Lesson date: February 24, 2025.

  • Understand and use the Net Change Theorem.

Assignment

  • Vocabulary and teal boxes
  • p332 1, 4–6, 10, 11, 14, 15, 17, 18 22, 25–27

Additional Resources

Net Change Theorem

Let’s get the definition out of the way first.

The Net Change Theorem

\[\begin{align} \int_a^b F'(x) \, dx = F(b) - F(a) \end{align}\]

This should look familiar, because it is just the definite integral part of the fundamental theorem with $F’$ instead of $f$, which we’ve already discussed is the same damn thing.

\[\begin{align} \int_a^b F'(x) \, dx = F(b) - F(a) &&\text{Net Change Theorem}\\ \int_a^b f(x) \, dx = F(b) - F(a) &&\text{FToC Part Deux} \end{align}\]

This reframing of the fundamental theorem is helpful for understanding what the area under a curve tells you, particularly when the curve in question is a rate. For example, integrating over miles per hour will give the miles traveled (more on that later).

With that said, this section focuses on applying the net change theorem, mostly in the form of problems where you are given a rate and you want to find either accumulation or displacement.

Chemical Flow

You have a chemical flowing into a tank at a rate of $180 + 3t$ liters per minute. How many liters are in the tank after 20 minutes?

Before hopping into this one, let’s look at derivatives and units. When we dealt with position, velocity and acceleration, we saw the units change from meters to meters per second, and then to meters per second per second (meters per second squared). The degree of the independent variable, which will always be in the denominator, increased.

The antiderivative goes the other way. So if we have liters per minute, the antiderivative would be just liters and integrating our function over the given interval would give us the change in liters we are looking for.

\[\begin{align} \int_0^{20} 180 + 3t \, dt = \left[ 180t + \frac{3}{2}t^2 \right]_0^{20} = 4200 \end{align}\]

Particle Motion

If you remember when we first looked at particle motion, there is a difference between moving left and moving right. One will be regarded as positive (typically moving right) and the other negative.

With this in mind, there is a difference between finding the displacement and total distance traveled. Displacement is how far away from the start. Two steps forward and one step back is just one step forward. Total distance, on the other hand, would be three steps.

So, let’s look at a particle’s velocity in meters per second over the interval $[1,5]$.

\[\begin{align} v(t) = t^3 -10t^2 + 29t - 20 \end{align}\]

Finding displacement is the easier of the two. Integrating can produce negative results, so any time spent moving in the negative direction will be accounted for. A straight integration over the interval will give us displacement. I’m skipping most of the steps here, but I’ll address why a bit further down.

\[\begin{align} \int_1^5 t^3 -10t^2 + 29t - 20 \, dt &= \left[ \frac{1}{4}t^4 - \frac{10}{3}t^3 +\frac{29}{2}t^2 -20t \right]_1^5 \\[1em] &= \frac{32}{3} \, \text{meters} \end{align}\]

At the end of the interval, our particle is $32/3$ meters to the right of where it was at the beginning of the interval. Note that is not from its original position. If you were looking for that, the interval would have to start at $0$.

Total distance traveled is more complicated as we need to split our function into intervals based on where it crosses the $x$-axis. The fact that integrals produce negatives has to be accounted for if we want total area. This means finding zeros and then flipping the negative portions to positive. In this case, we’ll give you the extra piece of information that the function crosses at $x=4$.

\[\begin{align} \int_1^4 t^3 -10t^2 + 29t - 20 \, dt + \int_4^5 -\left( t^3 -10t^2 + 29t - 20\right) \, dt \end{align}\]

If you prefer sticking with positives, your integral properties from section 4.3 allow you to flip the interval to flip the sign.

\[\begin{align} &\int_1^4 t^3 -10t^2 + 29t - 20 \, dt + \int_5^4 t^3 -10t^2 + 29t - 20 \, dt \\ &= \left[\frac{t^4}{4} - \frac{10t^3}{3}+\frac{29t^2}{2} - 20t \right]_1^4 + \left[\frac{t^4}{4} - \frac{10t^3}{3}+\frac{29t^2}{2} - 20t \right]_5^4 \\ &= \left[ \left(\frac{(4)^4}{4} - \frac{10(4)^3}{3}+\frac{29(4)^2}{2} - 20(4) \right) - \left(\frac{(1)^4}{4} - \frac{10(1)^3}{3}+\frac{29(1)^2}{2} - 20(1) \right)\right]_1^4 \\ &\qquad+ \left[ \left(\frac{(4)^4}{4} - \frac{10(4)^3}{3}+\frac{29(4)^2}{2} - 20(4) \right) - \left(\frac{(5)^4}{4} - \frac{10(5)^3}{3}+\frac{29(5)^2}{2} - 20(5) \right)\right]_1^4 \\ \end{align}\]

And I’m stopping at this point. Obviously, this is a lot of arithmetic. The one video provided for an exercise this section is 10 minutes long … and the interval on that problem starts at zero which eliminates a huge chunk of work.

So, use a calculator once you get beyond setting up the problem. The AP questions that don’t involve a calculator are more analytical than computational.

And with the lid off of the calculator thing, using absolute value will give you total distance traveled without needing to find the zeros and splitting the integral.

\[\begin{align} \int_1^5 |t^3 -10t^2 + 29t - 20| \, dt &= 11.8\overline{3} \end{align}\]