# Area Under a Curve using Integration

## Distances from Velocity Graphs

Suppose a car travels at a constant positive velocity $80 \text{ km h}^{-1}$ for $2$ hours.

\begin{align} \displaystyle \text{distance travelled} &= \text{speed} \times \text{time} \\ &= 80 \text{ km h}^{-1} \times 2 \text{ h} \\ &= 160 \text{ km} \end{align}
We we sketch the graph velocity against time, the graph is a horizontal line, and we can see that the distance travelled is the area shaded in green.
Therefore the distance travelled is found by the definite integral:
\begin{align} \displaystyle \int_{0}^{2}{80}dt &= \big[80t\big]_{0}^{2} \\ &= 80 \times 2 – 80 \times 0 \\ &= 160 \text{ km} \end{align}
Now the speed decreased at a constant rate so taht the car, initially travelling at $80 \text{ km h}^{-1}$, stops in 2 hours. In this case, the average speed is $40 \text{ km h}^{-1}$.
\begin{align} \displaystyle \text{distance travelled} &= \text{speed} \times \text{time} \\ &= 40 \text{ km h}^{-1} \times 2 \text{ h} \\ &= 80 \text{ km} \end{align}

\begin{align} \displaystyle \text{area of the triangle} &= \dfrac{1}{2} \times \text{based} \times \text{height} \\ &= \dfrac{1}{2} \times 2 \times 80 \\ &= 80 \text{ km} \end{align}
Again, the area shaded in green is the distance travelled, and we can find it using the difinite integral:
\displaystyle \begin{align} \int_{0}^{2}{(80-40t)}dt &= \big[80t-20t^2\big]_{0}^{2} \\ &= (80 \times 2 – 20 \times 2^2) – (80 \times 0 – 20 \times 0^2) \\ &= 80 \text{ km} \end{align}
$$\displaystyle \therefore \text{distance travelled} = \int_{t_1}^{t_2}{v(t)}dt$$

### Example 1

The velocity-time graph for a car journey is shown in the graph. Find the total distance travelled by the car, where $t$ is in hours and $v$ is in $\text{ km h}^{-1}$.

## Displacement and Velocity Functions

For some displacement $s(t)$, the velocity function is $v(t)=s'(t)$.
So, given a velocity function it is determined that the displacement function by the integral:
$$s(t) = \int{v(t)}dt$$
The constant of integration determines where on the line the object begins, called the initial potition. Using the displacement function we can determine the change in displacement in a time interval $t_1 \le t \le t_2$.
\displaystyle \begin{align} \text{Displacement} &= s(t_2) – s(t_1) \\ &= \int_{t_1}^{t_2}{v(t)}dt \end{align}
To find the total distance travelled given a velocity function $v(t)=s'(t)$ on $t_1 \le t \le t_2$:

• Draw an accurate sign diagram for $v(t)$ to determine any changes of direction
• Find $s(t)$ by integration, including a constant of integration
• Find $s(t)$ at each time the direction chagnes.
• Draw a motion diagram.
• Calculate the total distance travelled from the motion diagram.

### Example 2

Find the distance travelled by a car with a velocity $v(t) = 2t+5$, where $t$ is in hours and $v$ is in $\text{km h}^{-1}$ for the first $5$ hours.

\begin{align} \displaystyle \text{distance travelled} &= \int_{0}^{5}{(2t+5)}dt \\ &= \big[t^2+5t\big]_{0}^{5} \\ &= (5^2 + 5 \times 5) – (0^2 + 5 \times 0) \\ &= 50 \text{ km} \end{align}

## Displacement and Velocity Functions

The acceleration functin is the derivative of the velocity function, so $a(t) = v'(t)$. Given an acceleration function, we can determine the velocity function by integration.:
$$v(t) = \int{a(t)}dt$$

### Example 3

A particle moves in a straight line with velocity $v(t)=6t^2-18t+12 \text{ m s}^{-1}$. Find the total distance travelled that the particle moves in the first $3$ seconds of motion.

\begin{align} \displaystyle v(t) &= 0 \\ (t-1)(t-2) &= 0 \\ t &= 1 \text{ and } t=2 \\ x(t) &= \int{(6t^2-18t+12)}dt \\ &= 2t^2 -9t^2 + 12t +c \\ x(0) &= c \\ x(1) &= 5+c \\ x(2) &= 4+c \\ x(3) &= 9+c \\ \end{align}
Since the signs change, the particle reverses direction at $t=1$ and $t=2$ seconds.
Sign Diagram:

\begin{array}{|c|c|c|} \hline
\text{time} & \text{movement} & \text{distance travelled} \\ \hline
t=0 & \text{ moving positive direction} & 0 \\ \hline
0 \lt t \lt 1 & \text{moving from } x=c \text{ to } x=5+c & 5 \\ \hline
t=1 & \text{ the particle changes its direction} & 0 \\ \hline
1 \lt t \lt 2 & \text{moving from } x=5+c \text{ to } x=4+c & 1 \\ \hline
t=2 & \text{ the particle changes its direction} & 0\\ \hline
2 \lt t \lt 3 & \text{moving from } x=4+c \text{ to } x=9+c & 5 \\ \hline
t=3 & x=9+c & 0 \\ \hline
\end{array}
Motion Graph:

\begin{align} \displaystyle \text{total distance travelled} &= 5+1+5 \\ &= 11 \text{ m} \end{align}