Water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water has accumulated after $4$ minutes?

At a constant rate of $3$ liters per minute, the total amount of water added after $4$ minutes is

$3 \frac{L}{min} \times 4 min = 12 L$

Graphically, the constant rate $r (t) = 3$ forms a horizontal line. The total accumulation corresponds exactly to the area of the rectangle between $r (t)$ , the $x$ -axis, and the time interval $[0, 4]$ . Accumulated change is represented by the area under the rate curve.

Accumulation under a variable rate

Even for rates that aren’t constant, the same idea still applies: the accumulation of change over an interval is the area between the graph of the rate function and the $x$ -axis over that interval.

A particle’s velocity, in meters per second, is modeled by $v (t)$ and shown in the graph below.

Figure 6.1.2 Graph of velocity

a) How far is the particle from its starting position after the first $3$ seconds?

b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds

(spoiler)

This is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

Use geometry to calculate the areas of the trapezoid (in blue) and the rectangle (in green), then sum the areas. The area of a trapezoid is

$A_{trap} = \frac{h}{2} (b_{1} + b_{2})$

So the total area, or the distance accumulated over the first $3$ seconds, is

$A = \frac{2}{2} (2 + 4) + 1 (4)$

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds

(spoiler)

This is the area under the graph on $[2, 7]$ as shown:

Then the distance traveled is

$A = 3 (4) rectangle + \frac{2}{2} (4 + 3) trapezoid$

$= 19 meters$

AP tip:

Know your units! On free-response questions (FRQs) that require them, missing or incorrect units can cost you points.

Area “under” a curve

In calculus, “area under a curve” specifically means the bounded area between the graph of a function and the $x$ -axis over a given interval. When a region is:

Above the $x$ -axis: Total amount increases (positive accumulation)
Below the $x$ -axis: Total amount decreases (negative accumulation)

Consider the velocity graph below over the interval $[0, 6]$ , where velocity is measured in meters per second. The second figure shows the same graph with the relevant regions shaded.

Recall from section 4.2.1 that the sign of velocity indicates the direction of motion.

Positive accumulation: Over $0 < t < 2$ seconds, $v (t) > 0$ . The particle travels to the right (position increases) for a distance of:

$A_{1} = \frac{1}{2} (2) (2) = 2 meters$

Negative accumulation: Over $2 < t < 6$ seconds, $v (t) < 0$ . The particle moves left after changing direction at $t = 2$ seconds. The signed area of the trapezoid is:

$A_{2} = \frac{- 1}{2} (4 + 2) = - 3 meters$

The net change in position (displacement) is the net signed area, while the total distance traveled is the sum of the magnitudes of the changes.

Displacement	Total distance
$2 + (- 3) = - 1$	$∣2∣ + ∣ - 3∣ = 5$

This means the particle ends up $1$ meter to the left of its starting position but traveled a total distance of $5$ meters.

AP tip:

Use signed area when the problem asks for net change or displacement.
Use absolute values of area when the problem asks for total distance or total amount added or removed.

Accumulation as area under a rate function

Accumulated change = area under rate function graph
Units: (vertical units) × (horizontal units)
Works for both constant and variable rates

Signed area and net change

Area above $x$ -axis: positive (adds to total)
Area below $x$ -axis: negative (subtracts from total)
Net change (displacement) = sum of signed areas

Total distance vs net change

Net change/displacement: sum of signed areas (can be negative)
Total distance/amount: sum of absolute values of areas (always positive)
Use absolute values when asked for total traveled or accumulated

Units in accumulation problems

Always include units in final answers (e.g., meters, liters)
Rate × time (or other variable) gives accumulated quantity units

Area “under” a curve

Refers to area between function graph and $x$ -axis over interval
Negative area when function is below $x$ -axis (signed area)

AP exam tips

Label all answers with correct units
Use signed area for net change/displacement
Use absolute area for total distance/amount

Accumulation of change

What you’ll learn

Accumulation: Interpret the area under a rate function as a total accumulated amount.
Net vs. total area: Use signed area to find net change versus total distance or quantity.
Unit analysis: Determine and apply the correct units for accumulated quantities.

In the first part of AP Calculus, derivatives were used to find rates of change from a function that describes an amount.

Accumulation under a constant rate

When a rate of change is constant, total accumulation is simply the rate $\times$ the time. For example,

Water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water has accumulated after $4$ minutes?

At a constant rate of $3$ liters per minute, the total amount of water added after $4$ minutes is

$3 \frac{L}{min} \times 4 min = 12 L$

Accumulation under a variable rate

A particle’s velocity, in meters per second, is modeled by $v (t)$ and shown in the graph below.

a) How far is the particle from its starting position after the first $3$ seconds?

b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds

(spoiler)

This is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

Use geometry to calculate the areas of the trapezoid (in blue) and the rectangle (in green), then sum the areas. The area of a trapezoid is

$A_{trap} = \frac{h}{2} (b_{1} + b_{2})$

So the total area, or the distance accumulated over the first $3$ seconds, is

$A = \frac{2}{2} (2 + 4) + 1 (4)$

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds

(spoiler)

This is the area under the graph on $[2, 7]$ as shown:

Then the distance traveled is

$A = 3 (4) rectangle + \frac{2}{2} (4 + 3) trapezoid$

$= 19 meters$

AP tip:

Know your units! On free-response questions (FRQs) that require them, missing or incorrect units can cost you points.

Area “under” a curve

In calculus, “area under a curve” specifically means the bounded area between the graph of a function and the $x$ -axis over a given interval. When a region is:

Above the $x$ -axis: Total amount increases (positive accumulation)
Below the $x$ -axis: Total amount decreases (negative accumulation)

Consider the velocity graph below over the interval $[0, 6]$ , where velocity is measured in meters per second. The second figure shows the same graph with the relevant regions shaded.

Recall from section 4.2.1 that the sign of velocity indicates the direction of motion.

Positive accumulation: Over $0 < t < 2$ seconds, $v (t) > 0$ . The particle travels to the right (position increases) for a distance of:

$A_{1} = \frac{1}{2} (2) (2) = 2 meters$

Negative accumulation: Over $2 < t < 6$ seconds, $v (t) < 0$ . The particle moves left after changing direction at $t = 2$ seconds. The signed area of the trapezoid is:

$A_{2} = \frac{- 1}{2} (4 + 2) = - 3 meters$

The net change in position (displacement) is the net signed area, while the total distance traveled is the sum of the magnitudes of the changes.

Displacement	Total distance
$2 + (- 3) = - 1$	$∣2∣ + ∣ - 3∣ = 5$

This means the particle ends up $1$ meter to the left of its starting position but traveled a total distance of $5$ meters.

AP tip:

Use signed area when the problem asks for net change or displacement.
Use absolute values of area when the problem asks for total distance or total amount added or removed.

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.