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Net Change Versus Total Accumulated Amount

Distinguish cancellation in net change from total accumulated magnitude, and calculate both from a rate function.

Intermediate 9 min read

What You'll Learn

  • Define net change and total accumulated magnitude for a rate on an interval
  • Explain why positive and negative rate contributions can cancel in a signed integral
  • Calculate the integral of an absolute rate by identifying where the rate changes sign
  • Prove that total accumulated magnitude is at least the absolute value of net change
  • Identify when the two quantities are equal and avoid a common absolute-value error

Why One Integral May Not Tell the Whole Story

A rate can be positive during one part of an interval and negative during another. Its integral records the net change: positive and negative contributions can cancel. If instead we want to count all the change regardless of direction, we must prevent that cancellation.

These are different questions. For example, a quantity might increase by 2 units and later decrease by 2 units. Its net change is zero, but the total amount by which it changed is 4 units. The distinction is captured by comparing the integral of a rate with the integral of its absolute value.

Two Quantities From the Same Rate

Let \(r\) be a continuous rate function on an interval \([a,b]\). Its signed net change over the interval is the integral of the rate. Its total accumulated magnitude is the integral of the rate's absolute value. Continuity ensures that both integrals exist.

Definition: The net change associated with \(r\) on \([a,b]\) is \(\int_a^b r(t)\,dt\). The total accumulated magnitude is \(\int_a^b |r(t)|\,dt\).

The absolute value keeps the size of the rate but removes its direction: \(|r(t)|\) is nonnegative whether \(r(t)\) is positive or negative. Thus the net change retains signs, while the total accumulated magnitude counts contributions from both directions as nonnegative amounts.

$$ \text{net change}=\int_a^b r(t)\,dt, \qquad \text{total accumulated magnitude}=\int_a^b |r(t)|\,dt. $$

If \(r(t)\ge 0\) throughout the interval, these quantities are equal. If \(r(t)\le 0\) throughout, the net change is nonpositive and the total accumulated magnitude is its opposite. When the rate has both positive and negative values, the two quantities generally differ because the signed integral includes cancellation.

Worked Example: Equal Increase and Decrease

Worked Example: Equal Increase and Decrease

Let \(r(t)=t-2\) on \([0,4]\). Find the net change and the total accumulated magnitude.

The rate is zero when \(t-2=0\), so it changes sign at \(t=2\). For \(0\le t<2\), \(r(t)<0\); for \(2<t\le4\), \(r(t)>0\).

First calculate the net change using an antiderivative, \(F(t)=\frac{t^2}{2}-2t\):

$$ \int_0^4 (t-2)\,dt = \left[\frac{t^2}{2}-2t\right]_0^4 = (8-8)-(0-0) = 0. $$

For total accumulated magnitude, use \(|t-2|=2-t\) on \([0,2]\) and \(|t-2|=t-2\) on \([2,4]\):

$$ \begin{aligned} \int_0^4 |t-2|\,dt &=\int_0^2 (2-t)\,dt+\int_2^4(t-2)\,dt\\ &=\left[2t-\frac{t^2}{2}\right]_0^2 +\left[\frac{t^2}{2}-2t\right]_2^4\\ &=(4-2-0)+(0-(-2))\\ &=2+2\\ &=4. \end{aligned} $$

Check: The signed contribution from \([0,2]\) is \(-2\), and the signed contribution from \([2,4]\) is \(2\), so the net change is \(-2+2=0\). Their magnitudes add to \(2+2=4\), as calculated.

Why Total Accumulated Magnitude Cannot Be Smaller

There is a useful general relationship between these two quantities. The total accumulated magnitude is always at least as large as the absolute value of the net change. To see why, separate the positive and negative contributions.

Define \(r_+(t)=\max(r(t),0)\) and \(r_-(t)=\max(-r(t),0)\). At every \(t\), these functions satisfy \(r(t)=r_+(t)-r_-(t)\) and \(|r(t)|=r_+(t)+r_-(t)\). Let \(P=\int_a^b r_+(t)\,dt\) and \(M=\int_a^b r_-(t)\,dt\). Both \(P\) and \(M\) are nonnegative.

Theorem: For a continuous rate \(r\) on \([a,b]\), \(\left|\int_a^b r(t)\,dt\right|\le \int_a^b |r(t)|\,dt\). Equality holds exactly when \(r\) does not take both positive and negative values on \([a,b]\).

Proof. By the definitions of \(P\) and \(M\), the net change is \(P-M\), and the total accumulated magnitude is \(P+M\). Since \(P,M\ge0\), the triangle inequality gives \(|P-M|\le P+M\). Substituting the two integral expressions proves the inequality.

$$ \left|\int_a^b r(t)\,dt\right| =|P-M| \le P+M =\int_a^b |r(t)|\,dt. $$

For the equality condition, if \(P=0\) or \(M=0\), then \(|P-M|=P+M\). If both are positive, then \(|P-M|<P+M\): when \(P\ge M>0\), \(P-M<P+M\); when \(M>P>0\), \(M-P<P+M\). For continuous \(r\), \(P=0\) means there is no interval on which \(r\) is positive, and \(M=0\) means there is no interval on which it is negative. Thus equality holds exactly when the rate has one sign throughout, allowing zeros. This completes the proof.

Worked Example: A Rate That Changes from Positive to Negative

Worked Example: A Rate That Changes from Positive to Negative

Let \(r(t)=4-2t\) on \([0,3]\). Find both quantities.

The zero of the rate satisfies \(4-2t=0\), so \(t=2\). The rate is positive on \([0,2)\) and negative on \((2,3]\). An antiderivative is \(F(t)=4t-t^2\). Therefore, the net change is

$$ \int_0^3 (4-2t)\,dt = \left[4t-t^2\right]_0^3 = (12-9)-(0-0) = 3. $$

To calculate total accumulated magnitude, integrate the rate on \([0,2]\) and its opposite on \([2,3]\):

$$ \begin{aligned} \int_0^3 |4-2t|\,dt &=\int_0^2(4-2t)\,dt+\int_2^3(2t-4)\,dt\\ &=\left[4t-t^2\right]_0^2+\left[t^2-4t\right]_2^3\\ &=(8-4-0)+(9-12-(4-8))\\ &=4+1\\ &=5. \end{aligned} $$

Check: The positive contribution is \(4\) and the negative signed contribution is \(-1\). Their signed sum is \(4+(-1)=3\), matching the net change; their magnitudes sum to \(4+1=5\), matching the total accumulated magnitude. Also, \(5\ge|3|\), as the theorem requires.

Worked Example: A Rate That Is Negative Throughout

Worked Example: A Rate That Is Negative Throughout

Let \(r(t)=-(t+1)\) on \([0,2]\). Find the net change and total accumulated magnitude.

Since \(t+1>0\) on this interval, \(r(t)<0\), so \(|r(t)|=t+1\). An antiderivative of \(r(t)\) is \(-\frac{t^2}{2}-t\). Hence

$$ \int_0^2 -(t+1)\,dt = \left[-\frac{t^2}{2}-t\right]_0^2 = (-2-2)-0 = -4. $$

The absolute value of the rate is \(t+1\), whose antiderivative is \(\frac{t^2}{2}+t\). Thus

$$ \int_0^2 |-(t+1)|\,dt = \int_0^2(t+1)\,dt = \left[\frac{t^2}{2}+t\right]_0^2 = (2+2)-0 = 4. $$

Check: The rate is negative everywhere, so its signed integral must be the negative of its total magnitude. The results \(-4\) and \(4\) have exactly that relationship.

A Common Mistake: Taking the Absolute Value After Integrating

A frequent error is to assume that the absolute value of an integral is the integral of the absolute value. The first worked example gives a direct counterexample:

$$ \left|\int_0^4(t-2)\,dt\right| =|0| =0, \qquad \int_0^4|t-2|\,dt =4. $$

The incorrect calculation replaces \(\int_a^b |r(t)|\,dt\) with \(\left|\int_a^b r(t)\,dt\right|\). That operation is valid only when the integral has no cancellation between positive and negative contributions; for a continuous rate, this means the rate does not take both signs on the interval. In general, first determine the sign of the rate, then integrate its absolute value on the portions where its sign is fixed.

When the Distinction Matters

Use the signed integral when the question asks for a net change, because increases and decreases must retain their directions. Use the integral of the absolute value when the question asks for the total amount of change or the accumulated magnitude without regard to direction.

Before calculating, ask what is being counted. If opposing contributions should cancel, integrate the rate. If every contribution should count positively, integrate the absolute value of the rate. A graph or formula can help identify where the rate is positive or negative, but the essential distinction is in the quantity being asked for.

Check Your Understanding

For each question, decide whether it concerns net change, total accumulated magnitude, or both.

  1. State the formula for net change from a continuous rate \(r\) on \([a,b]\).
  2. State the formula for total accumulated magnitude on the same interval.
  3. If a rate is negative throughout an interval and its net change is \(-7\), what is its total accumulated magnitude?
  4. For \(r(t)=t-1\) on \([0,2]\), at what input does the rate change sign?
  5. Explain why \(\left|\int_a^b r(t)\,dt\right|\) cannot exceed \(\int_a^b|r(t)|\,dt\).

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