The Net Change Theorem

The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

Subtracting \(F \left(\right. a \left.\right)\) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.

We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure 5.32.

Figure 5.32 The graph shows speed versus time for the given motion of a car.

Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by

Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by

Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.

To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.

Example 5.24

Finding Net Displacement

Given a velocity function \(v \left(\right. t \left.\right) = 3 t - 5\) (in meters per second) for a particle in motion from time \(t = 0\) to time \(t = 3 ,\) find the net displacement of the particle.

Solution

Applying the net change theorem, we have

\[\begin{aligned} \int_{0}^{3} \left(\right. 3 t - 5 \left.\right) d t & = \frac{3 t^{2}}{2} - 5 t \left|\right._{0}^{3} \\ \\ & = \left[\right. \frac{3 \left(\right. 3 \left.\right)^{2}}{2} - 5 \left(\right. 3 \left.\right) \left]\right. - 0 \\ & = \frac{27}{2} - 15 \\ & = \frac{27}{2} - \frac{30}{2} \\ & = - \frac{3}{2} . \end{aligned}\]

The net displacement is \(- \frac{3}{2}\) m (Figure 5.33).

Figure 5.33 The graph shows velocity versus time for a particle moving with a linear velocity function.