Integrate a rate to find change
A rate of change describes how a quantity changes per unit of time. Integrating it over an interval accumulates the signed change and recovers the quantity units
\[\Delta Q=\int_a^b R(t)\,dt=Q(b)-Q(a)\]
- Integrate over the full interval Use the lower and upper times in a definite integral of the rate \(\displaystyle \Delta Q=\int_a^b R(t)\,dt\)
- Keep the sign A negative rate contributes a decrease, so do not replace a signed contribution by its absolute value when finding net change \(\displaystyle \int_a^b v(t)\,dt=\Delta x\)
- Accumulate the units Multiplying a rate by a time interval leaves the units of the changing quantity \(\displaystyle \left(\text{litres per minute}\right)\times\text{minutes}=\text{litres}\)
- Split distance at direction changes For distance travelled, split the velocity integral where the velocity changes sign and add the magnitudes \(\displaystyle D=\left|\int_a^c v(t)\,dt\right|+\left|\int_c^b v(t)\,dt\right|\)
Worked example
The rate of change of a quantity is given below. Find the net change over the interval
\[R(t)=3t-2\ \text{litres per minute},\quad 0\le t\le2\]
- \(\displaystyle \int_0^2(3t-2)\,dt=\left[\frac{3}{2}t^{2}-2t\right]_0^2\)Integrate the rate over the stated time interval
- \(\displaystyle \left(\frac{3}{2}\times4-4\right)-0=2\)Evaluate the primitive at the upper and lower limits to obtain the signed change
- \(\displaystyle \Delta Q=2\ \text{litres}\)The net change has the units of the quantity, so it is 2 litres
Common mistakes
Evaluating instead of integrating
✕\(\displaystyle \Delta Q=R(4)=6\ \text{litres per minute}\)✓\(\displaystyle \Delta Q=\int_0^4(t+2)\,dt=16\ \text{litres}\)
The integral accumulates the rate throughout the interval, whereas R(4) gives the rate at one instantKeeping rate units after integration
✕\(\displaystyle \int_0^4(t+2)\,dt=16\ \text{litres per minute}\)✓\(\displaystyle \int_0^4(t+2)\,dt=16\ \text{litres}\)
Multiplying a rate by a time interval removes the per-minute part and leaves quantity unitsAdding the initial quantity to a change
✕\(\displaystyle Q(0)=80\ \text{litres},\quad\Delta Q=80+\int_0^2(t+1)\,dt=84\ \text{litres}\)✓\(\displaystyle Q(0)=80\ \text{litres},\quad\Delta Q=\int_0^2(t+1)\,dt=4\ \text{litres}\)
The integral is the change itself; add the initial quantity only to find the final quantityDifficulty levels
introFind the net change from a constant or linear rate over a stated interval
standardFind net change or displacement from a polynomial rate and interpret the result in context
challengeFind change from a transcendental rate or recover a quantity in general units
examInterpret a multi-part rate problem, including the difference between change, final amount and distance
Questions about this topic
What is the difference between net change and final quantity?+
The definite integral of a rate gives the net change. Add the initial quantity to that change only when finding the final quantity
Why can net change be negative?+
A negative rate contributes a decrease, and the signed integral records the total increase or decrease
How is distance different from displacement?+
Displacement is the signed integral of velocity. Distance is the sum of the magnitudes of the signed contributions after splitting at direction changes
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration
In this chapter
Approximate area by an endpoint or trapezium ruleEvaluate and transform a definite integralElementary primitives and exact definite integralsFind complete families of polynomial anti-derivativesThe area function and the fundamental theorem of calculusSigned integrals, geometric area and motionFind the area between two curvesAverage value of a functionTotal change and kinematics from a rateDetermine a polynomial from its derivative and a boundary conditionRecover motion from acceleration