Integration as Accumulated Change

Intro and Net Change Theorem	Integration as Accumulated Change Problems
Integration as Accumulated Change Hints

Integration as Accumulated Change

As the result of the Fundamental Theorems of Calculus from the Definite Integration section, we can now use integration to solve Accumulated Rate of Change, or Net Change problems. Here is the Net Change Theorem (which is basically a reinstatement of the Fundamental Theorem of Calculus):

Net Change Theorem

The definite integral of the rate of change gives the total accumulated change, or net change, of the quantity in interval $ \left[ {a,b} \right]$:

$ \displaystyle \int\limits_{a}^{b}{{{F}’\left( x \right)}}\,dt=F\left( b \right)-F\left( a \right)$

The Accumulated Rate of Change can be measured by the area under the graph of a function over a certain interval. Thus, this can be represented by the definite integral of the function. We must be careful though, since the area below the $ \boldsymbol {x}$-axis is considered to be negative in measuring accumulated change.

(Note that we also addressed Position, Velocity, and Acceleration with Derivatives here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section, and here in the Antiderivatives and Indefinite Integration section.)

Important Hint for Definite Integration Applications: If you’re not sure about whether to integrate, or what to integrate, remember this: the area under the curve is the integral. That area will represent different things, based on what the units of the axes are. In general, you just need to multiply the units of the $ y$-axis and the units of the $ x$-axis to get the units of the area under the curve. For example, If the $ y$-axis represents velocity and the $ x$-axis represents time, the integral represents total distance. As another example, if the $ y$-axis represents calories/day and the $ x$-axis represents days, then the area under the curve would represent the total calories over the days specified.

Integration as Accumulated Change Hints

When working these problems, remember the following:

Integration as Accumulated Change Hints

For Integration as Accumulated Change problems, we typically have rate (velocity) on the $ y$-axis and time on the $ x$-axis. The net change is the area under the curve, or the integral of the velocity function. For example, we may have: $ \require{cancel} \displaystyle \frac{{\text{miles}}}{{\cancel{{\text{hours}}}}}\text{(}y\text{-axis)}\times \cancel{{\text{hours}}}\text{(}x\text{-axis)}=\text{miles (the area, which is the integral)}$.
Amount changed from $ a$ to $ b$ is $ \displaystyle \int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$, where $ {f}’\left( x \right)$ is the rate of change of the amounts. $ f\left( x \right)$ represents the actual amounts (for example, miles, dollars or temperature).
Ending amount is the beginning amount plus the change. At $ b$, the amount is what it was at $ a$ (Initial Condition), plus the change from $ a$ to $ b$: $ \displaystyle f\left( b \right)=f\left( a \right)+\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$ (from the First Fundamental Theorem of Calculus). If the change is negative (amount decreasing), then $ \displaystyle f\left( b \right)=f\left( a \right)-\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$.
The Average Velocity can be obtained by the “Integral Over Interval” formula: Average Velocity from time $ a$ to time $ b$ is $ \displaystyle \frac{{\int\limits_{a}^{b}{{f\left( x \right)dx}}}}{{\left( {b-a} \right)}}=\frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)dx}}$, where $ f\left( x \right)$ is a function of the velocity versus time.
The total distance traveled (how far we go back and forth) is $ \displaystyle \int\limits_{a}^{b}{{\left| {\,v\left( x \right)} \right|}}\,dx$, whereas the total displacement (where we are on a line, compared to where we started) is $ \displaystyle \int\limits_{a}^{b}{{v\,\left( x \right)}}\,dx$.
The Definite Integral of a function’s population growth rate in a time interval gives the total change in population in that interval.
The Definite Integral of a density of a population in a distance interval (for example, cars per mile) gives the total amount of that population in that interval (for example, total number of cars).

Note that if the function is totally above the $ y$-axis in the given interval, this calculation is the area between the function and the $ x$-axis. If the function is totally below the $ x$-axis in this interval, this calculation is the opposite (negative) of the area between the $ x$-axis and the function. If this function is both above and below the $ x$-axis, then this calculation is the area above the $ x$-axis subtracted by the area below the $ x$-axis.

Remember these rules with position, velocity, and acceleration:

Definite Integral of a function’s derivative gives the accumulated change.
Definite Integral of a function’s velocity gives the total change in position.
Definite Integral of a function’s acceleration gives the total change in velocity.
When the velocity is positive, an article is moving to the right, when it’s negative, it’s moving to the left, and when it’s 0, it’s standing still. This makes sense since the area under the curve of a velocity graph is distance, and when distance is positive, it’s accumulating (article is moving to right), and when it’s negative, it’s lessening (article moving to left).

Integration as Accumulated Change Problems

First, here’s a problem that uses the area under a curve:

Problem: The following graph depicts the speed of a car over a period over 6 hours. Estimate how far it traveled during that time, and the average speed between 10am and 4pm.

Solution: Since we can think of distance as $ \text{rate}\,\times \,\text{time}$, we can get the distance by the area under the curve. Add up (as best we can) all the squares under the curve: $ \left( {20\times 11} \right)+18+14+8+8+8+7+5+18+8+16=330\,\,\text{km}$. To get the average speed between 10am and 4pm,take the total distance and divide by the total time: $ \displaystyle \frac{{330\,\text{km}}}{{6\,\text{hours}}}=55\,\text{km/hour}$.

Here’s another type of problem you may see:

Accumulating Change Problem:

A particle is moving along the $ x$-axis, with the particle’s velocity (units/sec) for time $ t$ ($ 0\le t\le 9$) as shown:

a. Find the total displacement of the particle in the 9-second interval.

b. Find the particle’s total distance traveled in the 9-second interval.

c. If the position of the particle is $ p\left( t \right)$, and $ p\left( 4 \right)=3$, find the position of the particle at $ t=9$.

d. Find the average value of the velocity of the particle during this 9-second time interval.

e. What is the displacement of the particle from the 2^nd second to the 6^th second? (from $ t=1$ to $ t=6$)?

Solution:

a. The displacement is the value of the integral of the function in that interval, with the areas above the $ x$-axis positive and under the $ x$-axis negative. This displacement is $ 8-8=0$.

b. The distance is the absolute value of the integral across this interval. The distance traveled is $ 8+8=16$.

c. If the position at $ t=4$ is 3, the position at $ t=9$ would be $ \displaystyle 3+\frac{1}{2}-8=-4\frac{1}{2}$. This is because the ending value is the beginning value (3) plus the change, which is the integral. We start out at 3, then have above the $ x$-axis (from $ t=4$ to $ t=5$) and 8 below the $ x$-axis (from $ t=5$ to $ t=9$).

d. The average value is integral (displacement) over interval, so we have $ \displaystyle \frac{0}{9}=0$ (which is the same as $ \displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( t \right)dt}}=\frac{1}{{\left( {9-0} \right)}}\int\limits_{0}^{9}{{f\left( t \right)dt}}=\frac{0}{9}=0$).

e. The displacement from the 2^nd second to the 6^th second is the area (positive or negative) from where $ t=1$ to $ t=6$, which is $ 6 -.5=5.5$. This is the same as $ \displaystyle \int\limits_{1}^{6}{{f\left( t \right)dt}}=-5.5$. Note that we started the interval with $ t=1$ instead of $ t=2$, since the 2^nd second starts when $ t=1$.

Here are more Integration as Accumulated Change problems. Note that in some problems, we will use the fnInt( function on our calculator to integrate. (We will learn how to integrate exponents here in the Exponential and Logarithmic Integration section.)

Integration as Accumulated Change Problem	Solution
In a certain year, the average value of an American’s income was changing at a rate (dollars per month) that is modeled by the function $ c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$, where $ t$ is months since January 1 of that year. What change in income can the average American expect by the end of September, of that same year?	For the change in income, use the formula $ \displaystyle \int\limits_{a}^{b}{{{f}’\left( t \right)}}\,dt$, where $ f\left( t \right)$ is the actual amount of income. The rate of change formula is $ c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$; integrate that function from 0 (January) to 9 (September) (using the calculator): $ \displaystyle \int\limits_{0}^{9}{{{c}\left( t \right)}}\,dt=\int\limits_{0}^{9}{{40{{{\left( {1.002} \right)}}^{t}}}}\,dt=\$363.26$.
Water is leaking out of a bucket at a rate of $ r\left( t \right)=6{{e}^{{-.1t}}}$ liters per minute, where $ t$ is the number of minutes since the leak started. If the bucket holds 500 liters of water when the leak began, how much water does the bucket hold an hour later?	Ending amount is the beginning amount (initial condition) plus the change. At $ b$, the amount is what it was at $ a$, plus the change from $ a$ to $ b$. (Amount changed from $ a$ to $ b$ is $ \displaystyle \int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$, where $ \displaystyle {f}’\left( x \right)$ is the rate of change of the amounts). Since water is leaking out of the bucket (negative change), its amount is decreasing: $ \displaystyle f\left( b \right)=f\left( a \right)-\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$. Thus, $ \displaystyle f\left( b \right)=500-\int\limits_{0}^{{60}}{{r\left( t \right)}}\,dt=500-\int\limits_{0}^{{60}}{{6{{e}^{{-.1t}}}}}\,dt=500-59.851=440.149$ liters. (Use calculator to integrate and remember that there are 60 minutes in an hour.)
The density of cars (cars per mile) on a 10-mile stretch can be modeled by $ d\left( x \right)=200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,$, where $ x$ is the distance in miles from the starting point. To the nearest car, what is the total number of cars on the 10-mile stretch?	Remember that the total amount of a certain population in an interval can be obtained by integrating the density of that amount in that integral (see how $ \displaystyle \require{cancel} \frac{{\text{cars}}}{{\cancel{{\text{mile}}}}}\,\,\times \cancel{{\text{miles}}}=\text{cars}$)? The total number of cars then on the 10-mile stretch is: $ \displaystyle \int\limits_{0}^{{10}}{{d\left( x \right)\,dx}}=\int\limits_{0}^{{10}}{{200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,dx}}\approx 5716$ cars.
The velocity in meters per second of a particle is moving along the $ x$-axis at $ v\left( t \right)=2\cos \left( t \right)$ in the interval $ 0\le t\le 2\pi $. Determine when the particle is moving to the left, right, and stopped. Find the particle’s displacement and total distance over the time interval.	When the velocity is positive, an article is moving to the right, when it’s negative, it’s moving to the left, and when it’s 0, it’s standing still. Thus, when the article is moving to the right, $ 2\cos \left( t \right)\,>0$, or $ \displaystyle 0<t<\frac{\pi }{2}$ and $ \displaystyle \frac{{3\pi }}{2}<t<2\pi $. When the article is moving to the left, $ 2\cos \left( t \right)\,<0$, or $ \displaystyle \frac{\pi }{2}<t<\frac{{3\pi }}{2}$. The particle is stopped at $ \displaystyle t=\frac{\pi }{2}$ and $ \displaystyle t=\frac{{3\pi }}{2}$. The particle’s displacement is $ \displaystyle \int\limits_{0}^{{2\pi }}{{2\cos \left( t \right)}}\,dt=0$ and the particle’s distance is $ \displaystyle \int\limits_{0}^{{2\pi }}{{\left\| {2\cos \left( t \right)} \right\|}}\,dt=8$.
The rate at which customers arrive at a water park in the summer can be modeled by the function $ \displaystyle A\left( t \right)=\frac{{16000}}{{{{t}^{2}}-22t+160}}$, and the rate at which they leave later that day can be modeled by the function $ \displaystyle L\left( t \right)=\frac{{10000}}{{{{t}^{2}}-40t+350}}$ (both functions in people per hour, $ t$ is hours after midnight). At $ t=10$ (right before the park opens), there are no people. At $ t=23$ (when park closes), everyone has left the park. (The park is open from 10am to 11pm.) a. What is the number of people who have entered the park by 4pm ($ t=16$)? b. The price of a ticket to the park is $20 until 4pm, and $10 after 4pm. How much did the water park make this day? c. At what time is the number of people in the water park at a maximum?	a. The definite integral of a function’s population growth rate in a time interval gives the total change in population in that interval: $ \displaystyle \int\limits_{{10}}^{{16}}{{A\left( t \right)}}=\int\limits_{{10}}^{{16}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}\approx 2136$ people entered the park by 4pm. b. We want $20 times the number of people entering the park between 10am and 4pm, and $10 times the number of people entering the park between 4pm and closing: $ \displaystyle 20\int\limits_{{10}}^{{16}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}+10\int\limits_{{16}}^{{23}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}=\$53383.94$. c. Let $ \displaystyle H\left( t \right)=\int{{\left( {A\left( t \right)-L\left( t \right)} \right)}}$ equal the number of people in the park at time $ t$: $ \displaystyle H\left( t \right)=\int\limits_{{}}^{{}}{{\left( {\frac{{16000}}{{{{t}^{2}}-22t+160}}-\frac{{10000}}{{{{t}^{2}}-40t+350}}} \right)\,}}dt$, which is the number of people entering minus the number of people leaving. To get the time for the maximum number of people in the park, take the derivative of this function, which is just $ \displaystyle {H}’\left( t \right)=\frac{{16000}}{{{{t}^{2}}-22t+160}}-\frac{{10000}}{{{{t}^{2}}-40t+350}}$, and set to 0 (using graphing calculator and the intersect function; you can find examples here in the Introduction to Quadratics section): $ t=11.37$, or between 11am and 11:30am). Tricky!

Integration as Accumulated Change Problem

Solution

In a certain year, the average value of an American’s income was changing at a rate (dollars per month) that is modeled by the function $ c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$, where $ t$ is months since January 1 of that year.

What change in income can the average American expect by the end of September, of that same year?

For the change in income, use the formula $ \displaystyle \int\limits_{a}^{b}{{{f}’\left( t \right)}}\,dt$, where $ f\left( t \right)$ is the actual amount of income.

The rate of change formula is $ c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$; integrate that function from 0 (January) to 9 (September) (using the calculator): $ \displaystyle \int\limits_{0}^{9}{{{c}\left( t \right)}}\,dt=\int\limits_{0}^{9}{{40{{{\left( {1.002} \right)}}^{t}}}}\,dt=\$363.26$.

Water is leaking out of a bucket at a rate of $ r\left( t \right)=6{{e}^{{-.1t}}}$ liters per minute, where $ t$ is the number of minutes since the leak started.

If the bucket holds 500 liters of water when the leak began, how much water does the bucket hold an hour later?

Ending amount is the beginning amount (initial condition) plus the change. At $ b$, the amount is what it was at $ a$, plus the change from $ a$ to $ b$. (Amount changed from $ a$ to $ b$ is $ \displaystyle \int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$, where $ \displaystyle {f}’\left( x \right)$ is the rate of change of the amounts). Since water is leaking out of the bucket (negative change), its amount is decreasing: $ \displaystyle f\left( b \right)=f\left( a \right)-\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$.

Thus, $ \displaystyle f\left( b \right)=500-\int\limits_{0}^{{60}}{{r\left( t \right)}}\,dt=500-\int\limits_{0}^{{60}}{{6{{e}^{{-.1t}}}}}\,dt=500-59.851=440.149$ liters. (Use calculator to integrate and remember that there are 60 minutes in an hour.)

The density of cars (cars per mile) on a 10-mile stretch can be modeled by $ d\left( x \right)=200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,$, where $ x$ is the distance in miles from the starting point.

To the nearest car, what is the total number of cars on the 10-mile stretch?

Remember that the total amount of a certain population in an interval can be obtained by integrating the density of that amount in that integral (see how $ \displaystyle \require{cancel} \frac{{\text{cars}}}{{\cancel{{\text{mile}}}}}\,\,\times \cancel{{\text{miles}}}=\text{cars}$)?

The total number of cars then on the 10-mile stretch is: $ \displaystyle \int\limits_{0}^{{10}}{{d\left( x \right)\,dx}}=\int\limits_{0}^{{10}}{{200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,dx}}\approx 5716$ cars.

The velocity in meters per second of a particle is moving along the $ x$-axis at $ v\left( t \right)=2\cos \left( t \right)$ in the interval $ 0\le t\le 2\pi $.

Determine when the particle is moving to the left, right, and stopped.

Find the particle’s displacement and total distance over the time interval.

When the velocity is positive, an article is moving to the right, when it’s negative, it’s moving to the left, and when it’s 0, it’s standing still.

Thus, when the article is moving to the right, $ 2\cos \left( t \right)\,>0$, or $ \displaystyle 0<t<\frac{\pi }{2}$ and $ \displaystyle \frac{{3\pi }}{2}<t<2\pi $. When the article is moving to the left, $ 2\cos \left( t \right)\,<0$, or $ \displaystyle \frac{\pi }{2}<t<\frac{{3\pi }}{2}$. The particle is stopped at $ \displaystyle t=\frac{\pi }{2}$ and $ \displaystyle t=\frac{{3\pi }}{2}$. The particle’s displacement is $ \displaystyle \int\limits_{0}^{{2\pi }}{{2\cos \left( t \right)}}\,dt=0$ and the particle’s distance is $ \displaystyle \int\limits_{0}^{{2\pi }}{{\left| {2\cos \left( t \right)} \right|}}\,dt=8$.

The rate at which customers arrive at a water park in the summer can be modeled by the function $ \displaystyle A\left( t \right)=\frac{{16000}}{{{{t}^{2}}-22t+160}}$, and the rate at which they leave later that day can be modeled by the function $ \displaystyle L\left( t \right)=\frac{{10000}}{{{{t}^{2}}-40t+350}}$ (both functions in people per hour, $ t$ is hours after midnight).

At $ t=10$ (right before the park opens), there are no people. At $ t=23$ (when park closes), everyone has left the park. (The park is open from 10am to 11pm.)

a. What is the number of people who have entered the park by 4pm ($ t=16$)?

b. The price of a ticket to the park is $20 until 4pm, and $10 after 4pm. How much did the water park make this day?

c. At what time is the number of people in the water park at a maximum?

a. The definite integral of a function’s population growth rate in a time interval gives the total change in population in that interval: $ \displaystyle \int\limits_{{10}}^{{16}}{{A\left( t \right)}}=\int\limits_{{10}}^{{16}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}\approx 2136$ people entered the park by 4pm.

b. We want $20 times the number of people entering the park between 10am and 4pm, and $10 times the number of people entering the park between 4pm and closing: $ \displaystyle 20\int\limits_{{10}}^{{16}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}+10\int\limits_{{16}}^{{23}}{{\frac{{16000}}{{{{t}^{2}}-22t+160}}}}=\$53383.94$.

c. Let $ \displaystyle H\left( t \right)=\int{{\left( {A\left( t \right)-L\left( t \right)} \right)}}$ equal the number of people in the park at time $ t$: $ \displaystyle H\left( t \right)=\int\limits_{{}}^{{}}{{\left( {\frac{{16000}}{{{{t}^{2}}-22t+160}}-\frac{{10000}}{{{{t}^{2}}-40t+350}}} \right)\,}}dt$, which is the number of people entering minus the number of people leaving. To get the time for the maximum number of people in the park, take the derivative of this function, which is just $ \displaystyle {H}’\left( t \right)=\frac{{16000}}{{{{t}^{2}}-22t+160}}-\frac{{10000}}{{{{t}^{2}}-40t+350}}$, and set to 0 (using graphing calculator and the intersect function; you can find examples here in the Introduction to Quadratics section): $ t=11.37$, or between 11am and 11:30am). Tricky!

On to Exponential and Logarithmic Integration – you’re ready!