Applications of Integration

Area Under/Between Curves (done previously)

Rectillinear Motion #2

Remember previously the anti-differentiation rules for rectillinear motion:

Well, now knowing about anti-differentiation, we can put together a few more rules:

Moreso, something special happens when we use definite integrals for the above formulas, as the constant of integration is negated, we can easily find the change in displacement: (\(s\)), velocity: (\(v\)), and acceleration: (\(a\)), by integrating their derivative functions over a range of \(x\). This can take various forms of questions:

1) A particle is accelerating such that \(a(t) = 3t\). Given the particle starts at the origin, what is the change of velocity between \(t=2\) and \(t=5\)?

\(\displaystyle \begin{aligned} v(t) &= \int_{2}^{5} 4t dt \\[5pt] &= \left[2t^2 \right]_{2}^{5} \\[5pt] &= 2 \cdot 25 - 2 \cdot 4 \\[5pt] &= 50 - 8 \\[5pt] &= 42 \ ms^{-1} \end{aligned} \)

2) Given a car starts 2 metres behind the start line, and the car is moving at velocity, \(v(t) = 6t^2+2t\), where would the car be after 5 seconds.

\(s(0) = -2\)

\(\displaystyle \begin{aligned} s(t) &= \int 6t^2 + 2t \ dt \\[5pt] &= 2t^3 + t^2 + c \\[15pt] -2 &= 2(0)^3 + (0)^2 + c \\[5pt] c &= -2 \\[15pt] s(t) &= 2t^3 + t^2 - 2 \\[5pt] s(15) &= 2(5)^3 + (5)^2 -2 \\[5pt] &= 2 \times 125 + 25 -2 \\[5pt] &= 250 + 23 \\[5pt] &= 273 \ \text{metres} \end{aligned} \)

3) Given a bodys acceleration is expressed by \(v(t) = 4t\), what is the average velocity between \(t=5,7\) if the bodys starts at the origin.

Note that the question is asking for the 'average' velocity. This means we must ultimately divide a change of \(s\) by a change of \(t\).

\(s(0) = 0\)

\(\displaystyle \begin{aligned} s(t) &= \int 4t \ dt \\[5pt] &= 2t^2 + c \\[5pt] 0 &= 0 + c \\[5pt] c &= 0 \\[15pt] s(t) &= 2t^2 \\[5pt] s(5) &= 50 \ \ \ \ \ \ \ s(7) = 98 \\[15pt] Avg. Velocity &= \frac{\Delta s}{\Delta t} \\[5pt] \frac{\Delta s}{\Delta t} &= \frac{98-50}{7-5} \\[5pt] &= \frac{48}{2} \\[5pt] &= 24 \ \text{ms}^{-1} \end{aligned} \)

Calculating Total Change

A question may ask you to calculate the change of a certain metric using derivitive function. This itself may not be involved with rectillinear motion:

4) A bucket can hold a maximum volume of 65 litres. Given a pipe adds water to the bucket at a rate of \(\frac{dV}{dt} = {(2-2t)^2}\) litres per second, how long until the bucket is full?

Let's say this question allows use of the Classpad Graphical Calculator. You can do this without, but unless it's like 7 marks it won't be in calc-free.

\(\begin{aligned} \text{|CP| Solve(} \int_{0}^{y} (2-2t)^2 \ dy &= 65 \text{,} \ y \text{)} \\[5pt] y &= 4.6279 \\[5pt] &= 4.63 \ \text{seconds} \end{aligned} \)

Hence, it will take \(4.63\) seconds to fill the bucket

5) Ozone (\(O_3\)) is a naturally gaseous molecule within the atmosphere that shields us from ultra violet radiation that causes skin cancers. A group of scients run experiments on the stability of ozone. To do this, they construct a sealed, steel vat which contains 1600 Litres of ozone. A scientist opens the vat such that gas escapes at \(\frac{dV}{dt} = x^3 + 5 \ dt \). How much gas escaped between \(t = 3\), \(t = 4\).

Some questions will give you a scenario. The question is still pretty simple. Assuming use of calculators:

\(\begin{aligned} V &= \int_{2}^{3} x^3 + 5 \ dt \\[5pt] &= 61.25 \ \text{litres} \end{aligned} \)

61.25 litres of ozone will leave the vat between \(t = 3\) and \(t = 4\).