From a Consumption Rate to Total Fuel
A vehicle’s fuel consumption can change during a trip. The rate may depend on distance traveled, elapsed time, or conditions represented by a model. A single constant rate may therefore not describe the whole trip.
A rate model lets us estimate how much fuel is used over the trip by adding the fuel used on many small portions. In the limit, this sum is a definite integral. The essential practical choice is to match the integration variable to the denominator of the rate: a rate per kilometre is integrated over kilometres, while a rate per hour is integrated over hours.
Rates and Their Units
A consumption rate describes how quickly fuel is used. The independent variable is part of the model: it tells us whether the rate changes with distance or with time.
- A distance-based rate might be measured in litres per kilometre (L/km). If \(x\) is measured in kilometres, then \(r(x)\) has units L/km.
- A time-based rate might be measured in litres per hour (L/h). If \(t\) is measured in hours, then \(q(t)\) has units L/h.
- A rate may instead be reported in litres per 100 kilometres (L/100 km). This is a scaled distance-based unit and must be converted appropriately when it is integrated with respect to kilometres.
The units show why the matching matters. Multiplying a rate in L/km by a small distance in km gives fuel in litres. Multiplying a rate in L/h by a small time in hours also gives litres. Multiplying L/h by kilometres does not give litres.
The continuity condition ensures the rate has no unresolved jumps that would obstruct the usual definite-integral model. A model defined by separate formulas on several intervals can be integrated on each interval and the results added.
Why Integrating a Rate Gives Total Fuel
Consider a trip described by a distance-based rate \(r(x)\), measured in L/km. Divide the trip from \(a\) to \(b\) into short distance intervals. On a small interval of width \(\Delta x_i\), the rate is approximately \(r(x_i^*)\) for some point \(x_i^*\) in that interval. The fuel used there is approximately rate multiplied by distance:
Adding these estimates gives an approximation to the trip’s total fuel. As the intervals become finer, the sum approaches the definite integral. This is precisely the Riemann-sum definition of the integral.
Divide the distance interval \([a,b]\) into small pieces of widths \(\Delta x_i\).
Use the rate at a sample point multiplied by the piece’s distance: \(r(x_i^*)\Delta x_i\).
The sums \(\sum_i r(x_i^*)\Delta x_i\) approach \(\int_a^b r(x)\,dx\) as the largest piece width approaches zero.
The units follow the same reasoning. In each sum, \((\text{L/km})(\text{km})=\text{L}\), so the limiting total is measured in litres. For a time-based rate, the corresponding small quantity is \(q(t_i^*)\Delta t_i\), whose units are \((\text{L/h})(\text{h})=\text{L}\).
Worked Example: A Rate Given in Litres per Kilometre
Worked Example: A Rate Given in Litres per Kilometre
A model gives the fuel-consumption rate after \(x\) kilometres as \(r(x)=0.06+0.0002x\) L/km. Estimate the fuel used from \(x=0\) km to \(x=100\) km.
The rate is per kilometre, so integrate with respect to distance in kilometres:
An antiderivative is \(0.06x+0.0001x^2\). Evaluating at the bounds gives
The units are consistent: each rate value in L/km is multiplied through the integral by a distance in km, leaving litres. To check the antiderivative, differentiate \(0.06x+0.0001x^2\); the result is \(0.06+0.0002x\), the stated rate. Thus the model estimates 7 L over the 100 km trip.
Worked Example: A Rate Given in Litres per Hour
Worked Example: A Rate Given in Litres per Hour
During a three-hour trip, fuel use is modelled by \(q(t)=0.4+0.02t\) L/h, where \(t\) is elapsed time in hours. Find the total fuel predicted by the model.
Because the rate is per hour, use the time interval from \(0\) to \(3\):
An antiderivative is \(0.4t+0.01t^2\). Therefore,
For the check, \(\frac{d}{dt}(0.4t+0.01t^2)=0.4+0.02t\), which matches \(q(t)\). Also, \((\text{L/h})(\text{h})=\text{L}\). The model predicts total fuel use of 1.29 L over the three hours.
Rates Reported per 100 Kilometres
Fuel consumption is often reported in L/100 km. This does not mean the rate can be integrated over kilometres without adjustment. If \(c(x)\) is the numerical rate in L/100 km and \(x\) is measured in km, then the fuel used over a small distance \(\Delta x\) is approximately \(c(x)\Delta x/100\) litres. The factor of \(1/100\) converts the reported rate to fuel per kilometre.
Worked Example: A Rate Given in Litres per 100 Kilometres
Worked Example: A Rate Given in Litres per 100 Kilometres
A distance-based model gives \(c(x)=6+0.01x\) L/100 km, with \(x\) measured in kilometres. Estimate fuel use from \(0\) km to \(250\) km.
Convert the rate to L/km by dividing by \(100\), then integrate over the distance:
An antiderivative is \(0.06x+0.00005x^2\). Thus,
Differentiate the antiderivative: \(\frac{d}{dx}(0.06x+0.00005x^2)=0.06+0.0001x=(6+0.01x)/100\), as required. The conversion makes the integrand L/km, and multiplication by km leaves litres. The predicted fuel use is 18.125 L.
Choosing Bounds and Interpreting the Result
The bounds should describe the portion of the trip represented by the question. For a distance-based model, \(a\) and \(b\) are the starting and ending distance readings. They need not be zero: if a model is given in terms of the odometer reading and the trip runs from 120 km to 200 km, the distance interval is \([120,200]\), not \([0,80]\), unless the model’s input was explicitly defined as distance since the trip began.
For a time-based model, the bounds are the starting and ending times measured using the same time variable as the rate. If \(t\) is hours since departure and a trip lasts 2.5 hours, the bounds are \(0\) and \(2.5\). A time rate should not be integrated over a distance interval, even if the trip’s distance is known.
The integral gives the fuel predicted by the specified model over the specified interval. It is an estimate of actual use when the rate model is an approximation to real driving. The integral calculation itself is exact for the mathematical model, but model assumptions and measurement accuracy can affect how closely the result represents a real trip.
A Common Unit Mistake
Suppose a rate is \(6\) L/100 km and the trip is 100 km. A tempting but incorrect calculation is \(6(100)=600\) L. This treats the numerical value \(6\) as though its units were L/km. But the rate means 6 litres for each 100 kilometres, so over 100 km the fuel use is \(6\) L, not \(600\) L.
The correction is to divide by 100 before multiplying by the distance:
The same check applies when integrating: if the rate is in L/100 km and the variable is in km, include the factor \(1/100\). More generally, write down the units of the rate and differential before evaluating the integral. Their product should have units of fuel.
Check Your Understanding
For each question, identify the correct variable and keep the rate units visible as you set up the calculation.
- A model is \(r(x)=0.05+0.0001x\) L/km for \(0\le x\le 200\). What definite integral gives the total fuel use?
- A fuel-use rate is \(q(t)=0.3+0.04t\) L/h for a 2-hour interval. What units does \(\int_0^2q(t)\,dt\) have?
- A rate is reported as \(8\) L/100 km over a 150 km trip. What conversion factor should be used before integrating with respect to kilometres?
- Why is it not consistent to integrate a rate in L/h with respect to distance in km without an additional relationship between time and distance?
- What derivative check can be used after evaluating a fuel-rate integral using an antiderivative?
