Skip to main content

Section 6.1 Integrals as Net Change

We have some rudimentary ideas of what an integral is, but we want to challenge and expand those ideas by examining the object at the root of the definition of the Definite Integral: a Riemann Sum. We want to generalize a bit more our notion of what a Riemann sum is. So for now, let’s think about how we can use a Riemann sum to think about a measurement that will not be an area. That’s been our only real context so far, so let’s try to stretch that thinking.

Subsection Estimating Movement

Activity 6.1.1 Estimating Movement.

We’re observing an object traveling back and forth in a straight line. Throughout a 5 minute interval, we get the following information about the velocity (in feet/second) of the object.
Table 6.1.2. Velocity of an Object
\(t\) \(v(t)\)
0 0
30 2
60 4.25
90 5.75
120 3.5
150 0.75
180 -1.25
210 -3.5
240 -2.75
270 -0.5
300 -0.25

(a)

Describe the motion of the object in general.
Hint.
How do we interpret the different values of velocity? How do we interpret the sign of velocity? What about how velocity changes from one of the 30-second time points to the next?

(b)

When was the acceleration of the object the greatest? When was it the least?
Hint.
You can decide how to interpret the "least" acceleration: it is either where the acceleration is closest to 0, or it is the most negative value of the acceleration. These are interpreted differently, but it’s a bit ambiguous what we might mean when we say "least acceleration."

(c)

Estimate the total displacement of the object over the 5 minute interval. What is the overall change in position from the start to the end?
Hint.
How do we use velocity and some time interval to estimate the distance traveled? How do we estimate/assume the velocity on each 30-second time interval?

(d)

Is this different than the total distance that the object traveled over the 5 minute interval? Why or why not?
Hint.
How do we think about (or ignore) the direction of the object? Why is this important here?

(e)

If we know the initial position of the object, how could we find the position of the object at some time, \(t\text{,}\) where \(t\) is a multiple of 30 between 0 and 300?
Hint.
Can we limit the time intervals that we use to calculate the object’s displacement? How do we use displacement and a starting point to find an ending point?
So what are the big ideas in this short activity? There are a lot, and many of them already are things we know, at least to some level. So we are really focusing on adding depth to our understanding of these big ideas. Let’s list them in the order that they showed up in this activity:
  1. We interpret the velocity as the derivative of the position of the object. So when we interpret the value of the velocity of the object (large vs small, positive vs negative, etc.) we are interpreting these through the lens of a rate of change.
  2. Acceleration is the derivative of the velocity function. While we don’t have the full picture of the velocity function at any value of \(t\text{,}\) we still were interested in the rates at which velocity changes with regard to time.
  3. We can estimate the total displacement of the object by predicting how far it traveled in each 30-second time interval. We might pick the starting velocity for each 30-second interval and multiply that by 30 seconds. We could alternatively pick the ending velocity of each 30-second interval. Then, we can add all of these products of velocity and time together to approximate a total change in position! Doesn’t this feel like a Riemann sum?
  4. When we calculate displacement, the negative velocities get multiplied out to get negative changes in position for the object -- that’s because a negative velocity means that the object is moving backwards. If we wanted to calculate the distance traveled, then we need to not account for negative velocities. We can just disregard the sign of the velocity on each time interval and repeat the process above. So, another Riemann sum then?
  5. In order to forecast some position at time \(t\text{,}\) we just need to start with the initial position, and then calculate (or approximate) the displacement from \(t=0\) to whatever time \(t\leq 300\) we care about, and then add the displacement to the initial position.
Ok, now let’s formalize those results!

Subsection Position, Velocity, and Acceleration

We know that the velocity of an object is really a rate of change of the position of that object with regard to time. Similarly, the acceleration of an object is the rate of change of the velocity of the object with regard to time. So we’re really thinking about derivatives!

Definition 6.1.3 Position, Velocity, and Acceleration Functions.

For an object moving along a straight line, if \(s(t)\) represents the position of that object at time \(t\text{,}\) then the velocity of the object at time \(t\) is \(v(t) = s'(t)\) and the acceleration of the object at time \(t\) is \(a(t) = v'(t) = s''(t)\text{.}\)
Once we establish this relationship, we can answer questions about movement of an object using the same interpretations of derivatives that we practiced in Chapter 3 of this text.

Activity 6.1.4 A Friendly Jogger.

Consider a jogger running along a straight-line path, where their velocity at \(t\) hours is \(v(t)=2t^2-8t+6\text{,}\) and velocity is measured in miles per hour. We begin observing this jogger at \(t=0\) and observe them over a course of 3 hours.

(a)

When is the jogger’s acceleration equal to 0 mi/hr\(^2\text{?}\)
Hint.
Solve \(a(t) = v'(t) = 0\text{.}\)

(b)

Does this time represent a maximum or minimum velocity for the jogger?
Hint.
You can use the First Derivative Test or the Second Derivative Test here!

(c)

When is the jogger’s velocity equal to 0 mi/hr?

(d)

Describe the motion of the jogger, including information about the direction that they travel and their top speeds.

Subsection Displacement, Distance, and Speed

Let’s revisit Activity 6.1.1. When we approximated the displacement of the object, we built a Riemann sum:
\begin{equation*} \sum_{k=1}^{10} v(t_k^*) \Delta t \end{equation*}
We chose our \(t_k^*\) as either the time at the beginning of each 30-second interval or the time at the end of the 30-second interval, but that was only because of the limited information that we had about different values of \(v(t)\text{.}\) If we had information about the \(v(t)\) function at any values of \(t\) (\(0\leq t\leq 300\)), then we could pick any time in each 30-second time interval for our Riemann sum! We might note, though, that if we did have this kind of information about the velocity at any time in the 5-minute interval, then we also would build a more precise approximation by subdividing the time interval into smaller/shorter pieces. So maybe the Riemann sum \(\displaystyle\sum_{k=1}^{100} v(t_k^*)\Delta t\) (where we are dividing up the 5 minute interval into 100, 3-second intervals) would do a better job! But why stop there? If we have the definition of the velocity function, and so we can truly obtain the velocity of the object at any time in the 5 minute interval, then we can use the definition of the definite integral as the limit of a Riemann sum:
\begin{equation*} \lim_{n\to\infty} \sum_{k=1}^n v(t_k^*) \Delta t = \int_{t=0}^{t=300} v(t)\;dt \end{equation*}
This should work out well with our first understanding of displacement: the displacement of an object is just the difference in position from the starting time to the ending time. So we could say that if \(s(t)\) is the position function, then we might expect to represent displacement from \(t=a\) to \(t=b\) as \(s(b)-s(a)\text{.}\) But isn’t this just the Fundamental Theorem of Calculus, since \(s'(t) = v(t)\text{?}\)

Definition 6.1.5 Displacement of an Object.

If an object is moving along a straight line with velocity \(v(t)\) and position \(s(t)\text{,}\) then the displacement of the object from time \(t=a\) to \(t=b\) is
\begin{equation*} \int_{t=a}^{t=b} v(t)\;dt = s(b)-s(a) \end{equation*}
Let’s keep revisiting the same activity. We also noticed that when we looked at the distance compared to the displacement, the only difference was that we were integrating the absolute value of the velocity function, since we didn’t care about the sign of the velocity (the direction that the object was traveling) on each interval.

Definition 6.1.6 Distance Traveled.

If an object is moving along a straight line with velocity \(v(t)\text{,}\) then the distance traveled by the object from time \(t=a\) to \(t=b\) is:
\begin{equation*} \int_{t=a}^{t=b}|v(t)|\;dt \end{equation*}
Here, we call \(|v(t)|\) the speed of the object (instead of the velocity).
We should note that we don’t have any quick and easy ways of dealing with the integral of the absolute value of a function.
\begin{equation*} |v(t)| = \begin{cases} -v(t) \amp \text{when } v(t)\lt 0\\ \phantom{-}v(t) \amp \text{when } v(t)\geq 0 \end{cases} \end{equation*}
So, in order for us to integrate \(|v(t)|\text{,}\) we need to think about where the velocity passes through 0, so that we can see where it might change from positive to negative.

Activity 6.1.7 Tracking our Jogger.

Let’s revisit our jogger from Activity 6.1.4.

(a)

Calculate the total displacement of the jogger from \(t=0\) to \(t=3\text{.}\)
Hint.
Set up and evaluate a definite integral here, using the velocity function.

(b)

Think back to our description of the jogger’s movement: when is this jogger moving backwards? Split up the time interval from \(t=0\) (the start of their run) to \(t=c\) (where \(c\) is the time that the jogger changed direction) to \(t=3\text{.}\) Calculate the displacements on each of these two intervals.

(c)

Calculate the total distance that the jogger traveled in their 3 hour run.
Hint.
Remember that we’re really calculating:
\begin{equation*} \left\vert \int_{t=0}^{t=c} v(t)\;dt\right\vert + \left\vert \int_{t=c}^{t=3} v(t)\;dt\right\vert \end{equation*}

Subsection Finding the Future Value of a Function

We can again think back to Activity 6.1.1 and build our last result of this section. Remember when we were looking to predict the location of our object at different times: we said it was reasonable to start at our initial position, and then add the displacement of the object from that initial time up to the time that we were interested in. So, to estimate the object’s position after 150 seconds, we would calculate:
\begin{equation*} s(0)+\int_{t=0}^{t=150} v(t)\;dt\text{.} \end{equation*}
But we said we could do this to estimate the object’s position at any value for time, \(t\text{.}\)
We can note that this relationship between velocity and position can exist in many other contexts: any pair of functions that are derivatives/antiderivatives of each other can have this relationship!

Practice Problems Practice Problems

1.

Explain the following terms in reference to an object moving along a straight path from time \(t=a\) to time \(t=b\text{.}\)

(a)

Position of the object at time \(t\text{.}\)

(b)

Displacement of the object.

(c)

Distance traveled by the object.

(d)

Velocity of the object at time \(t\text{.}\)

(e)

Speed of the object at time \(t\text{.}\)

2.

Consider the graph of a velocity function, \(v(t)\text{,}\) of some object moving along a line on the time interval \([0,7]\text{.}\)
A polynomial function that starts at (0,0), decreases down to some negative output around t=2, increases to (4,0) and then continues to some positive output before decreasing and ending at (7,0).
Figure 6.1.10.

(a)

Do you expect the displacement of the object from \(t=0\) to \(t=7\) to be positive, negative, or 0?
Solution.
Displacement will likely be negative, since it looks like the size of the integral between \(t=0\) and \(t=4\) is larger than the size of the integral from \(t=4\) up to \(t=7\text{.}\) Another way to say this is that there is more negative velocity accumulated than positive velocity.

(b)

Write two different expressions that represent the total displacement of the object from \(t=0\) to \(t=7\text{.}\)
Solution.
There are plenty of ways to do this! Here are two:
  1. \(\displaystyle \displaystyle \int_{t=0}^{t=7} v(t)\;dt\)
  2. \(\displaystyle \displaystyle \int_{t=0}^{t=4} v(t)\;dt + \int_{t=4}^{t=7} v(t)\;dt\)

(c)

Do you expect the distance traveled by the object from \(t=0\) to \(t=7\) to be positive, negative, or 0?
Solution.
Distance is always non-negative, and since the velocity was not just constant at 0 (with no movement), then we expect the distance to be positive. It is the integral of speed, which is \(|v(t)|\text{,}\) velocity without taking into account the direction.

(d)

Write two different expressions that represent the total distance traveled by the object from \(t=0\) to \(t=7\text{.}\)
Solution.
There are plenty of ways to do this! Here are two:
  1. \(\displaystyle \displaystyle \int_{t=0}^{t=7} |v(t)|\;dt\)
  2. \(\displaystyle \displaystyle -\int_{t=0}^{t=4} v(t)\;dt \int_{t=4}^{t=7} v(t)\;dt\)

3.

Let’s consider an animal running along a straight path with the velocity function:
\begin{align*} v(t) \amp = \frac{t^4}{10} - t^3 + \frac{27t^2}{10} - \frac{9t}{5}\\ \amp = \frac{t}{10}(t-1)(t-3)(t-6) \end{align*}
on the time interval \([0,6]\text{.}\)

(a)

What is the total displacement of the animal on the time interval \([0,1]\text{?}\)
Solution.
\begin{align*} \int_{t=0}^{t=1} v(t)\;dt \amp = \int_{t=0}^{t=1} \frac{t^4}{10} - t^3 + \frac{27t^2}{10} - \frac{9t}{5} \;dt\\ \amp\left( \frac{t^5}{50} - \frac{t^4}{4}+\frac{27t^3}{30} - \frac{9t^2}{10}\right)\bigg|_{t=0}^{t=1}\\ \amp = -\frac{23}{100} \end{align*}

(b)

What is the total displacement of the animal on the time interval \([1,3]\text{?}\)
Solution.
\begin{align*} \int_{t=1}^{t=3} v(t)\;dt \amp = \int_{t=1}^{t=3} \frac{t^4}{10} - t^3 + \frac{27t^2}{10} - \frac{9t}{5} \;dt\\ \amp\left( \frac{t^5}{50} - \frac{t^4}{4}+\frac{27t^3}{30} - \frac{9t^2}{10}\right)\bigg|_{t=1}^{t=3}\\ \amp = \frac{26}{25} \end{align*}

(c)

What is the total displacement of the animal on the time interval \([3,6]\text{?}\)
Solution.
\begin{align*} \int_{t=3}^{t=6} v(t)\;dt \amp = \int_{t=3}^{t=6} \frac{t^4}{10} - t^3 + \frac{27t^2}{10} - \frac{9t}{5} \;dt\\ \amp\left( \frac{t^5}{50} - \frac{t^4}{4}+\frac{27t^3}{30} - \frac{9t^2}{10}\right)\bigg|_{t=3}^{t=6}\\ \amp = -\frac{729}{100} \end{align*}

(d)

What is the total displacement of the animal on the time interval \([0,6]\text{?}\)
Solution.
We can integrate from \(t=0\) to \(t=6\text{,}\) or just add up the previous displacements.
\begin{align*} \int_{t=0}^{t=6} v(t)\;dt \amp = \int_{t=0}^{t=6} \frac{t^4}{10} - t^3 + \frac{27t^2}{10} - \frac{9t}{5} \;dt\\ \amp\left( \frac{t^5}{50} - \frac{t^4}{4}+\frac{27t^3}{30} - \frac{9t^2}{10}\right)\bigg|_{t=0}^{t=6}\\ \amp = -\frac{162}{25} \end{align*}

(e)

What is the total distance traveled by the animal on the time interval \([0,6]\text{?}\)
Solution.
\begin{align*} \int_{t=0}^{t=6} |v(t)|\;dt \amp = \left|\int_{t=0}^{t=1} v(t)\;dt\right|\\ \amp + \left|\int_{t=1}^{t=3} v(t)\;dt\right|\\ \amp + \left|\int_{t=3}^{t=6} v(t)\;dt\right|\\ \amp = \frac{214}{25} \end{align*}

(f)

Write a short summary of the animal’s movement, including notes about direction, speed, and where the animal travels.
Solution.
From \(t=0\) to \(t=1\text{,}\) the animal is moving backwards, slowly. It speeds up until around \(t\approx 0.4\) and then slows down until it stops and turns around at \(t-1\text{.}\) From then until \(t=3\) it moves forwards. It speeds up until around \(t\approx 2.1\text{,}\) and then slows down until it stops to turn back around at \(t=3\text{.}\) Then, it speeds up quickly in the negative direction until \(t=\approx 5\) when it begins to slow down to turn around at \(t=6\text{.}\) It accelerates very quickly in the positive direction until \(t=7\text{.}\)

4.

Consider an object with velocity function \(v(t)=t^2-4t+2\) on the interval \([0,100]\) with the initial position \(s(0)=3\text{.}\)

(a)

Determine the position function, \(s(t)\text{,}\) for \(0\leq t\leq 100\) using the Future Position of an Object.
Solution.
\begin{align*} s(t) \amp = s(0) + \int_{x=0}^{x=t} v(x)\;dx \\ \amp = 3 + \int_{x=0}^{x=t} x^2-4x+2\;dx\\ \amp = 3 + \left(\frac{x^3}{3}-2x^2+2x\right)|_{x=0}^{x=t}\\ \amp =\frac{t^3}{3}-2t^2+2t+3 \amp \text{on the interval }[0,100] \end{align*}

(b)

Determine the position function, \(s(t)\text{,}\) for \(0\leq t\leq 100\) using the Solving Initial Value Problems strategy.
Solution.
We’ll find the family of antiderivatives, of which \(s(t)\) is one.
\begin{align*} \int v(t)\;dt \amp = \int t^2-4t+2 \; dt \\ \amp = \frac{t^3}{3}-2t^2+2t + C \end{align*}
Now, \(s(t)\) is the antiderivative where \(s(0)=3\text{.}\)
\begin{align*} 3 \amp \frac{0^3}{3}-2(0)^2 + 2(0)+C\\ 3 \amp = C \end{align*}
So \(s(t) =\dfrac{t^3}{3}-2t^2+2t +3\text{.}\)

(c)

Compare the results from both methods. Explain why these are equivalent.

5.

Consider an object with an acceleration function \(a(t) = t+\sin(2\pi t)\) for \(t\geq 0\) with \(v(0)=5\text{.}\)

(a)

Determine the velocity function, \(v(t)\text{,}\) for \(t\geq 0\) using the Future Position of an Object.
Solution.
\begin{align*} a(t) \amp = v(0)+\int_{x=0}^{x=t} a(x)\;dx\\ \amp = 5 + \int_{x=0}^{x=t} \left(x+\sin(2\pi x)\right)\;dx\\ \amp = 5 + \left(\frac{x^2}{2} - \frac{\cos(2\pi x)}{2\pi} \right)\bigg|_{x=0}^{x=t}\\ \amp = \frac{t^2}{2}- \frac{\cos(2\pi t)}{2\pi} + 5 + \frac{1}{2\pi} \amp \text{for } t\geq 0 \end{align*}

(b)

Determine the velocity function, \(v(t)\text{,}\) for \(t\geq 0\) using the Solving Initial Value Problems strategy.
Solution.
\begin{align*} \int a(t)\;dt \amp = \int t+\sin(2\pi t)\;dt \\ \amp = \frac{t^2}{2}- \frac{\cos(2\pi t)}{2\pi} +C \end{align*}
We know that \(v(t)\) is one of these antiderivatives, so we can use \(v(0)=5\) to solve for \(C\text{.}\)
\begin{align*} 5 \amp = \frac{0^2}{2}-\frac{\cos(0)}{2\pi} + C \\ 5 + \frac{1}{2\pi} \amp = C \end{align*}
So \(v(t)=\dfrac{t^2}{2}- \dfrac{\cos(2\pi t)}{2\pi} + 5 + \dfrac{1}{2\pi}\) for \(t\geq 0\text{.}\)

(c)

Can you obtain the position function, \(s(t)\text{?}\) Explain why or why not, based on the information given.
Solution.
No, we cannot obtain the position function \(s(t)\text{,}\) since we do not know an initial (or any) position. We can only obtain the family of possible position functions.

6.

During a brake test for a heavy truck, the truck decelerates from an initial velocity of 88 ft/s with the acceleration function \(a(t) = -17\) ft/s². Assume that the initial position of the truck is \(s(0)=0\text{.}\)

(a)

Find the velocity function for the truck.
Solution.
\begin{align*} v(t) \amp = v(0)+\int_{x=0}^{x=t} a(x)\;dx \\ \amp 88 + \int_{x=0}^{x=t} -17\;dx\\ \amp 88-17t \end{align*}

(b)

When does the truck stop? In this situation, the truck won’t have a negative velocity (since it’s just braking and not eventually going in reverse). What time interval is the velocity function relevant on?
Solution.
The truck stops when \(v(t)=0\text{:}\)
\begin{align*} 0 \amp = 88-17t \\ t \amp \frac{88}{17} \end{align*}
This means that the velocity function is only relevant on the interval \(\left[0, \frac{88}{17}\right]\text{.}\)

(c)

What is the total displacement of the truck on this time interval?
Solution.
\begin{align*} \int_{t=0}^{t=88/17} v(t)\;dt \amp = \int_{t=0}^{t=88/17} 88-17t\;dt\\ \amp = \left(88t-\frac{17t^2}{2}\right)\bigg|_{t=0}^{t=88/17}\\ \amp \frac{3872}{17} \end{align*}
So the truck moved approximately \(227.76\) feet.

(d)

Safety standards say that for a truck like this, it needs to be able to stop (from a speed of 88ft/s) in, at most, 200 feet.
Do we need to make changes to the braking mechanism, in order to have the acceleration function change? If so, what does the acceleration need to be (assuming it is constant and we are just replacing it with a new negative number)?
Solution.
Note that for an acceleration function \(a(t) = -k\) for some positive number \(k\text{,}\) the velocity function is \(v(t)=88-kt\) and the truck stops at \(t=\frac{88}{k}\text{.}\)
Now we can set the displacement integral equal to 200 in order to solve for \(k\text{.}\)
\begin{align*} 200 \amp = \int_{t=0}^{t=88/k} 88-kt\;dt \\ \amp = \left( 88t - \frac{kt^2}{2}\right)\bigg|_{t=0}^{t=88/k}\\ \amp = \frac{88^2}{k} - \frac{88^2}{2k}\\ 200 \amp = \frac{88^2}{2k} \\ 400k \amp 88^2 \\ k \amp = \frac{484}{25} = 19.36 \end{align*}
So the truck needs to have an acceleration function of \(a(t) = -19.36\) ft/s\(^2\) or more negative for it to stop in time.
PrevTopNext