Accumulation Problems

Let's say you're the mayor of a small town. Although you know the rate at which your town grows, you want to know your town's population over the four years that you have served as mayor.

You could go knocking on the door of every house in your town asking how long each resident has lived there. Or, you could simply use a definite integral to interpret the accumulation of people over the time you've served as mayor.

The Fundamental Theorem of Calculus says that for a continuous function \(f\) on the interval \([a,b] \),

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

where \(F\) is the antiderivative of \(f\).

In terms of our population example, to find the town's population from your first year as mayor to your fourth, the Fundamental Theorem of Calculus says you should compute

\[ \int_0^4 f(x)\, dx = F(4) - F(0) \]

where \(f\) is a function of population of your town. In other words, all you have to do is subtract the population from your first year as mayor from the population from your fourth year as mayor.

The Meaning of Accumulation Problems

Let's start by defining what an accumulation problem is.

Suppose there is a small leak below your sink such that it pours water at a constant rate of 1 milliliter per minute, and you place a bucket below the sink overnight, so it does not make a mess until a plumber comes to fix it. Here, the water will start to accumulate in the bucket, and since you know that the rate of change is constant, you can actually calculate how much water will be in the bucket!

For example, let's say that you place the bucket at 10:00 pm in the night and the plumber will come at 8:00 am the next morning. This means that the bucket will be filling for \(10\) hours, which in minutes is

\[ 10(60\text{ min})=600\text{ min.}\]

Next, you multiply this time by the rate at which the bucket fills, that is

\[600\text{ min} \left( \frac{1\text{ ml}}{\text{min}} \right) = 600\text{ ml}\]

so there will be 600ml (about 20 ounces) in the bucket before the sink gets fixed.

This calculation was simple because the rate of change was constant, but what happens if the rate of change is not constant? These problems are typically solved with definite integrals, so Calculus has the answer!

Accumulation Problems in Calculus

You can see an accumulation problem as a situation where you are stacking a quantity that changes over time. The Net Change Theorem, derived from the Fundamental Theorem of Calculus, states that the integral of a rate of change is the net change,

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

Going back to the town population example, remember that you need to know how much the town's population has increased since you were chosen mayor. Let's say you were chosen \( 4\) years ago. Knowing a function \( F(x) \) that tells you the current population of the town would make this a trivial task, as you would only need to subtract the current population, \( F(4),\) from the town's population when you started as a mayor, \( F(0).\) That is

\[ F(4) - F(0).\]

As mentioned before, this would require a survey and would be very time-consuming. However, remember that you know the rate at which the town grows! Suppose this rate is given by the function

\[ f(x) = 1920e^{0.01x}.\]

Since the above function describes a rate of change, you can apply the Net Change Theorem to find the accumulation of population \(A\) in the four years of your government, so

\[ \begin{align} A &= \int_0^4 f(x) \, \mathrm{d}x \\ &= \int_0^4 1920 e^{0.01x}\,\mathrm{d}x. \end{align}\]

\[ \begin{align} A &= \int_0^4 1920e^{0.01x}\,\mathrm{d}x \\ &= \left(\frac{1920}{0.01}e^{0.01(4)}\right)-\left(\frac{1920}{0.01}e^{0.01(0)}\right) \\ &= 192\,000e^{0.04}-192\,000 \\ &= 7835.6686, \end{align}\]

so the growth is of about \(7835\) people.

Figure 1. The area below the curve in the given interval represents the increase in population

This means that over the four years that you have served as mayor, your town has grown by about \(7835\) people. This is the net change of the population, which can also be seen as the accumulation of the population over those four years.

Accumulation Problem Methods

Solving accumulation problems can be summarized as solving a definite integral. However, usually accumulation problems are word problems, so it is important to understand exactly what function you should be integrating, or if you should be integrating at all!

The Net Change Theorem states that the net change of a function is the definite integral of the rate of change of the function. So, if you are provided with a rate of change, you need to integrate it. Otherwise, you just need to find the difference of the function at two different values.

If the function represents a rate of change, and in this case you should integrate, the problem will use phrases like,

• Rate

• Derivative

• Speed

• Velocity

Here is an example.

Sometimes you will not need to integrate! You have to play close attention to the words used in the problem.

Always remember to check if the function represents a change of rate or not!