Relating velocity, displacement, antiderivatives and areas

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We give an alternative interpretation of the definite integral and make a connection between areas and antiderivatives.

We have a geometric interpretation of the derivative as the slope of a tangent line at a point. We have not yet found a geometric interpretation of antiderivatives.

1 More than one perspective

We’ll start with a question:

Suppose you are in slow traffic moving at $4$ mph from $2$pm to $5$pm. How far have you traveled?

\[ \text {displacement}= \answer {12}\ \text {miles}. \]

Since the (constant) velocity over a time interval is given by

\[ \text {velocity}=\frac {\text {displacement}}{\text {elapsed time}}, \]

it follows that the displacement of the object over an interval of time is given by

\[ \text {displacement}=\text {velocity}\cdot (\text {elapsed time}), \]

The graph of the (constant) velocity function $v$ on the interval $[2,5]$ is given in the figure.

So, the displacement of an object is given by the product of the velocity and elapsed time, and is represented by the area of the shaded rectangle in the figure.

Since the displacement of an object moving at constant velocity over a time interval can be represented by the area of a rectangle, we can approximate displacement of an object moving at nonconstant velocity using Riemann sums. We explain this process in detail in our next example.

Assume an object is moving along a straight line with the velocity $v(t)=\frac {t^2}{4}+1$, for $0\le t\le 4$, where $t$ is measured in seconds, and $v$ in $\frac {\mathrm {ft}}{\mathrm {s}}$.

Find the displacement of the object over the time interval $[0,4]$.

Since we know how to compute the displacement when the velocity is constant, we will start by asuming that the velocity is constant and equal to $v(0)$, the initial velocity, for the entire four-second time interval, $\Delta t=4$.

This scenario will give us a very rough approximation of the displacement, but it is a good starting point.

\[ \text {displacement}\approx v(0)\cdot \Delta t=1\cdot (4-0)=4 \mathrm {ft}/\mathrm {s} \]

This scenario is illustrated in the figure below.

Now it should be clear how to proceed. We will keep approximating the velocity with a constant over smaller and smaller intervals of time.

In our next step we will assume that the velocity is constant during two-second time intervals $[0,2]$ and $[2,4]$. For each of these two intervals, we assume that the velocity is equal to the velocity at the beginning of the time interval.

In this case, the object would move by $v(0)\cdot 2 \mathrm {ft}$ during the time interval $[0,2]$, and during the next two seconds, the object would move by $v(2)\cdot 2 \mathrm {ft}$.

If we let $\Delta t$ denote the length of the time interval, we can approximate the displacement and write

\[ \text {displacement}\approx v(0)\cdot \Delta t+v(2)\cdot \Delta t=1\cdot 2+2\cdot 2=\answer [given]{6} \mathrm {ft}/\mathrm {s} \]

Using sigma notation, we write

\[ \text {displacement}\approx \sum _{k=1}^2v((k-1)\cdot \Delta t)\Delta t \]

Since we evaluate the velocity at the sample points $t_{k}^*=(k-1)\cdot \Delta t$, $k=1,2$, we can also write

\[ \text {displacement}\approx \sum _{k=1}^2v(t_{k}^*)\Delta t. \]

This is a left Riemann sum for the function $v$ on the interval $[0,4]$, when $n=2$. This scenario is illustrated in the figure below.

Next, we approximate the displacement of the object assuming that the velocity is constant during one-second time intervals, and that velocity is equal to the velocity at the beginning of each time interval.

Under this assumption, during the first second, the object would move by $v(0)\cdot 1 \mathrm {ft}$; during the second time interval of one second, the object would move by $v(1)\cdot 1 \mathrm {ft}$, during the third second, the object would move by $v(2)\cdot 1 \mathrm {ft}$, and during the last second, it would move by $v(3)\cdot 1 \mathrm {ft}$.

During the entire time interval, the object would move by \begin{align*} \text {displacement} &\approx v(0)\cdot \Delta t+v(1)\cdot \Delta t+v(2)\cdot \Delta t+v(3)\cdot \Delta t\\ &=\frac {\answer [given]{30}}{4} \mathrm {ft}/\mathrm {s} \end{align*}

Using sigma notation and the fact that $\Delta t=\frac {4-0}{4}=1$, we can write

\[ \text {displacement}\approx \sum _{k=1}^4v((k-1)\cdot \Delta t)\Delta t \]

Again, we denote the sample points by $t_{k}^*=(k-1)\cdot \Delta t$, $k=1,2,3,4$, and write

\[ \text {displacement}\approx \sum _{k=1}^4v(t_{k}^*)\Delta t \]

This is a left Riemann sum for the function $v$ on the interval $[0,4]$, when $n=4$. This scenario is illustrated in the figure below.

Now, we approximate the displacement of the object assuming that the velocity is constant during half-second time intervals, and that velocity is equal to the velocity at the beginning of each time interval.

Under this assumption, during the first half-second, the object would move by $v(0)\cdot \frac {1}{2} \mathrm {ft}$; during the second time interval of half a second, the object would move by $v\left (\frac {1}{2}\right )\cdot \frac {1}{2} \mathrm {ft}$, during the third half- second interval, the object would move by $v(1)\cdot \frac {1}{2} \mathrm {ft}$,etc.

During the entire time interval, the object would move by \begin{align*} \text {displacement} &\approx v(0)\Delta t+v\left (\frac {1}{2}\right )\Delta t+ \cdots +v\left (\frac {7}{2}\right )\Delta t\\ &=\answer [given]{8.375} \mathrm {ft} \end{align*}

Using sigma notation and the fact that $\Delta t=\frac {4-0}{8}=\frac {1}{2}$, we can write

\[ \text {displacement}\approx \sum _{k=1}^8v((k-1)\cdot \Delta t)\Delta t \]

Using the standard notation for sample points, we can write

\[ \text {displacement}\approx \sum _{k=1}^8v(t_{k}^*)\Delta t \]

This is a left Riemann sum for the function $v$ on the interval $[0,4]$,when $n=8$. This scenario is illustrated in the figure below.

We pause here to illustrate the $k$th rectangle when computing a left Riemann sum of the function $v$ on the interval $[0,4]$.

We continue by approximating the displacement of the object assuming that the velocity is constant during $4/n$-second time intervals, with velocity equal to the velocity at the beginning of each time interval.

Under this assumption, during the first $\frac {4}{n}$-second interval, the object would move by $v(0)\cdot \frac {4}{n} \mathrm {ft}$; during the second such time interval, the object would move by $v\left (\frac {4}{n}\right )\cdot \frac {4}{n} \mathrm {ft}$, during the third time interval, the object would move by

$v\left (2\frac {4}{n}\right )\cdot \frac {4}{n} \mathrm {ft}$, etc.

During the entire time interval, the object would move by \begin{align*} \text {displacement} \approx v(0)\cdot \Delta t &+v\left (\frac {4}{n}\right )\cdot \Delta t+v\left (2\frac {4}{n}\right )\cdot \Delta t+\cdots \\ &+v\left ((n-1)\frac {4}{n}\right )\cdot \Delta t \end{align*}

Using sigma notation, notation for the length of each time interval, $\Delta t=\frac {4-0}{n}=\frac {4}{n}$, and notation for sample pints, we write

\[ \text {displacement}\approx \sum _{k=1}^nv((k-1)\cdot \Delta t)\Delta t \]

and

\[ \text {displacement}\approx \sum _{k=1}^nv(t_{k}^*)\Delta t. \]

This is a left Riemann sum for the function $v$ on the interval $[0,4]$.

What happens when we let $n\to \infty $?

Approximations get better and better, because we approximate the velocity with a constant over smaller and smaller time intervals. Therefore, it seems reasonable to write

\[ \text {displacement}=\lim _{n\to \infty }\sum _{k=1}^nv(t_k^*)\Delta t \]

By the definition of the definite integral, we can write

\[ \text {displacement}=\int _{0}^{4}v(t) dt. \]

This gives us another interpretation of the definite integral. Namely,

“definite integral of the velocity over $[a,b]$= the displacement over $[a,b]$.”

On the other hand, the displacement of an object during the time interval $[0,4]$ is given by

\[ \text {displacement}=\text {change in position} \]

The change in position can be written in terms of $s$, the position function of the object

\[ \text {displacement}=s(4)-s(0). \]

But, we know that $s$ is an antiderivative of $v$. Therefore,

\[ s(t)=\int v(t) dt=\int \left (\frac {t^2}{4}+1\right ) dt=\frac {t^3}{\answer [given]{12}}+\answer [given]{t}+C \]

If $t=0$, we get

\[ s(0)=C \]

So, the position function is given by

\[ s(t)=\frac {t^3}{\answer [given]{12}}+\answer [given]{t}+s(0). \]

Now we can compute the displacement

\[ \text {displacement}=s(4)-s(0)=\frac {\answer [given]{28}}{3}. \]

Much more important than this computation is the fact that we were able to express the displacement as a definite integral and compute the definite integral of $v$ using the antiderivative of $v$.

\[ \int _{0}^{4}v(t) dt=s(4)-s(0) \]

Is this always so?

Is it true that for any continuous function $f$ and its antiderivative $F$, we can compute the definite integral of $f$ on the interval $[a,b]$ by this simple formula

\[ \int _{a}^{b}f(t) dt=F(b)-F(a)? \]

2025-01-06 18:52:30