6.1 Accumulation of change

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Accumulation of change

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What you’ll learn:

Interpreting accumulation as the area under a rate function’s graph
Using signed area to find net change (displacement) vs total area
Understanding units in accumulation problems

In the first part of AP Calculus, you were usually given a function (modeling position, population, cost, etc.) and asked to find its rate of change using derivatives.

Now we reverse that idea: if you know the rate of change of a quantity (speed, growth rate, inflow/outflow rate, and so on), you can determine how much the original quantity changes over time.

For example:

Suppose water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water will be in the bucket after $4$ minutes?

When the rate is constant, the total change is just

$3 \times 4 = 12 liters .$

This total change (in the amount of water) is called the accumulation of change.

The graph of the rate function $r (t) = 3$ is a horizontal line. Visually, the product $3 \times 4$ is the area of a rectangle between the line $y = 3$ and the $x$ -axis from $t = 0$ to $t = 4$ .

The units work the same way: area units come from (vertical units) $\times$ (horizontal units).

$\frac{L}{min} \times min = L$

So when you multiply a rate by a time, you can also interpret the accumulated change as the area under the rate graph.

If the rate is not 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 $3$ seconds?
b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

The shaded region is made of:

a trapezoid (blue)
a rectangle (green)

So the total area (and therefore the distance traveled) is

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

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds is the area under the curve on that interval, as shown:

The shaded region is made of:

a rectangle (green)
a trapezoid (blue)

So the distance traveled is

$A = 3 (4) + \frac{1}{2} \cdot 2 (4 + 3)$

$= 19 meters$

AP tip:

Know your units! Always label your final answer with correct units (e.g., meters, liters, etc.). On free-response questions (FRQs), missing or incorrect units can cost you points.

Area “under” a curve

When calculus refers to the “area under a curve,” it means the area between the graph of a function and the $x$ -axis over a given interval.

A key detail: if the function is below the $x$ -axis, that area counts as negative when you’re finding accumulation (net change). This is often called signed area.

Consider the velocity graph below, where velocity is measured in meters per second:

Recall from section 4.2.1 that the sign of velocity tells you direction. We can split the area into two parts:

the triangle (blue) above the $x$ -axis
the trapezoid (green) below the $x$ -axis

The particle travels right (positive velocity) over $0 < t < 2$ seconds, for a displacement of

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

$= 2 meters$

At $t = 2$ seconds, it changes direction and travels left (negative velocity) over $2 < t < 6$ seconds. The (unsigned) area of that trapezoid is

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

$= 3 meters$

Because this region is below the $x$ -axis, its signed area is negative. So the net change in position (displacement) is the net signed area:

$2 + (- 3) = - 1$

This means the particle ends up $1$ meter to the left of its starting position.

If you want total distance traveled, you add the magnitudes of the changes instead:

$∣2∣ + ∣ - 3∣ = 5 meters$

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.

Examples

A container is slowly leaking liquid throughout the day, but it’s also being refilled at certain times. The rate at which the liquid is entering or leaving the container (in liters per hour) varies over time and is recorded in the graph below as function $R (t)$ , where positive values represent liquid being added and negative values represent leakage.

Figure 6.1.7 Graph of R(t)

How much water has the container gained or lost in the first $5$ hours?

Note: The curve between the points $(1, - 1)$ and $(3, - 1)$ is a semi-circle.

Solution

(spoiler)

Multiplying the rate by time corresponds to finding area. Here, that area represents the change in the amount of water (in liters).

Split the region into the geometric shapes $A_{1}$ through $A_{5}$ as shown:

Regions $A_{1}$ to $A_{4}$ are below the $x$ -axis, so they have negative signed area.

$A_{1}$ is a triangle with signed area

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

$= - \frac{1}{2}$

$A_{2}$ is a rectangle with signed area

$A_{2} = (2) (- 1) = - 2$

$A_{3}$ is a semi-circle with radius $1$ , so its signed area is

$A_{3} = - \frac{1}{2} π (1)^{2}$

$= - \frac{π}{2}$

$A_{4}$ is a trapezoid with height $1$ and one base length $1$ . To find the other base length, we need the $x$ -intercept of the line through $(4, - 1)$ and $(5, 2)$ . Don’t assume the graph is drawn to scale or rely on visual estimates (like “about halfway”).

Using algebra, the equation of the line is

$y = 3 x - 13$

The $x$ -intercept occurs when $y = 0$ , so the intercept is $(\frac{13}{3}, 0)$ .

That makes the top base length $1 \frac{1}{3}$ , so the signed area is

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

$= - \frac{7}{6}$

$A_{5}$ is a triangle above the $x$ -axis with height $2$ and base $\frac{2}{3}$ . Its signed area is

$A_{5} = \frac{1}{2} \cdot \frac{2}{3} \cdot (2)$

$= \frac{2}{3}$

The net signed area (the accumulation of change) is

$A = - \frac{1}{2} - 2 - \frac{π}{2} - \frac{7}{6} + \frac{2}{3}$

$= - \frac{π}{2} - 3$

$\approx - 4.6 liters$

Because the result is negative, the container has lost approximately $4.6$ liters over the first $5$ hours.

The temperature in a room is affected by a heating and cooling system that turns on and off during the day. Let $T (t)$ be the temperature of the room at time $t$ .

The graph below shows its derivative, $T^{'} (t)$ , the rate of the change in temperature (in $° C$ per hour) at time $t$ , with $t = 0$ corresponding to $12$ AM midnight.

Figure 6.1.9 Graph of T'(t)

What is the change in room temperature between $1$ AM and $6$ AM?

Solution

(spoiler)

The change in temperature over an interval is the net signed area under $T^{'} (t)$ on that interval.

Split the region into the geometric shapes $A_{1}$ through $A_{4}$ as shown:

$A_{1}$ is a triangle with signed area

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

$= - 0.5$

$A_{2}$ is a trapezoid with height $1$ , one base length $2$ , and the other base length found from the line through $(3, - 1)$ and $(4, 1)$ .

The equation of that line is

$y = 2 x - 7$

Its $x$ -intercept is $(3.5, 0)$ , so the longer base length is $2.5$ . Then the signed area is

$A_{2} = \frac{1}{2} (- 1) (2 + 2.5)$

$= - 2.25$

$A_{3}$ is a triangle with signed area

$A_{3} = \frac{1}{2} (0.5) (1)$

$= 0.25$

Lastly, find $A_{4}$ by subtracting the area of the semicircle from the area of the rectangle from $x = 4$ to $x = 6$ .

Rectangle: height $1$ , width $2$ , so area $2$ .
Semi-circle: radius $1$ , so area $0.5 π (1)^{2} = 0.5 π$ .

$A_{4} = 2 - 0.5 π$

The net signed area is

$- 0.5 - 2.25 + 0.25 + (2 - 0.5 π)$

$\approx - 2.1$

This means the room’s temperature decreased overall from $1$ AM to $6$ AM by $2.1° C$ .

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)

Example applications

Water flow: net liters gained/lost = signed area under rate graph
Velocity: displacement = signed area; total distance = sum of absolute areas
Temperature change: net change = signed area under $T^{'} (t)$

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:

Interpreting accumulation as the area under a rate function’s graph
Using signed area to find net change (displacement) vs total area
Understanding units in accumulation problems

In the first part of AP Calculus, you were usually given a function (modeling position, population, cost, etc.) and asked to find its rate of change using derivatives.

Now we reverse that idea: if you know the rate of change of a quantity (speed, growth rate, inflow/outflow rate, and so on), you can determine how much the original quantity changes over time.

For example:

Suppose water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water will be in the bucket after $4$ minutes?

When the rate is constant, the total change is just

$3 \times 4 = 12 liters .$

This total change (in the amount of water) is called the accumulation of change.

The graph of the rate function $r (t) = 3$ is a horizontal line. Visually, the product $3 \times 4$ is the area of a rectangle between the line $y = 3$ and the $x$ -axis from $t = 0$ to $t = 4$ .

The units work the same way: area units come from (vertical units) $\times$ (horizontal units).

$\frac{L}{min} \times min = L$

So when you multiply a rate by a time, you can also interpret the accumulated change as the area under the rate graph.

If the rate is not 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.

a) How far is the particle from its starting position after $3$ seconds?
b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

The shaded region is made of:

a trapezoid (blue)
a rectangle (green)

So the total area (and therefore the distance traveled) is

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

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds is the area under the curve on that interval, as shown:

The shaded region is made of:

a rectangle (green)
a trapezoid (blue)

So the distance traveled is

$A = 3 (4) + \frac{1}{2} \cdot 2 (4 + 3)$

$= 19 meters$

AP tip:

Know your units! Always label your final answer with correct units (e.g., meters, liters, etc.). On free-response questions (FRQs), missing or incorrect units can cost you points.

Area “under” a curve

When calculus refers to the “area under a curve,” it means the area between the graph of a function and the $x$ -axis over a given interval.

A key detail: if the function is below the $x$ -axis, that area counts as negative when you’re finding accumulation (net change). This is often called signed area.

Consider the velocity graph below, where velocity is measured in meters per second:

Recall from section 4.2.1 that the sign of velocity tells you direction. We can split the area into two parts:

the triangle (blue) above the $x$ -axis
the trapezoid (green) below the $x$ -axis

The particle travels right (positive velocity) over $0 < t < 2$ seconds, for a displacement of

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

$= 2 meters$

At $t = 2$ seconds, it changes direction and travels left (negative velocity) over $2 < t < 6$ seconds. The (unsigned) area of that trapezoid is

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

$= 3 meters$

Because this region is below the $x$ -axis, its signed area is negative. So the net change in position (displacement) is the net signed area:

$2 + (- 3) = - 1$

This means the particle ends up $1$ meter to the left of its starting position.

If you want total distance traveled, you add the magnitudes of the changes instead:

$∣2∣ + ∣ - 3∣ = 5 meters$

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.

Examples

A container is slowly leaking liquid throughout the day, but it’s also being refilled at certain times. The rate at which the liquid is entering or leaving the container (in liters per hour) varies over time and is recorded in the graph below as function $R (t)$ , where positive values represent liquid being added and negative values represent leakage.

How much water has the container gained or lost in the first $5$ hours?

Note: The curve between the points $(1, - 1)$ and $(3, - 1)$ is a semi-circle.

Solution

(spoiler)

Multiplying the rate by time corresponds to finding area. Here, that area represents the change in the amount of water (in liters).

Split the region into the geometric shapes $A_{1}$ through $A_{5}$ as shown:

Regions $A_{1}$ to $A_{4}$ are below the $x$ -axis, so they have negative signed area.

$A_{1}$ is a triangle with signed area

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

$= - \frac{1}{2}$

$A_{2}$ is a rectangle with signed area

$A_{2} = (2) (- 1) = - 2$

$A_{3}$ is a semi-circle with radius $1$ , so its signed area is

$A_{3} = - \frac{1}{2} π (1)^{2}$

$= - \frac{π}{2}$

Using algebra, the equation of the line is

$y = 3 x - 13$

The $x$ -intercept occurs when $y = 0$ , so the intercept is $(\frac{13}{3}, 0)$ .

That makes the top base length $1 \frac{1}{3}$ , so the signed area is

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

$= - \frac{7}{6}$

$A_{5}$ is a triangle above the $x$ -axis with height $2$ and base $\frac{2}{3}$ . Its signed area is

$A_{5} = \frac{1}{2} \cdot \frac{2}{3} \cdot (2)$

$= \frac{2}{3}$

The net signed area (the accumulation of change) is

$A = - \frac{1}{2} - 2 - \frac{π}{2} - \frac{7}{6} + \frac{2}{3}$

$= - \frac{π}{2} - 3$

$\approx - 4.6 liters$

Because the result is negative, the container has lost approximately $4.6$ liters over the first $5$ hours.

The temperature in a room is affected by a heating and cooling system that turns on and off during the day. Let $T (t)$ be the temperature of the room at time $t$ .

The graph below shows its derivative, $T^{'} (t)$ , the rate of the change in temperature (in $° C$ per hour) at time $t$ , with $t = 0$ corresponding to $12$ AM midnight.

What is the change in room temperature between $1$ AM and $6$ AM?

Solution

(spoiler)

The change in temperature over an interval is the net signed area under $T^{'} (t)$ on that interval.

Split the region into the geometric shapes $A_{1}$ through $A_{4}$ as shown:

$A_{1}$ is a triangle with signed area

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

$= - 0.5$

$A_{2}$ is a trapezoid with height $1$ , one base length $2$ , and the other base length found from the line through $(3, - 1)$ and $(4, 1)$ .

The equation of that line is

$y = 2 x - 7$

Its $x$ -intercept is $(3.5, 0)$ , so the longer base length is $2.5$ . Then the signed area is

$A_{2} = \frac{1}{2} (- 1) (2 + 2.5)$

$= - 2.25$

$A_{3}$ is a triangle with signed area

$A_{3} = \frac{1}{2} (0.5) (1)$

$= 0.25$

Lastly, find $A_{4}$ by subtracting the area of the semicircle from the area of the rectangle from $x = 4$ to $x = 6$ .

Rectangle: height $1$ , width $2$ , so area $2$ .
Semi-circle: radius $1$ , so area $0.5 π (1)^{2} = 0.5 π$ .

$A_{4} = 2 - 0.5 π$

The net signed area is

$- 0.5 - 2.25 + 0.25 + (2 - 0.5 π)$

$\approx - 2.1$

This means the room’s temperature decreased overall from $1$ AM to $6$ AM by $2.1° C$ .

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Accumulation of change

12 min read

Font

Discuss

Feedback

What you’ll learn:

Interpreting accumulation as the area under a rate function’s graph
Using signed area to find net change (displacement) vs total area
Understanding units in accumulation problems

In the first part of AP Calculus, you were usually given a function (modeling position, population, cost, etc.) and asked to find its rate of change using derivatives.

Now we reverse that idea: if you know the rate of change of a quantity (speed, growth rate, inflow/outflow rate, and so on), you can determine how much the original quantity changes over time.

For example:

Suppose water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water will be in the bucket after $4$ minutes?

When the rate is constant, the total change is just

$3 \times 4 = 12 liters .$

This total change (in the amount of water) is called the accumulation of change.

The graph of the rate function $r (t) = 3$ is a horizontal line. Visually, the product $3 \times 4$ is the area of a rectangle between the line $y = 3$ and the $x$ -axis from $t = 0$ to $t = 4$ .

The units work the same way: area units come from (vertical units) $\times$ (horizontal units).

$\frac{L}{min} \times min = L$

So when you multiply a rate by a time, you can also interpret the accumulated change as the area under the rate graph.

If the rate is not 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 $3$ seconds?
b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

The shaded region is made of:

a trapezoid (blue)
a rectangle (green)

So the total area (and therefore the distance traveled) is

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

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds is the area under the curve on that interval, as shown:

The shaded region is made of:

a rectangle (green)
a trapezoid (blue)

So the distance traveled is

$A = 3 (4) + \frac{1}{2} \cdot 2 (4 + 3)$

$= 19 meters$

AP tip:

Know your units! Always label your final answer with correct units (e.g., meters, liters, etc.). On free-response questions (FRQs), missing or incorrect units can cost you points.

Area “under” a curve

When calculus refers to the “area under a curve,” it means the area between the graph of a function and the $x$ -axis over a given interval.

A key detail: if the function is below the $x$ -axis, that area counts as negative when you’re finding accumulation (net change). This is often called signed area.

Consider the velocity graph below, where velocity is measured in meters per second:

Recall from section 4.2.1 that the sign of velocity tells you direction. We can split the area into two parts:

the triangle (blue) above the $x$ -axis
the trapezoid (green) below the $x$ -axis

The particle travels right (positive velocity) over $0 < t < 2$ seconds, for a displacement of

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

$= 2 meters$

At $t = 2$ seconds, it changes direction and travels left (negative velocity) over $2 < t < 6$ seconds. The (unsigned) area of that trapezoid is

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

$= 3 meters$

Because this region is below the $x$ -axis, its signed area is negative. So the net change in position (displacement) is the net signed area:

$2 + (- 3) = - 1$

This means the particle ends up $1$ meter to the left of its starting position.

If you want total distance traveled, you add the magnitudes of the changes instead:

$∣2∣ + ∣ - 3∣ = 5 meters$

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.

Examples

A container is slowly leaking liquid throughout the day, but it’s also being refilled at certain times. The rate at which the liquid is entering or leaving the container (in liters per hour) varies over time and is recorded in the graph below as function $R (t)$ , where positive values represent liquid being added and negative values represent leakage.

Figure 6.1.7 Graph of R(t)

How much water has the container gained or lost in the first $5$ hours?

Note: The curve between the points $(1, - 1)$ and $(3, - 1)$ is a semi-circle.

Solution

(spoiler)

Multiplying the rate by time corresponds to finding area. Here, that area represents the change in the amount of water (in liters).

Split the region into the geometric shapes $A_{1}$ through $A_{5}$ as shown:

Regions $A_{1}$ to $A_{4}$ are below the $x$ -axis, so they have negative signed area.

$A_{1}$ is a triangle with signed area

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

$= - \frac{1}{2}$

$A_{2}$ is a rectangle with signed area

$A_{2} = (2) (- 1) = - 2$

$A_{3}$ is a semi-circle with radius $1$ , so its signed area is

$A_{3} = - \frac{1}{2} π (1)^{2}$

$= - \frac{π}{2}$

Using algebra, the equation of the line is

$y = 3 x - 13$

The $x$ -intercept occurs when $y = 0$ , so the intercept is $(\frac{13}{3}, 0)$ .

That makes the top base length $1 \frac{1}{3}$ , so the signed area is

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

$= - \frac{7}{6}$

$A_{5}$ is a triangle above the $x$ -axis with height $2$ and base $\frac{2}{3}$ . Its signed area is

$A_{5} = \frac{1}{2} \cdot \frac{2}{3} \cdot (2)$

$= \frac{2}{3}$

The net signed area (the accumulation of change) is

$A = - \frac{1}{2} - 2 - \frac{π}{2} - \frac{7}{6} + \frac{2}{3}$

$= - \frac{π}{2} - 3$

$\approx - 4.6 liters$

Because the result is negative, the container has lost approximately $4.6$ liters over the first $5$ hours.

The temperature in a room is affected by a heating and cooling system that turns on and off during the day. Let $T (t)$ be the temperature of the room at time $t$ .

The graph below shows its derivative, $T^{'} (t)$ , the rate of the change in temperature (in $° C$ per hour) at time $t$ , with $t = 0$ corresponding to $12$ AM midnight.

Figure 6.1.9 Graph of T'(t)

What is the change in room temperature between $1$ AM and $6$ AM?

Solution

(spoiler)

The change in temperature over an interval is the net signed area under $T^{'} (t)$ on that interval.

Split the region into the geometric shapes $A_{1}$ through $A_{4}$ as shown:

$A_{1}$ is a triangle with signed area

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

$= - 0.5$

$A_{2}$ is a trapezoid with height $1$ , one base length $2$ , and the other base length found from the line through $(3, - 1)$ and $(4, 1)$ .

The equation of that line is

$y = 2 x - 7$

Its $x$ -intercept is $(3.5, 0)$ , so the longer base length is $2.5$ . Then the signed area is

$A_{2} = \frac{1}{2} (- 1) (2 + 2.5)$

$= - 2.25$

$A_{3}$ is a triangle with signed area

$A_{3} = \frac{1}{2} (0.5) (1)$

$= 0.25$

Lastly, find $A_{4}$ by subtracting the area of the semicircle from the area of the rectangle from $x = 4$ to $x = 6$ .

Rectangle: height $1$ , width $2$ , so area $2$ .
Semi-circle: radius $1$ , so area $0.5 π (1)^{2} = 0.5 π$ .

$A_{4} = 2 - 0.5 π$

The net signed area is

$- 0.5 - 2.25 + 0.25 + (2 - 0.5 π)$

$\approx - 2.1$

This means the room’s temperature decreased overall from $1$ AM to $6$ AM by $2.1° C$ .

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)

Example applications

Water flow: net liters gained/lost = signed area under rate graph
Velocity: displacement = signed area; total distance = sum of absolute areas
Temperature change: net change = signed area under $T^{'} (t)$

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:

Interpreting accumulation as the area under a rate function’s graph
Using signed area to find net change (displacement) vs total area
Understanding units in accumulation problems

In the first part of AP Calculus, you were usually given a function (modeling position, population, cost, etc.) and asked to find its rate of change using derivatives.

Now we reverse that idea: if you know the rate of change of a quantity (speed, growth rate, inflow/outflow rate, and so on), you can determine how much the original quantity changes over time.

For example:

Suppose water is flowing into a bucket at a steady rate of $3$ liters per minute. How much water will be in the bucket after $4$ minutes?

When the rate is constant, the total change is just

$3 \times 4 = 12 liters .$

This total change (in the amount of water) is called the accumulation of change.

The graph of the rate function $r (t) = 3$ is a horizontal line. Visually, the product $3 \times 4$ is the area of a rectangle between the line $y = 3$ and the $x$ -axis from $t = 0$ to $t = 4$ .

The units work the same way: area units come from (vertical units) $\times$ (horizontal units).

$\frac{L}{min} \times min = L$

So when you multiply a rate by a time, you can also interpret the accumulated change as the area under the rate graph.

If the rate is not 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.

a) How far is the particle from its starting position after $3$ seconds?
b) How far has the particle traveled between $t = 2$ and $t = 7$ seconds?

Solutions

a) The distance traveled after $3$ seconds is the area under the graph from $t = 0$ to $t = 3$ , as shown below.

The shaded region is made of:

a trapezoid (blue)
a rectangle (green)

So the total area (and therefore the distance traveled) is

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

$= 10 meters$

b) The distance traveled from $t = 2$ to $t = 7$ seconds is the area under the curve on that interval, as shown:

The shaded region is made of:

a rectangle (green)
a trapezoid (blue)

So the distance traveled is

$A = 3 (4) + \frac{1}{2} \cdot 2 (4 + 3)$

$= 19 meters$

AP tip:

Know your units! Always label your final answer with correct units (e.g., meters, liters, etc.). On free-response questions (FRQs), missing or incorrect units can cost you points.

Area “under” a curve

When calculus refers to the “area under a curve,” it means the area between the graph of a function and the $x$ -axis over a given interval.

A key detail: if the function is below the $x$ -axis, that area counts as negative when you’re finding accumulation (net change). This is often called signed area.

Consider the velocity graph below, where velocity is measured in meters per second:

Recall from section 4.2.1 that the sign of velocity tells you direction. We can split the area into two parts:

the triangle (blue) above the $x$ -axis
the trapezoid (green) below the $x$ -axis

The particle travels right (positive velocity) over $0 < t < 2$ seconds, for a displacement of

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

$= 2 meters$

At $t = 2$ seconds, it changes direction and travels left (negative velocity) over $2 < t < 6$ seconds. The (unsigned) area of that trapezoid is

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

$= 3 meters$

Because this region is below the $x$ -axis, its signed area is negative. So the net change in position (displacement) is the net signed area:

$2 + (- 3) = - 1$

This means the particle ends up $1$ meter to the left of its starting position.

If you want total distance traveled, you add the magnitudes of the changes instead:

$∣2∣ + ∣ - 3∣ = 5 meters$

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.

Examples

A container is slowly leaking liquid throughout the day, but it’s also being refilled at certain times. The rate at which the liquid is entering or leaving the container (in liters per hour) varies over time and is recorded in the graph below as function $R (t)$ , where positive values represent liquid being added and negative values represent leakage.

How much water has the container gained or lost in the first $5$ hours?

Note: The curve between the points $(1, - 1)$ and $(3, - 1)$ is a semi-circle.

Solution

(spoiler)

Multiplying the rate by time corresponds to finding area. Here, that area represents the change in the amount of water (in liters).

Split the region into the geometric shapes $A_{1}$ through $A_{5}$ as shown:

Regions $A_{1}$ to $A_{4}$ are below the $x$ -axis, so they have negative signed area.

$A_{1}$ is a triangle with signed area

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

$= - \frac{1}{2}$

$A_{2}$ is a rectangle with signed area

$A_{2} = (2) (- 1) = - 2$

$A_{3}$ is a semi-circle with radius $1$ , so its signed area is

$A_{3} = - \frac{1}{2} π (1)^{2}$

$= - \frac{π}{2}$

Using algebra, the equation of the line is

$y = 3 x - 13$

The $x$ -intercept occurs when $y = 0$ , so the intercept is $(\frac{13}{3}, 0)$ .

That makes the top base length $1 \frac{1}{3}$ , so the signed area is

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

$= - \frac{7}{6}$

$A_{5}$ is a triangle above the $x$ -axis with height $2$ and base $\frac{2}{3}$ . Its signed area is

$A_{5} = \frac{1}{2} \cdot \frac{2}{3} \cdot (2)$

$= \frac{2}{3}$

The net signed area (the accumulation of change) is

$A = - \frac{1}{2} - 2 - \frac{π}{2} - \frac{7}{6} + \frac{2}{3}$

$= - \frac{π}{2} - 3$

$\approx - 4.6 liters$

Because the result is negative, the container has lost approximately $4.6$ liters over the first $5$ hours.

The temperature in a room is affected by a heating and cooling system that turns on and off during the day. Let $T (t)$ be the temperature of the room at time $t$ .

The graph below shows its derivative, $T^{'} (t)$ , the rate of the change in temperature (in $° C$ per hour) at time $t$ , with $t = 0$ corresponding to $12$ AM midnight.

What is the change in room temperature between $1$ AM and $6$ AM?

Solution

(spoiler)

The change in temperature over an interval is the net signed area under $T^{'} (t)$ on that interval.

Split the region into the geometric shapes $A_{1}$ through $A_{4}$ as shown:

$A_{1}$ is a triangle with signed area

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

$= - 0.5$

$A_{2}$ is a trapezoid with height $1$ , one base length $2$ , and the other base length found from the line through $(3, - 1)$ and $(4, 1)$ .

The equation of that line is

$y = 2 x - 7$

Its $x$ -intercept is $(3.5, 0)$ , so the longer base length is $2.5$ . Then the signed area is

$A_{2} = \frac{1}{2} (- 1) (2 + 2.5)$

$= - 2.25$

$A_{3}$ is a triangle with signed area

$A_{3} = \frac{1}{2} (0.5) (1)$

$= 0.25$

Lastly, find $A_{4}$ by subtracting the area of the semicircle from the area of the rectangle from $x = 4$ to $x = 6$ .

Rectangle: height $1$ , width $2$ , so area $2$ .
Semi-circle: radius $1$ , so area $0.5 π (1)^{2} = 0.5 π$ .

$A_{4} = 2 - 0.5 π$

The net signed area is

$- 0.5 - 2.25 + 0.25 + (2 - 0.5 π)$

$\approx - 2.1$

This means the room’s temperature decreased overall from $1$ AM to $6$ AM by $2.1° C$ .