Net Change and Total Change

What You’ll Learn

In this lesson you’ll learn the difference between net change and total change, and when you need to use absolute values with integrals.

The Concept

The definite integral always gives net signed change. That means positive and negative contributions cancel each other out.

Sometimes that’s what you want. Sometimes it isn’t.

• Net change: the integral as-is. Positive and negative parts cancel. This tells you the overall result.
• Total change: you need the absolute value of each piece. Nothing cancels. This tells you the total amount of activity.

The classic example: velocity.

• Integrating velocity gives net displacement (where you ended up relative to where you started). If you walk 5 miles east then 3 miles west, your net displacement is 2 miles east.
• Integrating the absolute value of velocity gives total distance traveled. You walked 8 miles total.

positive area = 4.5 negative area = -2 net = 2.5, total = 6.5

For f(x) = x on [-2, 3]: the green triangle above the axis has area 4.5, the red triangle below has area 2. The net change (integral) is 4.5 - 2 = 2.5. The total change (sum of absolute values) is 4.5 + 2 = 6.5.

Worked Example

Example 1: Net displacement vs total distance

A particle has velocity v(t) = t - 2 on [0, 5].

The velocity is negative on [0, 2] (moving backward) and positive on [2, 5] (moving forward).

Net displacement:

$$\int_0^5 (t - 2)\,dt = \left[\frac{t^2}{2} - 2t\right]_0^5 = \frac{25}{2} - 10 = 2.5$$

Total distance: split at t = 2 where v changes sign.

$$\left|\int_0^2 (t-2)\,dt\right| + \int_2^5 (t-2)\,dt = |-2| + 4.5 = 2 + 4.5 = 6.5$$

The particle ended up 2.5 units from where it started, but it traveled 6.5 units total.

Example 2: Simple net change

For f(x) = x on [-2, 3]:

$$\int_{-2}^3 x\,dx = \left[\frac{x^2}{2}\right]_{-2}^3 = \frac{9}{2} - \frac{4}{2} = 2.5$$

That’s the net change. The total area is 4.5 + 2 = 6.5 (as shown in the visual above).

Example 3: Rate of change context

A company’s profit rate is P’(t) = 100 - 20t (thousands per month) on [0, 8].

P’ = 0 at t = 5. Positive on [0, 5], negative on [5, 8].

Net profit change:

$$\int_0^8 (100 - 20t)\,dt = \left[100t - 10t^2\right]_0^8 = 800 - 640 = 160$$

The company gained 160 thousand net. But it gained more than that in the first 5 months and lost some in months 5 through 8. The total activity (gains + losses) would be larger.

Real-World Application

• Physics: net displacement vs total distance from velocity
• Finance: net profit vs total revenue + total costs
• Game design: net health change vs total damage taken + healing received
• Biology: net population change vs total births + total deaths

You’ve Got This

The rule is simple: the integral gives net change. If you want total change, split at zeros and add absolute values. Ask yourself: “Do I care about the final result, or the total amount of activity?” Net change answers the first question. Total change answers the second.

Quiz

The definite integral (without absolute values) gives:

• A.Total unsigned change
• B.Net signed change
• C.The derivative
• D.The average value

Integrating velocity gives:

• A.Total distance
• B.Average speed
• C.Net displacement
• D.Acceleration

To find total distance from velocity, you:

• A.Integrate velocity directly
• B.Take the derivative
• C.Integrate the absolute value of velocity
• D.Find the maximum velocity

For f(x) = x on [-2, 3], the net change is 2.5. The total change is:

• A.2.5
• B.4.5
• C.6.5
• D.9

Net change is most useful when you care about:

• A.Total amount of activity
• B.The final result relative to the start
• C.Only the maximum value
• D.Only the minimum value