Why a Changing Rate Requires a Careful Total
A rate may follow one rule during one part of an observation and a different rule later. A machine might produce at a steady rate during startup, increase its output once it reaches normal operation, and then slow down near shutdown. To find the total change over the full observation, integrate the rate across all the relevant time intervals.
The key is to match each part of the integral with the formula that applies there. A single formula cannot be used across a point where the rate’s definition changes. Instead, split the interval at every change point, evaluate each contribution, and add the contributions with their signs intact.
Piecewise Rates and Their Change Points
A piecewise-defined rate is a function specified by different formulas on different parts of its domain. For example, a rate \(r(t)\) could be given by one formula for \(0\le t<2\), another for \(2\le t<4\), and a third for \(4\le t\le 5\). The values \(2\) and \(4\) are the change points or breakpoints of the formula.
On each interval, use only the formula assigned to that interval. The choice of which formula applies at the breakpoint itself is normally stated with inequalities such as \(t<2\) or \(t\ge 2\). For an ordinary definite integral of a piecewise-continuous rate, changing the value at finitely many individual endpoints does not change the integral. The interval limits still matter: they determine which formula to use over each stretch of positive length.
The integral in this formula is signed. If the rate is negative on an interval, its integral over that interval is negative and reduces the net total. When rates have units, the rate’s units multiplied by the time units give the units of each integral contribution. Every term in the sum must therefore represent the same kind of quantity before the terms are combined.
Why Splitting the Integral Works
The justification is the additivity property of the definite integral. If \(f\) is integrable on \([a,b]\) and \(c\) lies between \(a\) and \(b\), then $$ \int_a^b f(t)\,dt=\int_a^c f(t)\,dt+\int_c^b f(t)\,dt. $$ Applying this property repeatedly splits an interval at any finite list of interior points.
For a piecewise rate, the function \(r\) is the same rate over the whole interval, but its formula is different on each piece. On \([a,c_1]\), evaluate the integral using the first formula; on \([c_1,c_2]\), use the second; continue until the last piece. Additivity then says that these separate integrals sum to the integral over the full interval. The split is not an approximation: it is an exact rewriting of the total.
This result requires the rate to be integrable on the intervals being used. Piecewise-continuous rates with finitely many formula changes on a bounded interval satisfy this condition. In an actual calculation, the practical task is to identify the relevant pieces and make sure no change point inside the requested interval is skipped.
A Reliable Calculation Process
Write down its starting and ending times, and identify which portions of the piecewise rate lie inside it.
List the endpoints in increasing order so that each subinterval uses just one formula.
Keep the original subinterval bounds attached to the corresponding formula.
Add positive and negative results using their signs, then check the units and arithmetic.
Worked Examples
Worked Example: A Production Rate with Three Phases
A production rate \(r(t)\), measured in liters per hour, is $$ r(t)= \begin{cases} 3t, & 0\le t<2,\\ 6, & 2\le t<4,\\ -2, & 4\le t\le 5. \end{cases} $$ Find the net amount produced from \(t=0\) to \(t=5\).
The formula changes at \(t=2\) and \(t=4\), so split the total into three integrals: $$ \int_0^5 r(t)\,dt = \int_0^2 3t\,dt+\int_2^4 6\,dt+\int_4^5(-2)\,dt. $$
For the first interval, an antiderivative of \(3t\) is \(\frac{3}{2}t^2\): $$ \int_0^2 3t\,dt = \left[\frac{3}{2}t^2\right]_0^2 = \frac{3}{2}(2^2)-\frac{3}{2}(0^2) = 6-0 = 6\text{ L}. $$
For the second interval, an antiderivative of \(6\) is \(6t\): $$ \int_2^4 6\,dt = [6t]_2^4 = 6(4)-6(2) = 24-12 = 12\text{ L}. $$
For the third interval, an antiderivative of \(-2\) is \(-2t\): $$ \int_4^5(-2)\,dt = [-2t]_4^5 = (-2)(5)-(-2)(4) = -10+8 = -2\text{ L}. $$
Combine the signed contributions: $$ 6\text{ L}+12\text{ L}+(-2\text{ L})=16\text{ L}. $$ The negative rate contributes \(-2\) liters to the net total, so the net amount produced is \(16\) liters.
Check: Differentiating the antiderivatives gives \(\frac{d}{dt}(\frac{3}{2}t^2)=3t\), \(\frac{d}{dt}(6t)=6\), and \(\frac{d}{dt}(-2t)=-2\). The intervals cover \(0\) through \(5\) without a gap, and all contributions have units of liters.
Worked Example: A Rate That Becomes Negative
A quantity changes at the rate \(v(t)\), measured in meters per minute: $$ v(t)= \begin{cases} 5, & 0\le t<2,\\ 8-t, & 2\le t<5,\\ -1, & 5\le t\le 6. \end{cases} $$ Find the net change in the quantity from \(t=0\) to \(t=6\).
The rate changes formula at \(2\) and \(5\), so $$ \int_0^6v(t)\,dt = \int_0^2 5\,dt+\int_2^5(8-t)\,dt+\int_5^6(-1)\,dt. $$
The first contribution is $$ \int_0^2 5\,dt=[5t]_0^2=5(2)-5(0)=10\text{ m}. $$ For the middle piece, an antiderivative of \(8-t\) is \(8t-\frac{t^2}{2}\): $$ \begin{aligned} \int_2^5(8-t)\,dt &=\left[8t-\frac{t^2}{2}\right]_2^5\\ &=\left(8(5)-\frac{5^2}{2}\right)-\left(8(2)-\frac{2^2}{2}\right)\\ &=\left(40-\frac{25}{2}\right)-(16-2)\\ &=\frac{55}{2}-14 =\frac{27}{2}\text{ m}. \end{aligned} $$ The final contribution is $$ \int_5^6(-1)\,dt=[-t]_5^6=-6-(-5)=-1\text{ m}. $$
Therefore the net change is $$ 10+\frac{27}{2}-1 = 9+\frac{27}{2} = \frac{18}{2}+\frac{27}{2} = \frac{45}{2}\text{ m} = 22.5\text{ m}. $$
Check: The derivatives of the antiderivatives \(5t\), \(8t-\frac{t^2}{2}\), and \(-t\) are \(5\), \(8-t\), and \(-1\), respectively. The interval contributions sum to \(10+13.5-1=22.5\) meters, matching the exact value \(\frac{45}{2}\) meters.
Worked Example: Starting and Ending Inside Formula Intervals
A rate \(q(t)\), in units per hour, is defined by $$ q(t)= \begin{cases} t+1, & 0\le t<1,\\ 3, & 1\le t<3,\\ 2-t, & 3\le t\le 4. \end{cases} $$ Find the net accumulated quantity from \(t=\frac12\) to \(t=\frac72\).
The requested interval begins inside the first formula’s interval and ends inside the third. The relevant split points are \(\frac12\), \(1\), \(3\), and \(\frac72\). Thus $$ \int_{1/2}^{7/2}q(t)\,dt = \int_{1/2}^{1}(t+1)\,dt+\int_1^3 3\,dt+\int_3^{7/2}(2-t)\,dt. $$
For the first term, an antiderivative of \(t+1\) is \(\frac{t^2}{2}+t\): $$ \begin{aligned} \int_{1/2}^{1}(t+1)\,dt &=\left[\frac{t^2}{2}+t\right]_{1/2}^{1}\\ &=\left(\frac{1}{2}+1\right)-\left(\frac{(1/2)^2}{2}+\frac12\right)\\ &=\frac32-\left(\frac18+\frac12\right) =\frac32-\frac58 =\frac78. \end{aligned} $$ The second term is $$ \int_1^3 3\,dt=[3t]_1^3=3(3)-3(1)=9-3=6. $$ For the third term, an antiderivative of \(2-t\) is \(2t-\frac{t^2}{2}\): $$ \begin{aligned} \int_3^{7/2}(2-t)\,dt &=\left[2t-\frac{t^2}{2}\right]_3^{7/2}\\ &=\left(7-\frac{49}{8}\right)-\left(6-\frac{9}{2}\right)\\ &=\frac{7}{8}-\frac{3}{2} =\frac{7}{8}-\frac{12}{8} =-\frac{5}{8}. \end{aligned} $$
Combining the pieces gives $$ \frac78+6-\frac58 = \frac28+6 = \frac14+6 = \frac{25}{4}. $$ The net accumulated quantity is \(\frac{25}{4}\) units.
Check: Differentiating the three antiderivatives gives \(t+1\), \(3\), and \(2-t\), the stated rate formulas. The interval lengths are \(\frac12\), \(2\), and \(\frac12\), adding to \(3=\frac72-\frac12\), so the split covers exactly the requested interval.
Common Mistake: Using One Formula Across a Change Point
Suppose a rate is \(r(t)=t\) for \(0\le t<1\) and \(r(t)=4\) for \(1\le t\le2\). The total is not \(\int_0^2 t\,dt\), because \(t\) is not the rate formula on the whole interval. That incorrect calculation gives \(\left[\frac{t^2}{2}\right]_0^2=2\).
The correct calculation splits at \(t=1\): $$ \int_0^2r(t)\,dt = \int_0^1t\,dt+\int_1^2 4\,dt = \left[\frac{t^2}{2}\right]_0^1+[4t]_1^2 = \frac12+(8-4) = \frac92. $$ The mistake was extending the first formula past the point where it applies. The correction is to identify every change point before integrating.
Combining Contributions Without Losing Their Signs
After evaluating each interval, retain the sign of every result. A negative integral is not an arithmetic error to erase; it represents a negative contribution to the net change. For example, a sum such as \(7+(-3)+2\) equals \(6\), not \(12\). In rate problems, the direction of the rate and the requested quantity determine what the sign means, so follow the stated quantity carefully.
A useful final check is to ask whether the computed total is plausible in sign and size. Confirm that every subinterval is included exactly once, that each integral uses the correct formula, and that the rate-times-time units match the units of the accumulated quantity. When evaluating an integral with an antiderivative, also verify that differentiating the antiderivative recovers the integrand used on that interval.
Check Your Understanding
For each question, identify the intervals and formulas before combining any contributions.
- A rate is \(2t\) for \(0\le t<1\) and \(3\) for \(1\le t\le3\). Write, but do not evaluate, the integral for its total from \(t=0\) to \(t=3\).
- A rate is \(4\) on \([0,2)\) and \(-2\) on \([2,5]\). What signed contribution does the second interval make to the net total?
- A rate has formula changes at \(t=1\) and \(t=4\). If the requested interval is \([2,5]\), which formulas and subintervals are needed?
- Explain why a formula valid on \([0,1)\) cannot automatically be used to evaluate the integral over \([0,2]\) when the rate has a new formula from \(1\) onward.
- If a rate is measured in kilograms per minute and time in minutes, what units should each interval integral have?
