Read signed area from a rate graph

A room that warms and then cools

The graph shows the rate \(T'(t)\) at which the temperature of a room changes, in degrees Celsius per hour. A positive rate means the room warms, and a negative rate means it cools.

Drag the orange edge to the right. Where the rate is positive, the region is teal and the lower graph rises. The lower graph shows the net change in temperature so far: the degrees gained minus the degrees lost.

After \(t = 4\) the rate is negative and the region is red. The lower graph falls, because red area removes degrees.

At \(t = 4\) the teal area is 8. At \(t = 8\) the red area is 6, so the net change is \(8 - 6 = 2\) degrees.

Teal adds, red subtracts

The rate graph crosses the axis, so part of it lies above the axis and part below.

Drag the edge across the crossing. The lower graph rises while the edge moves through teal and falls while it moves through red.

Each teal region adds its area and each red region subtracts its area, so the net change is the teal area minus the red area.

Notation

1. The rate is \(f(t)\), and the interval runs from \(a\) to \(\cy{b}\).

2. The teal area above the axis is \(\cb{P}\) and the red area below it is \(\cr{N}\), both positive numbers. A region above the axis counts as positive and a region below counts as negative, so the integral is \(\cb{P}\) minus \(\cr{N}\). An area taken with this sign is a signed area.

   \[\int_a^b f(t)\,dt = \cb{P} - \cr{N}\]

3. To find the integral, first find where the graph crosses the axis. Each piece between crossings lies on one side of the axis.

4. Find the area of each piece. A graph made of straight lines gives rectangles and triangles.

   \[\text{triangle} = \tfrac12\times\text{base}\times\text{height}\]

5. Add the areas of the teal pieces. Subtract the areas of the red pieces. The result is the net change.

Example 1: A reservoir

The graph shows the rate \(w'(t)\) at which the water level of a reservoir changes, in centimeters per day. Find the net change in the level from \(t = 0\) to \(t = 6\).

1. From \(t = 0\) to \(t = 4\) the graph is above the axis, so the level rises. The region is a rectangle with base 2 and height 3 up to \(t = 2\), then a triangle with base 2 and height 3.

   \[\cb{6} + \cb{3}\]

2. From \(t = 4\) to \(t = 6\) the graph is below the axis, so the level falls. The triangle has base 2 and height 2, so its area is 2 and it counts as negative.

   \[-\cr{2}\]

3. Add the signed areas. The level rises by 7 centimeters.

   \[6 + 3 - 2 = 7\]

Example 2: A phone battery

The graph shows the rate \(b'(t)\) at which a phone's charge changes, in percentage points per hour. Is the charge higher or lower at \(t = 8\) than at \(t = 0\)?

1. The graph crosses the axis once, at \(t = 5\). It is below the axis before 5 and above after.

2. A tempting wrong move: The graph is below the axis for 5 of the 8 hours, so the charge is lower.

   \[\text{5 h below} > \text{3 h above}\]

   The red region has area 9. The teal region covers fewer hours but is taller, with area 12. The net change depends on area, not on time.

3. The red part is a rectangle with base 4 and height 2, and a triangle with base 1 and height 2. The teal part is a triangle with base 3 and height 8.

   \[-(8 + 1) + \tfrac12\cdot3\cdot8 = -9 + 12\]

4. The net change is 3 percentage points, so the charge is higher at \(t = 8\).

Practice

The graph shows the rate \(c'(t)\) at which the number of cars in a lot changes, in cars per hour. Find the net change from \(t = 0\) to \(t = 8\).

Split the interval at the crossing. Each piece is a triangle.

Questions

The graph shows the rate \(d'(t)\) at which the depth of a pond changes, in centimeters per hour.

1. What is the net change in depth from \(t = 6\) to \(t = 10\)?

   • \(-6\) centimeters
   • \(6\) centimeters
   • \(-3\) centimeters
   Answer

   \(-6\) centimeters

2. What is the net change in depth from \(t = 0\) to \(t = 10\)?

   • \(6\) centimeters
   • \(18\) centimeters
   • \(12\) centimeters
   • \(-6\) centimeters
   Answer

   \(6\) centimeters

3. The rate of change of a water level is \(r(t) = 3 - t\) centimeters per hour for \(0 \le t \le 5\). What is the net change in the level?

   • \(\tfrac{5}{2}\) centimeters
   • \(\tfrac{9}{2}\) centimeters
   • \(\tfrac{13}{2}\) centimeters
   • \(-\tfrac{5}{2}\) centimeters
   Answer

   \(\tfrac{5}{2}\) centimeters

Summary

To find a net change from a rate graph, you split at the axis crossings, add the teal areas and subtract the red areas.

This works because a region below the axis is a negative rate times a time, which removes some of the quantity.

You use it when a graph shows a rate that is positive on some intervals and negative on others.

This is one of four skills in topic 6.1 of Which & Why: AP Calculus.