What the Sign of a Rate Tells Us
A rate describes how a quantity is changing, not how large the quantity itself is. If \(Q(t)\) is a quantity and \(r(t)=Q'(t)\) is its rate of change, then a positive rate means \(Q\) is increasing at that time. A negative rate means \(Q\) is decreasing. The definite integral of the rate keeps track of these directions: increases contribute positively, while decreases contribute negatively.
This sign convention matters whenever a rate changes direction during an interval. Adding the signed contributions gives the quantity’s net change: the final value minus the initial value. A decrease can therefore offset an earlier increase, or vice versa.
Signed Contributions on an Interval
Suppose \(r\) is continuous on \([a,b]\), and \(Q'(t)=r(t)\). By the Net Change Theorem, the change in \(Q\) over the interval is
If \(r(t)\geq 0\) throughout an interval, its integral there is nonnegative, and \(Q\) does not decrease on that interval. If \(r(t)\leq 0\), its integral is nonpositive, and \(Q\) does not increase there. When the rate changes sign, split the integral at the points where its sign changes. Each part then records a signed contribution in the appropriate direction.
For example, suppose \(r\) changes from positive to negative at \(c\), where \(a<c<b\). Additivity of definite integrals gives
The first term is positive if \(r\) is positive on \([a,c]\), and the second is negative if \(r\) is negative on \([c,b]\). Their sum, not either contribution by itself, gives the net change over the full interval.
Why the Signed Contributions Give Net Change
The reason for the signed sum follows directly from the Net Change Theorem and the additivity of definite integrals. The partition point can be any point in the interval; it need not be where the rate changes sign. Choosing sign-change points is useful because it makes the direction of each contribution clear.
Choose partition points \(a=t_0<t_1<\cdots<t_n=b\), for example at the zeros where the rate changes sign.
On every subinterval, \(Q(t_i)-Q(t_{i-1})=\int_{t_{i-1}}^{t_i}r(t)\,dt\).
The intermediate values cancel, leaving \(Q(b)-Q(a)\), while the integrals combine to \(\int_a^b r(t)\,dt\).
In detail, adding the changes on all subintervals gives
Each difference equals the integral of \(r\) on its subinterval. Additivity then gives
This proves why the signed contributions combine to give net change: the quantity’s changes on consecutive intervals add, and the definite integrals add in exactly the same way.
Worked Examples
Worked Example: A Rate That Changes from Negative to Positive
Suppose \(Q'(t)=t-2\) for \(0\leq t\leq 5\). Find the signed contributions before and after the rate changes sign, then determine the net change in \(Q\).
First find where the rate is zero:
For \(0\leq t<2\), \(t-2<0\), so the quantity decreases. For \(2<t\leq 5\), \(t-2>0\), so it increases. An antiderivative of \(t-2\) is \(F(t)=\frac{t^2}{2}-2t\).
The signed contribution from \(0\) to \(2\) is
The contribution is \(-2\), a decrease of 2. From \(2\) to \(5\), the signed contribution is
This is an increase of \(\frac{9}{2}\). Therefore, the net change over the full interval is
Check directly using the antiderivative on the full interval:
The direct calculation agrees with the sum of the signed contributions.
Worked Example: An Increase Followed by an Equal Decrease
Suppose \(Q'(t)=3-t\) on \(0\leq t\leq 6\), and \(Q(0)=20\). Find the increases and decreases on either side of the rate’s zero, and determine \(Q(6)\).
The rate is zero when \(3-t=0\), so \(t=3\). It is positive for \(0\leq t<3\) and negative for \(3<t\leq 6\). An antiderivative is \(F(t)=3t-\frac{t^2}{2}\).
From \(0\) to \(3\),
This is an increase of \(\frac{9}{2}\). From \(3\) to \(6\),
This is a decrease of \(\frac{9}{2}\). The net change is \(\frac{9}{2}-\frac{9}{2}=0\), so the final value equals the initial value:
Check using the antiderivative over the full interval: \(F(6)-F(0)=0-0=0\). Thus the net change is zero, consistent with \(Q(6)=20\).
Worked Example: Several Direction Changes
Let \(Q'(t)=(t-1)(t-3)\) on \(0\leq t\leq 4\). Find the signed change on each interval where the rate keeps its sign, and then find the net change.
The rate is zero at \(t=1\) and \(t=3\). Its sign is positive on \([0,1)\), negative on \((1,3)\), and positive on \((3,4]\). Expanding gives \(Q'(t)=t^2-4t+3\). An antiderivative is \(F(t)=\frac{t^3}{3}-2t^2+3t\).
On the first interval,
On the middle interval,
On the final interval,
The signed contributions sum to \(\frac{4}{3}-\frac{4}{3}+\frac{4}{3}=\frac{4}{3}\). Check directly: \(F(4)-F(0)=\frac{4}{3}-0=\frac{4}{3}\). The net change is positive even though the quantity decreases during the middle portion.
A Common Sign Mistake
A frequent error is to replace a negative rate contribution by its positive magnitude when finding net change. For the first example, this mistake would turn the decrease \(-2\) into \(+2\), then calculate
That result is incorrect for net change. The rate is negative from \(0\) to \(2\), so the quantity decreases there; its contribution must remain \(-2\). The correct sum is
A negative integral does not mean the quantity itself is negative. It means the quantity’s change over that interval is downward. Likewise, a positive net change does not imply that the quantity increased at every moment: it may have decreased on part of the interval, as the third example shows.
Why Direction and Net Change Matter
In applications, the direction of change can be as informative as the final net result. A rate may be positive for one part of an interval and negative for another, and the integral records how those changes combine. When solving a problem, identify where the rate changes sign, interpret each signed contribution in context, and then combine the contributions without changing their signs.
To check a calculation, compare the sum of the contributions with a direct evaluation of the full definite integral. If an initial value is provided, add the net change to that initial value to obtain the final value. A negative contribution represents a decrease from the value at the start of that subinterval; it is not a claim that the quantity itself is below zero.
Check Your Understanding
For each question, keep the sign of every rate integral and interpret it as a change in the quantity.
- If \(Q'(t)>0\) throughout \([a,b]\), what does the sign of \(\int_a^b Q'(t)\,dt\) tell you about \(Q(b)-Q(a)\)?
- A rate is negative on \([0,2]\) and positive on \([2,5]\). What point would you use to split the integral when calculating net change, and why?
- For \(Q'(t)=t-1\) on \([0,3]\), find the signed contribution on each side of the zero of the rate and add them to obtain the net change.
- A quantity increases by 7 on one interval and decreases by 10 on the next. What is the net change across both intervals?
- Why can a quantity have a positive net change even if its rate is negative during part of the interval?
