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## Calculus 2

### Unit 1: Lesson 1

Accumulations of change introduction

# Exploring accumulation of change

AP.CALC:
CHA‑4 (EU)
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CHA‑4.A (LO)
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CHA‑4.A.1 (EK)
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CHA‑4.A.2 (EK)
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CHA‑4.A.3 (EK)
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CHA‑4.A.4 (EK)
Definite integrals are interpreted as the accumulation of quantities. Learn why this is so and how this can be used to analyze real-world contexts.
The definite integral can be used to express information about accumulation and net change in applied contexts. Let's see how it's done.

## Thinking about accumulation in a real world context

Say a tank is being filled with water at a constant rate of start color #11accd, 5, start text, space, L, slash, m, i, n, end text, end color #11accd (liters per minute) for start color #ca337c, 6, start text, space, m, i, n, end text, end color #ca337c. We can find the volume of the water (in start text, L, end text) by multiplying the time and the rate:
\begin{aligned} \text{Volume}&=\maroonD{\text{Time}}\times\blueD{\text{Rate}} \\ &=\maroonD{6\,\text{min}}\cdot\blueD{5\,\dfrac{\text{L}}{\text{min}}} \\ &=30\dfrac{\cancel{\text{min}}\cdot\text{L}}{\cancel{\text{min}}} \\ &=30\,\text{L} \end{aligned}
Now consider this case graphically. The rate can be represented by the constant function r, start subscript, 1, end subscript, left parenthesis, t, right parenthesis, equals, 5:
Function r sub 1 is graphed. Time in minutes is on the x-axis, from 0 to 10. Rate, in liters per minute, is on the y-axis. The graph is a line. The line starts at (0, 5), extends horizontally to the right, and ends at (10, 5).
Each horizontal unit in this graph is measured in minutes and each vertical unit is measured in liters per minute, so the area of each square unit is measured in liters:
start underbrace, start text, m, i, n, end text, end underbrace, start subscript, start text, w, i, d, t, h, end text, end subscript, dot, start underbrace, start fraction, start text, L, end text, divided by, start text, m, i, n, end text, end fraction, end underbrace, start subscript, start text, h, e, i, g, h, t, end text, end subscript, equals, start underbrace, start text, L, end text, end underbrace, start subscript, start text, a, r, e, a, end text, end subscript
A square represents a unit on a graph. The horizontal width represents minutes and the vertical height represents liters per minute. The area within represents liters. The equation to calculate area is width times height = area, or minutes times liters per minute = liters.
Furthermore, the area of the rectangle bounded by the graph of r, start subscript, 1, end subscript and the horizontal axis between t, equals, 0 and t, equals, 6 gives us the volume of water after 6 minutes:
Function r sub 1 is graphed. A rectangular area under the line is shaded. The area extends from 0 to 6 minutes and from 0 to 5 liters per minute . The area of the rectangle is calculated as 6 minutes times 5 liters per minute = 30 liters.
Now say another tank is being filled, but this time the rate isn't constant:
r, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, equals, 6, sine, left parenthesis, 0, point, 3, t, right parenthesis
Function r sub 2 is graphed. Time in minutes is on the x-axis, from 0 to 10. Rate, in liters per minute, is on the y-axis. The graph is a curve. The curve starts at (0, 0), moves upward concave down to about (5.2, 6), moves downward concave down, and ends at about (10, 0.8).
How can we tell the volume of water in this tank after 6 minutes? To do that, let's think about the Riemann sum approximation of the area under this curve between t, equals, 0 and t, equals, 6. For the sake of convenience, let's use an approximation where each rectangle is 1 minute wide.
The previous function, r sub 2, is graphed. Six rectangular bars, each 1 unit, or 1 minute, wide rise vertically from the horizontal axis to the curve from 0 to 6 minutes. Each bar moves upward so that its top right vertex touches the curve. The top left vertex for the five rectangles from 0 to 5 are outside of the curve. Each rectangle has less outside of the curve than the previous. The sixth is completely within the curve. From left to right, the rectangles have the following approximate heights. 1.8, 3.4, 4.7, 5.6, 6, 5.9.
We saw how each rectangle represents a volume in liters. Specifically, each rectangle in this Riemann sum is an approximation of the volume of water that was added to the tank at each minute. When we add all the areas, i.e. when all the volumes are accumulated, we get an approximation for the total volume of water after 6 minutes.
As we use more rectangles with smaller widths, we will get a better approximation. If we take this to a limit of accumulating infinite rectangles, we will get the definite integral integral, start subscript, 0, end subscript, start superscript, 6, end superscript, r, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, d, t. This means that the exact volume of water after 6 minutes is equal to the area bounded by the graph of r, start subscript, 2, end subscript and the horizontal axis between t, equals, 0 and t, equals, 6 .
Function r sub 2 is graphed. The area between the curve and the t-axis, between t = 1 and t = 6, is shaded.
And so, integral calculus allows us to find the total volume after 6 minutes:
integral, start subscript, 0, end subscript, start superscript, 6, end superscript, r, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, d, t, approximately equals, 24, point, 5, start text, L, end text

### Definite integral of the rate of change of a quantity gives the net change in that quantity.

In the example we saw, we had a function that describes a rate. In our case, it was the rate of volume over time. The definite integral of that function gave us the accumulation of volume—that quantity whose rate was given.
Another important feature here was the time interval of the definite integral. In our case, the time interval was the beginning left parenthesis, t, equals, 0, right parenthesis and 6 minutes after that left parenthesis, t, equals, 6, right parenthesis. So the definite integral gave us the net change in the amount of water in the tank between t, equals, 0 and t, equals, 6.
These are the two ways we commonly think about definite integrals: they describe an accumulation of a quantity, so the entire definite integral gives us the net change in that quantity.

### Why "net change" in the quantity and not simply the quantity?

Using the above example, notice how we weren't told whether there was any amount of water in the tank prior to t, equals, 0. If the tank was empty, then integral, start subscript, 0, end subscript, start superscript, 6, end superscript, r, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, d, t, approximately equals, 24, point, 5, start text, L, end text is really the amount of water in the tank after 6 minutes. But if the tank already contained, say, 7 liters of water, then the actual volume of water in the tank after 6 minutes is:
start underbrace, 7, end underbrace, start subscript, start text, v, o, l, u, m, e, space, a, t, space, end text, t, equals, 0, end subscript, plus, start overbrace, integral, start subscript, 0, end subscript, start superscript, 6, end superscript, r, start subscript, 2, end subscript, left parenthesis, t, right parenthesis, d, t, end overbrace, start superscript, start text, c, h, a, n, g, e, space, i, n, space, v, o, l, u, m, e, space, f, r, o, m, space, end text, t, equals, 0, start text, space, t, o, space, end text, t, equals, 6, end superscript
This is approximately 7, plus, 24, point, 5, equals, 31, point, 5, start text, space, L, end text.
Remember: The definite integral always gives us the net change in a quantity, not the actual value of that quantity. To find the actual quantity, we need to add an initial condition to the definite integral.
Problem 1.A
Problem set 1 will walk you through the process of analyzing a context that involves accumulation:
At time t, a population of bacteria grows at the rate of r, left parenthesis, t, right parenthesis grams per day, where t is measured in days.
Function r is graphed. Time in days is on the x-axis, from 0 to 10. Growth rate, in grams per day, is on the y-axis. The graph is a curve. The curve starts at (0, 1), moves upward concave up through (8, 5), and ends at about (10, 7.3). The area between the curve and the x-axis, between t = 0 and t = 8, is shaded.
What are the units of the quantity represented by the definite integral integral, start subscript, 0, end subscript, start superscript, 8, end superscript, r, left parenthesis, t, right parenthesis, d, t?

### Common mistake: Using inappropriate units

As with all applied word problems, units play an important role here. Remember that if r is a rate function measured in start fraction, start color #11accd, start text, Q, u, a, n, t, i, t, y, space, A, end text, end color #11accd, divided by, start color #ca337c, start text, Q, u, a, n, t, i, t, y, space, B, end text, end color #ca337c, end fraction, then its definite integral is measured in start color #11accd, start text, Q, u, a, n, t, i, t, y, space, A, end text, end color #11accd.
For example, in Problem set 1, r was measured in start fraction, start color #11accd, start text, g, r, a, m, s, end text, end color #11accd, divided by, start color #ca337c, start text, d, a, y, end text, end color #ca337c, end fraction, and so the definite integral of r was measured in start color #11accd, start text, g, r, a, m, s, end text, end color #11accd.
Problem 2
Eden walked at a rate of r, left parenthesis, t, right parenthesis kilometers per hour (where t is the time in hours).
What does integral, start subscript, 2, end subscript, cubed, r, left parenthesis, t, right parenthesis, d, t, equals, 6 mean?

### Common mistake: Misinterpreting the interval of integration

For any rate function r, the definite integral integral, start subscript, a, end subscript, start superscript, b, end superscript, r, left parenthesis, t, right parenthesis, d, t describes the accumulation of values between t, equals, a and t, equals, b.
A common mistake is to disregard one of the boundaries (usually the lower one), which results in a wrong interpretation.
For example, in Problem 2, it would be a mistake to interpret integral, start subscript, 2, end subscript, cubed, r, left parenthesis, t, right parenthesis, d, t as the distance Eden walked in 3 hours. The lower boundary is 2, so integral, start subscript, 2, end subscript, cubed, r, left parenthesis, t, right parenthesis, d, t is the distance Eden walked between the 2, start superscript, start text, n, d, end text, end superscript hour and the 3, start superscript, start text, r, d, end text, end superscript hour. Furthermore, in cases like that where the time interval is exactly one unit, we usually say "during the 3, start superscript, start text, r, d, end text, end superscript hour."
Problem 3
Julia's revenue is r, left parenthesis, t, right parenthesis thousand dollars per month (where t is the month of the year). Julia had made 3 thousand dollars in the first month of the year.
What does 3, plus, integral, start subscript, 1, end subscript, start superscript, 5, end superscript, r, left parenthesis, t, right parenthesis, d, t, equals, 19 mean?

### Common mistake: Ignoring initial conditions

For a rate function f and an antiderivative F, the definite integral integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, t, right parenthesis, d, t gives the net change in F between t, equals, a and t, equals, b. If we add an initial condition, we will get an actual value of F.
For example, in Problem 3, integral, start subscript, 1, end subscript, start superscript, 5, end superscript, r, left parenthesis, t, right parenthesis, d, t represents the change in the amount of money Julia made between the 1, start superscript, start text, s, t, end text, end superscript and the 5, start superscript, start text, t, h, end text, end superscript months. But since we added 3, which is the amount Julia had at the 1, start superscript, start text, s, t, end text, end superscript month, the expression now represents the actual amount in the 5, start superscript, start text, t, h, end text, end superscript month.

## Connection with applied rates of change

In differential calculus, we learned that the derivative f, prime of a function f gives the instantaneous rate of change of f for a given input. Now we're going the other way! For any rate function f, its antiderivative F gives the accumulated value of the quantity whose rate is described by f.
QuantityRate
Differential calculusf, left parenthesis, x, right parenthesisf, prime, left parenthesis, x, right parenthesis
Integral calculusF, left parenthesis, x, right parenthesis, equals, integral, start subscript, a, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, tf, left parenthesis, x, right parenthesis
Problem 4
The function k, left parenthesis, t, right parenthesis gives the amount of ketchup (in kilograms) produced in a sauce factory by time t (in hours) on a given day.
What does integral, start subscript, 0, end subscript, start superscript, 4, end superscript, k, prime, left parenthesis, t, right parenthesis, d, t represent?