This scenario will give us a very rough approximation of the displacement, but it is a good starting point. $$\text {displacement}\approx v(0)\cdot \Delta t=1\cdot (4-0)=4 \unit {ft}/\unit {s}$$ This scenario is illustrated in the figure below.

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

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

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

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

During the entire time interval, the object would move by

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

During the entire time interval, the object would move by

$$\text {displacement}\approx \sum _{k=1}^8v((k-1)\cdot \Delta t)\Delta t$$ 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 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 $v(0)\cdot \frac {4}{n} \unit {ft}$; during the second such time interval, the object would move by $v\left (\frac {4}{n}\right )\cdot \frac {4}{n} \unit {ft}$, during the third time interval, the object would move by

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

During the entire time interval, the object would move by

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

$$\text {displacement}=\int _{0}^{4}v(t)\d t.$$

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, $$s(t)=\int v(t)\d t=\int \left (\frac {t^2}{4}+1\right )\d t=\frac {t^3}{\answer [given]{12}}+\answer [given]{t}+C$$ 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 $$\text {displacement}=s(4)-s(0)=\frac {\answer [given]{28}}{3}.$$