From signed rate to accumulated magnitude
A rate can be positive during one part of an interval and negative during another. Its signed accumulation counts positive and negative contributions with their signs. But sometimes the quantity of interest is the total magnitude accumulated: every contribution counts positively, regardless of whether the rate is above or below zero.
For example, if a rate describes flow into or out of a tank, the signed accumulation records the net change in volume. Accumulated magnitude instead measures the total amount of flow in either direction. To calculate it, we integrate the absolute value of the rate.
The absolute value changes a negative rate into a nonnegative one, so the integral counts magnitude rather than signed change. If \(r\) is measured in litres per minute and \(t\) in minutes, for instance, then \(M(x)\) is measured in litres.
Why the zeros matter
On an interval where \(r(t)\ge 0\), we have \(|r(t)|=r(t)\). On an interval where \(r(t)\le 0\), we have \(|r(t)|=-r(t)\). Thus the task is to identify intervals on which the rate has one sign. Zeros are the natural places to split the interval, because a continuous rate cannot change sign without passing through zero.
Once the interval is split this way, each piece can be evaluated using the signed rate or its negative, as appropriate. For piecewise-linear rates, the integral on each piece is also the area of a trapezoid or triangle, as in the earlier material on piecewise-linear rate graphs.
A general splitting result
The following result states exactly why splitting at zeros works. Assume that \(r\) is continuous on \([a,b]\) and has finitely many zeros there. List the zeros in the interval in increasing order, including an endpoint if it is a zero. Together with \(a\) and \(b\), these points divide \([a,b]\) into subintervals. On the interior of each subinterval, \(r\) has a constant sign.
This formula applies when the sign is constant on each piece. For a continuous rate, splitting at all its zeros gives such pieces: if the sign changed between two consecutive zeros, the intermediate value theorem would force another zero between them. If a rate has a jump or another specified breakpoint, include that point in the partition as well and check the sign on each piece.
Why the formula works
On any one subinterval \([c_{j-1},c_j]\), there are two possibilities. If \(r\ge 0\), then \(|r|=r\) throughout that piece, so its accumulated magnitude is \(\int r\). This integral is nonnegative, and therefore equals \(\left|\int r\right|\). If \(r\le 0\), then \(|r|=-r\), so its accumulated magnitude is \(-\int r\). In this case \(\int r\le 0\), and again \(-\int r=\left|\int r\right|\).
Adding these piece-by-piece equalities gives the result. The first equality uses additivity of the definite integral over adjacent intervals. The sign condition is essential: without it, a piece’s signed integral need not equal the area of its absolute value.
Worked Example: One zero in a linear rate
Worked Example: One zero in a linear rate
Find the accumulated magnitude of \(r(t)=2t-4\) on \([0,5]\).
Step 1: Find the zero and determine the signs. Solve \(2t-4=0\), giving \(t=2\). For \(0\le t<2\), \(2t-4<0\); for \(2<t\le5\), \(2t-4>0\). Therefore split the integral at \(2\).
Step 2: Evaluate the two pieces. On \([0,2]\), the absolute-value rate is a triangle with base \(2\) and height \(4\), so its area is \(\frac12(2)(4)=4\). Equivalently, its endpoint heights are \(4\) and \(0\), giving trapezoid area \(2(4+0)/2=4\).
On \([2,5]\), the absolute-value rate rises linearly from \(0\) to \(2(5)-4=6\). Its triangular area is \(\frac12(3)(6)=9\).
Check: The rate is negative only on \([0,2]\) and positive only on \([2,5]\), as used. The heights \(4,0,6\) follow from the rate at the interval endpoints, and the two triangle areas add to \(13\). Thus the accumulated magnitude is \(13\).
Worked Example: Several sign changes in a measured rate
Worked Example: Several sign changes in a measured rate
A flow rate, measured in litres per minute, is piecewise linear through the points \((0,3)\), \((2,-1)\), \((4,-1)\), and \((6,2)\). Find its accumulated magnitude over the first six minutes.
Step 1: Locate every zero. On \([0,2]\), the line through \((0,3)\) and \((2,-1)\) is \(r(t)=3-2t\), so its zero is \(t=3/2\). On \([2,4]\), the rate is \(-1\), so there is no zero inside that interval. On \([4,6]\), the line through \((4,-1)\) and \((6,2)\) is \(r(t)=\frac32t-7\). Its zero satisfies \(\frac32t=7\), so \(t=14/3\).
The rate is positive on \([0,3/2]\), negative on \([3/2,14/3]\), and positive on \([14/3,6]\). The accumulated magnitude is the sum of the areas on those pieces.
Step 2: Calculate each area. On \([0,3/2]\), the rate falls from \(3\) to \(0\), so the triangle area is \(\frac12(3/2)(3)=9/4\). From \(3/2\) to \(2\), its magnitude rises from \(0\) to \(1\), giving \(\frac12(1/2)(1)=1/4\). From \(2\) to \(4\), the magnitude is constantly \(1\), giving rectangle area \(2(1)=2\).
From \(4\) to \(14/3\), the magnitude falls from \(1\) to \(0\), giving \(\frac12(2/3)(1)=1/3\). From \(14/3\) to \(6\), it rises from \(0\) to \(2\), giving \(\frac12(4/3)(2)=4/3\).
Check: The endpoint values on the last segment are \(r(4)=-1\) and \(r(6)=2\), consistent with the zero at \(14/3\). The areas on the first segment add to \(9/4+1/4=5/2\); the middle rectangle contributes \(2\), and the last segment contributes \(1/3+4/3=5/3\). Their sum is \(5/2+2+5/3=37/6\) litres.
Worked Example: Accumulated magnitude up to an intermediate time
Worked Example: Accumulated magnitude up to an intermediate time
A piecewise-linear rate passes through \((0,2)\), \((3,-1)\), and \((5,3)\). Find the accumulated magnitude from \(t=0\) to \(t=4\).
Step 1: Find the zeros before \(t=4\). On \([0,3]\), the rate falls from \(2\) to \(-1\), with slope \(-1\), so \(r(t)=2-t\) and its zero is \(t=2\). On \([3,5]\), it rises from \(-1\) to \(3\), with slope \(2\), so \(r(t)=2t-7\) and its zero is \(t=7/2\). At \(t=4\), the rate is \(2(4)-7=1\).
Step 2: Split at both zeros and evaluate. From \(0\) to \(2\), the rate forms a positive triangle with base \(2\) and height \(2\), giving area \(\frac12(2)(2)=2\). From \(2\) to \(3\), the magnitude forms a negative-rate triangle with base \(1\) and height \(1\), giving area \(\frac12(1)(1)=1/2\). From \(3\) to \(7/2\), it falls from magnitude \(1\) to \(0\), giving area \(\frac12(1/2)(1)=1/4\). From \(7/2\) to \(4\), it rises from \(0\) to \(1\), giving area \(\frac12(1/2)(1)=1/4\).
Check: The rate values at \(t=3\) and \(t=4\) are \(-1\) and \(1\), respectively, so the zero \(7/2\) lies between them. The four areas total \(2+1/2+1/2=3\), as required.
A common mistake: using one sign on the whole interval
A frequent error is to treat a rate as though it kept its sign throughout the interval, even when it crosses zero. For \(r(t)=2t-4\) on \([0,5]\), the signed areas are \(-4\) on \([0,2]\) and \(9\) on \([2,5]\). Adding them gives a signed integral of \(-4+9=5\). That value is not the accumulated magnitude: it gives the negative part a negative contribution.
The correction is to split at \(t=2\), reverse the sign on the negative-rate piece, and add positive areas: \(4+9=13\). The absolute value belongs inside the integral; it cannot be omitted just because the rate is easy to integrate as a signed quantity.
Check Your Understanding
For each question, identify the sign intervals before calculating any accumulated magnitude.
- For \(r(t)=t-3\) on \([0,5]\), where must the integral of \(|r(t)|\) be split?
- For a continuous rate with zeros at \(c_1\) and \(c_2\), what additional fact is needed to use a single sign on each interval between consecutive zeros?
- A rate is linear from \(4\) at \(t=0\) to \(-2\) at \(t=3\). Find its zero and describe how you would split the accumulated-magnitude calculation.
- Why is the area on an interval where \(r\le0\) calculated using \(-r\), not \(r\)?
- A rate has a jump discontinuity but no zero at the jump. What should be checked before evaluating its accumulated magnitude piece by piece?
