Area Under a Curve using Integration

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 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 decreases at a constant rate so that 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 definite 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 car, where $t$ is in hours and $v$ is in $ \text{ km h}^{-1}$.

\( \begin{align} \displaystyle
&\text{Total distance travelled} \\
&= \text{total area under the curve} \\
&= A+B+C+D+E \\
&= \dfrac{1}{2} \times 1 \times 80 + 2 \times 80 + \dfrac{80+30}{2} \times 1 + 1 \times 30 + \dfrac{1}{2} \times 30 \times 1 \\
&= 40 + 160 + 55 + 15 \\
&= 270 \text{ km}
\end{align} \)

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 position. 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 in direction
  • Find $s(t)$ by integration, including a constant of integration
  • Find $s(t)$ at each time the direction changes.
  • Draw a motion diagram.
  • Calculate the total distance travelled from the motion diagram.

Example 2

Find the distance travelled by 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 function 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:

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

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