Absolute-Rate Integrals for Gross Throughput

Why a Signed Rate Needs Care

A rate can be positive during one part of an interval and negative during another. The sign records direction relative to a chosen positive direction. For instance, positive flow might mean movement into a reservoir, while negative flow means movement out.

The definite integral of a signed rate gives the net change: the amount added in the positive direction minus the amount moved in the negative direction. But if the question asks how much material moved altogether, cancellation is not appropriate. The amount moved in either direction contributes positively to the total. This total is called gross throughput.

The key is to integrate the absolute value of the rate, splitting the calculation wherever the rate changes sign. This tutorial develops that method and relates it to net change and to the separate totals in each direction.

Definition and Key Relationships

Let $r(t)$ be a continuous signed rate on the time interval $[a,b]$, measured relative to a fixed positive direction. Assume the rate and time units are such that integrating $r(t)$ gives an amount. A positive rate contributes movement in the positive direction; a negative rate contributes movement in the opposite direction.

Definition: The gross throughput over $[a,b]$ is the total amount moved in either direction, without cancellation. It is calculated by integrating the absolute value of the rate: $$ G=\int_a^b |r(t)|\,dt. $$ The net change is $$ N=\int_a^b r(t)\,dt. $$

For a more explicit description of the two directions, define the positive-direction rate $r_+(t)=\max(r(t),0)$ and the opposite-direction rate $r_-(t)=\max(-r(t),0)$. Both are nonnegative. The amounts moved in the positive and opposite directions are, respectively,

P=\int_a^b r_+(t)\,dt, \qquad M=\int_a^b r_-(t)\,dt.

At every time, $r(t)=r_+(t)-r_-(t)$ and $|r(t)|=r_+(t)+r_-(t)$. Integrating these equalities gives the useful relationships

N=P-M, \qquad G=P+M.

Thus gross throughput is never obtained by subtracting one directional amount from the other. The net change does that subtraction; gross throughput adds the two amounts.

Why the Absolute-Rate Integral Works

Over a short time interval of length $\Delta t$, a rate close to $r(t)$ corresponds to an amount of movement close to $r(t)\Delta t$, with the sign indicating direction. The amount moved over that short interval is therefore close to $|r(t)|\Delta t$. Adding such unsigned contributions over a partition and refining the partition leads to the integral of $|r|$.

Derivation: For a partition of $[a,b]$ into short intervals of widths $\Delta t_i$, choose a sample time $t_i^*$ in each interval. The sum $$ \sum_i |r(t_i^*)|\Delta t_i $$ approximates the total movement, because each term estimates the amount moved during one interval without regard to direction. Since $r$ is continuous, $|r|$ is also continuous and hence integrable. As the partition is refined, these Riemann sums converge to $$ \int_a^b |r(t)|\,dt. $$ Therefore this integral gives gross throughput. If the signs are retained instead, the sums approximate signed change and converge to $\int_a^b r(t)\,dt$, the net change.

The units also provide a useful check: rate units multiplied by time units give amount units. For example, a rate in litres per minute integrated over minutes gives litres.

Finding the Intervals of Each Direction

To evaluate gross throughput in practice, find where the rate is zero and determine its sign between those times. On an interval where $r(t)\geq 0$, $|r(t)|=r(t)$. Where $r(t)\leq 0$, $|r(t)|=-r(t)$. Each sign-changing point is therefore a natural place to split the integral.

Find candidate direction changes.
Solve $r(t)=0$ in the interval of interest and include the interval endpoints.

Determine the sign on each subinterval.
Use a test value or factorization to decide whether the rate is positive or negative between consecutive zeros.

Integrate the unsigned rate.
Use $r(t)$ where it is nonnegative and $-r(t)$ where it is negative, then add the resulting amounts.

Compare with net change.
Integrate $r(t)$ with its sign intact. Check that gross throughput is the sum of the directional totals and net change is their difference.

Worked Examples

Worked Example: Flow That Reverses Once

A signed flow rate is $r(t)=t-2$ cubic metres per minute for $0\leq t\leq 5$, with $t$ measured in minutes. Find the gross throughput, the net change, and the amounts moved in each direction.

First find the zero: $t-2=0$, so the rate changes sign at $t=2$. For $0\leq t<2$, $t-2<0$; for $2<t\leq5$, $t-2>0$. Thus the amount moved in the negative direction is

M=\int_0^2 (2-t)\,dt =\left[2t-\frac{t^2}{2}\right]_0^2 =(4-2)-0 =2\text{ m}^3.

The amount moved in the positive direction is

P=\int_2^5 (t-2)\,dt =\left[\frac{t^2}{2}-2t\right]_2^5 =\left(\frac{25}{2}-10\right)-(2-4) =\frac{5}{2}-(-2) =\frac{9}{2}\text{ m}^3.

Therefore, gross throughput is $G=M+P=2+\frac{9}{2}=\frac{13}{2}\text{ m}^3$. The net change is $N=P-M=\frac{9}{2}-2=\frac{5}{2}\text{ m}^3$.

Check: An antiderivative of $r(t)=t-2$ is $F(t)=\frac{t^2}{2}-2t$, since $F'(t)=t-2$. Hence the signed integral is $F(5)-F(0)=\frac{5}{2}-0=\frac{5}{2}\text{ m}^3$, matching the net change. The two directional amounts add to $\frac{13}{2}\text{ m}^3$, as required for gross throughput.

Worked Example: Current and Total Charge

An electric current is $i(t)=6-2t$ amperes for $0\leq t\leq5$ seconds. Find the total charge that passes through a point, regardless of direction, and the net charge in the positive direction.

The current is zero when $6-2t=0$, so $t=3$. It is positive on $[0,3]$ and negative on $[3,5]$. The positive-direction charge is

P=\int_0^3 (6-2t)\,dt =\left[6t-t^2\right]_0^3 =(18-9)-0 =9\text{ C}.

On $[3,5]$, the magnitude of the current is $2t-6$. The charge moving in the opposite direction is

M=\int_3^5 (2t-6)\,dt =\left[t^2-6t\right]_3^5 =(25-30)-(9-18) =-5-(-9) =4\text{ C}.

The gross throughput is $G=P+M=9+4=13\text{ C}$. The net charge is $N=P-M=9-4=5\text{ C}$.

Check: An antiderivative of the signed current is $I(t)=6t-t^2$, because $I'(t)=6-2t$. The signed integral is $I(5)-I(0)=(30-25)-0=5\text{ C}$, confirming the net charge. Adding the directional charges gives $9+4=13\text{ C}$, confirming the gross total.

Worked Example: Two Direction Changes

A signed flow rate is $q(t)=t^2-4t+3$ litres per minute on $0\leq t\leq4$, where $t$ is in minutes. Find the gross throughput and the net change.

Factor the rate: $q(t)=(t-1)(t-3)$. Its zeros are $t=1$ and $t=3$. The rate is positive on $[0,1]$, negative on $[1,3]$, and positive on $[3,4]$. An antiderivative is

F(t)=\frac{t^3}{3}-2t^2+3t, \qquad F'(t)=t^2-4t+3.

The first positive-direction amount is

\int_0^1 q(t)\,dt=F(1)-F(0) =\left(\frac13-2+3\right)-0 =\frac43\text{ L}.

The signed integral on $[1,3]$ is $F(3)-F(1)=(9-18+9)-\frac43=0-\frac43=-\frac43\text{ L}$, so the amount moved in the negative direction is $\frac43\text{ L}$. The final positive-direction amount is

\int_3^4 q(t)\,dt=F(4)-F(3) =\left(\frac{64}{3}-32+12\right)-0 =\left(\frac{64}{3}-20\right) =\frac43\text{ L}.

Thus the total positive-direction amount is $P=\frac43+\frac43=\frac83\text{ L}$, and the negative-direction amount is $M=\frac43\text{ L}$. Gross throughput is $G=P+M=\frac83+\frac43=4\text{ L}$, while net change is $N=P-M=\frac83-\frac43=\frac43\text{ L}$.

Check: The full signed integral is $F(4)-F(0)=\frac43-0=\frac43\text{ L}$, which matches the net change. Differentiating $F$ recovers $q$, and the three subinterval contributions to gross throughput add to $\frac43+\frac43+\frac43=4\text{ L}$.

A Common Mistake: Taking the Absolute Value After Integrating

A tempting but incorrect calculation is to integrate the signed rate over the whole interval and then take the absolute value of the result. That gives the magnitude of the net change, not the gross throughput. In the first example, the incorrect calculation would be

\left|\int_0^5 (t-2)\,dt\right| =\left|\frac52\right| =\frac52\text{ m}^3.

This is not the total movement: it allows the $2\text{ m}^3$ moved in the negative direction to cancel part of the $\frac92\text{ m}^3$ moved in the positive direction. The correction is to take the absolute value before integrating, or equivalently to split at the zero and add the unsigned contributions:

\int_0^5 |t-2|\,dt =\int_0^2 (2-t)\,dt+\int_2^5 (t-2)\,dt =2+\frac92 =\frac{13}{2}\text{ m}^3.

In general, $\left|\int_a^b r(t)\,dt\right|$ measures the magnitude of net change. It is not generally equal to $\int_a^b|r(t)|\,dt$. They agree when the rate has one sign throughout the interval, but can differ when movement reverses direction.

Interpreting the Results

Gross throughput answers a question about cumulative movement, such as how much liquid passed through a pipe in total when flow sometimes reversed. Net change answers a different question: how much the final amount differs from the initial amount, with direction taken into account. The directional totals provide both pieces of information separately.

These quantities can be checked together. If $P$ and $M$ are the amounts moved in the two directions, then $G=P+M$ and $N=P-M$. Therefore $G\geq |N|$: adding two nonnegative amounts cannot give less than the magnitude of their difference. If a calculation produces gross throughput smaller than the magnitude of net change, a sign or arithmetic error has occurred.

For a rate that stays nonnegative, gross throughput equals net change. For a rate that stays nonpositive, gross throughput equals the magnitude of net change. When the rate changes direction, locate the sign changes, integrate the rate’s magnitude on each sign interval, and add the results. Keep the signed integral separate whenever the question asks for net change.

Check Your Understanding

Use the distinction between unsigned total movement and signed net change to answer the following.

For a signed rate that is negative throughout an interval, how are gross throughput and net change related?
What equation defines gross throughput from a rate $r(t)$ on $[a,b]$?
A rate is positive for part of an interval and negative for the rest. Why does taking the absolute value of its signed integral fail to give gross throughput?
If the positive-direction amount is $12$ litres and the opposite-direction amount is $5$ litres, what are the gross throughput and net change?
What should you determine before splitting an absolute-rate integral into simpler integrals?

Absolute-Rate Integrals for Gross Throughput

What You'll Learn

Why a Signed Rate Needs Care

Definition and Key Relationships

Why the Absolute-Rate Integral Works

Finding the Intervals of Each Direction