• 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

Let’s start with a couple of things we know. From the first fundamental theorem

\[\int_a^b f(x) \, dx = F(b) - F(a)\]

And from the second we get

\[F'(x) = f(x)\]

Combining the two, we can write

\[\int_a^b F'(x) \, dx = F(b) - F(a)\]

which is the Net Change Theorem.

For what this actually means, let’s look at just the right side of the equation for now. If you have some function, $F$, and you subtract it’s value at $a$ from its value at $b$, you get the change of that function over that interval. For example, take a position function, $s(t)$, and subtract the position at 1 second from the position at 3 seconds. The result will be the distance traveled.

What the Net Change Theorem says is that this is equivalent to integrating over that function’s derivative on that interval. So with our position function

\[s(3) - s(1) = \int_1^3 v(t) \, dt\]

In other words, given a function and interval, integrating over that function and interval would give you the change of the antiderivative.

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 to meters per second2. The degree of the independent variable, which will always be in the denominator, increased.

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

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

Particle Motion

If you remember when we first looked at these, there is a difference between moving left and moving right. So, you’ll often be asked to find two things: the displacement and the total distance traveled. The displacement is how far away from the start, so two steps forward and one step back is just one step forward, but the total distance would be three steps.

So, here is the velocity of a particle with our interval being $[1,5]$.

\[v(t) = t^3 -10t^2 + 29t - 20\]

Displacement is the easier of the two, since integrating can produce negative results. Any time spent moving in the opposite direction will be accounted for, so integrating over the interval will give us what we need.

\[\int_1^5 t^3 -10t^2 + 29t - 20 \, dt = \left[ \frac{1}{4}t^4 - \frac{10}{3}t^3 +\frac{29}{2}t^2 -20t \right]_1^5 = \frac{32}{3}\]

Total distance is more complicated since we need to split our function into intervals based on where it crosses the $x$-axis, and the flip the negative portions to positive. In this case, it crosses over to negative at $x=4$, so

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