Rates of Change with Integration
Rates of change with integration recover a total quantity from its rate, since \(\displaystyle\int_a^b \dfrac{dQ}{dt}\,dt=Q(b)-Q(a)\).
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Applications of calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
If you know the rate of change, integrate to recover the quantity. This Year 12 Mathematics Advanced topic (MAV-12-06) uses an initial condition to fix the constant.
Given \(\dfrac{dQ}{dt}\), integrate: \(Q=\int\dfrac{dQ}{dt}\,dt+C\). Use an initial condition to find \(C\). The total change over \([a,b]\) is \(\int_a^b\dfrac{dQ}{dt}\,dt\).
Method
- Integrate the rate.
- Find \(C\) from an initial condition.
- For a total change, use a definite integral of the rate.
| \(Q\) | \(=\) | \(3t^{2}+C\) |
| \(C\) | \(=\) | \(5\) |
| \(Q\) | \(=\) | \(3t^{2}+5\) |
| \(x\) | \(=\) | \(t^{2}+C\) |
| \(x\) | \(=\) | \(t^{2}\) |
| \(\int_0^3 4t\,dt\) | \(=\) | \(\big[2t^{2}\big]_0^3=18\ \text{L}\) |
| \(Q\) | \(=\) | \(e^{t}+C\) |
| \(Q\) | \(=\) | \(e^{t}\) |
Common pitfalls
Frequently asked questions
How do you find a quantity from its rate of change?
Integrate the rate with respect to time and add C, then use a known value to find C.
What is the total change over an interval?
It is the definite integral of the rate over that interval, which equals the area under the rate graph.
Why do you need an initial condition?
To find the constant of integration and pin down the particular function.
What is the area under a rate-time graph?
It is the total change in the quantity over that time.