The **net change theorem ** gives you a way to place a value on a changing quantity. It is stated formally as [1]:

The definite integral of the rate of change of a quantity F′(x) gives the net change (or total change) for the quantity on the interval [a, b].

To put this another way, a function’s net change is the definite integral of it’s derivative.

A **special case** of the theorem relates an object’s position function s(t) and its rate of change v(t). Distance traveled is an object’s final position minus its initial position—the integral of velocity [2].

## Net Change Theorem Example

**Example question #1:** Water flows into a lake at a rate of 8 + 3t gallons per minute, where 0 ≤ t ≤ 60. How much water flows into the lake in the first 20 minutes?

Step 1: **Insert your values into the formula. **From the question, we know that the limits of integration are 0 to 20. We’re also given the function to put into the formula: 8 + 3t:

Step 2: **Integrate using the usual rules for integration. **This one uses the integral rule for power functions:

Step 3: **Evaluate the formula.** To do this:

- Plug your lower limit of integration (0) into the formula and solve. That equals zero for this problem.
- Plug your upper limit of integration (20) into the formula and solve. Plugging in 20 and solving gives 3600 + 600 = 4200.
- Subtract (1) from (2): 4200 – 0 = 4200.

The net change of water flowing into the lake in the first 20 minutes is 4200 gallons.

**Example question #2**: If an ball’s velocity function is v(t) = t^{2} – 6t + 8, what is the ball’s displacement between t = 0 and t = 5?

Step 1: **Insert your values into the formula. **

Step 2: **Integrate using the usual rules for integration. **

Step 3: **Evaluate the formula.**

(⅓(5^{3}) – 3(5^{2}) + 8(5)) – (⅓(0^{2}) + 8(0))

The solution is 20/3.

## References

[1] Larson, R. & Edwards, B. (2009), Calculus, 9th Edition. Cengage Learning.

[2] Net Change Theorem. Retrieved April 12, 2021 from: https://web.ma.utexas.edu/users/m408s/AS/LM5-4-6.html