Reading One Graph From Another
An accumulation function records the net contribution of a rate from a chosen starting point. Its graph is not the rate graph: the rate graph tells us how the accumulated value changes, while the accumulated value depends on all the signed area collected so far.
Let \(f\) be a rate and define the accumulation from \(a\) by
The variable \(t\) is the input to the rate being accumulated; \(x\) is the endpoint, and therefore the input to \(A\). In particular, \(A(a)=0\). The signed area above the horizontal axis contributes positively, and the signed area below it contributes negatively.
To sketch or describe \(A\) from the graph of \(f\), answer two different questions. First, is \(A\) rising or falling as the endpoint moves? This depends on whether the rate is positive or negative. Second, is \(A(x)\) itself positive, negative, or zero? This depends on the net signed area from the starting point up to \(x\), not just on the rate at \(x\).
Why the Sign of the Rate Gives the Direction
For two endpoints \(u<v\), the change in accumulation between them is the signed area contributed over just that interval:
This identity follows by subtracting the two accumulated areas: the contribution from \(a\) to \(u\) occurs in both and cancels, leaving only the contribution from \(u\) to \(v\). Thus, if the rate graph stays above the axis between \(u\) and \(v\), that interval contributes positive area and \(A(v)>A(u)\). If it stays below the axis, the contribution is negative and \(A(v)<A(u)\). This is the reason the accumulation rises on positive-rate intervals and falls on negative-rate intervals.
Above the axis means positive rate; below the axis means negative rate.
Positive contributions make \(A\) rise; negative contributions make \(A\) fall.
Start from \(A(a)=0\), or from another specified starting value, and add each signed region.
Compare the total positive and negative contributions collected so far.
A rate equal to zero at one point does not, by itself, tell us that the accumulation is zero there or that the accumulation stops rising or falling on a whole interval. What matters for a change over an interval is the signed contribution across that interval. A crossing of the horizontal axis on the rate graph is therefore not automatically a zero of \(A\).
Worked Example: Following Signed Regions
Worked Example: Following Signed Regions
Suppose \(A(x)=\int_0^x f(t)\,dt\), and the rate graph has these regions: from \(0\) to \(2\), a triangle above the axis with base \(2\) and height \(2\); from \(2\) to \(4\), a rectangle at height \(-1\); and from \(4\) to \(6\), a triangle above the axis with base \(2\) and height \(2\). Find the accumulation at the region endpoints and describe its direction and sign.
The first triangle contributes
Starting with \(A(0)=0\), this gives \(A(2)=0+2=2\). The rate is positive on this interval, so \(A\) rises from \(0\) to \(2\).
The rectangle contributes signed area
Therefore \(A(4)=A(2)-2=2-2=0\). The rate is negative from \(2\) to \(4\), so \(A\) falls. Since the rate is constantly \(-1\), at \(x=3\) the contribution from \(2\) to \(3\) is \((3-2)(-1)=-1\), giving \(A(3)=2-1=1\). The accumulation stays positive between \(2\) and \(4\), then reaches zero at \(4\).
The last triangle contributes
It follows that \(A(6)=A(4)+2=0+2=2\). The rate is positive from \(4\) to \(6\), so the accumulation rises again. Since this triangular region has positive area at every endpoint strictly between \(4\) and \(6\), \(A\) is positive there. At the endpoints, the computed values are \(A(4)=0\) and \(A(6)=2\).
Check: The three signed contributions total \(2+(-2)+2=2\), which agrees with the accumulated value \(A(6)=2\). The negative rectangle exactly cancels the first triangle; it does not make the accumulation negative along the way.
The Accumulated Value Is Not the Rate
Direction and sign answer different questions. A falling accumulation can still be positive if its earlier positive contributions exceed the negative area collected so far. It can also become negative and continue falling. Similarly, a rising accumulation can remain negative for some time if it started below zero and has not yet gained enough positive area.
In particular, the location where \(f(x)=0\) only marks a point where the rate graph meets the axis. To find where \(A(x)=0\), compare all the signed area from the starting point to \(x\). The accumulation is zero when the positive and negative contributions balance, not merely when the current rate is zero.
Worked Example: A Zero Located by Exact Area
Worked Example: A Zero Located by Exact Area
Let \(A(x)=\int_0^x f(t)\,dt\). Suppose the rate graph is a triangle above the axis from \(0\) to \(5\), with base \(5\) and height \(2\), followed by a constant rate \(f(x)=-2\) from \(5\) to \(9\). Find where the accumulation is zero on the second interval.
The positive triangle contributes
so \(A(5)=5\). From \(5\) onward, the rate is \(-2\). For \(5\leq x\leq9\), the signed rectangular contribution from \(5\) to \(x\) is \(-2(x-5)\). Therefore
Set this accumulated value equal to zero:
The graph-reading interpretation is that \(2.5\) units of the negative-rate interval contribute area \(-2(2.5)=-5\), exactly cancelling the earlier positive area \(5\).
Check: At \(x=7.5\), \(A(7.5)=5-2(7.5-5)=5-5=0\). At the end of the negative interval, \(A(9)=5-2(9-5)=5-8=-3\). Thus the accumulation falls from \(5\) to \(-3\) and is zero at \(7.5\).
Estimating from a Sketch: Keep Uncertainty Visible
Graphs are often not labelled with exact coordinates, and the regions may not have simple geometric shapes. In that case, estimate the signed areas and keep track of how accurate the estimates are. A numerical estimate can suggest whether the accumulation is positive or negative, but it does not establish the actual sign unless the available information is strong enough to rule out other possibilities.
Worked Example: An Estimate That Does Not Determine the Sign
Suppose \(A(x)=\int_0^x f(t)\,dt\). From a sketch, estimate the positive region from \(0\) to \(3\) as \(4.5\) square units, with an error of at most \(1\). Estimate the magnitude of the negative region from \(3\) to \(7\) as \(5.5\) square units, also with an error of at most \(1\). Estimate the endpoint value \(A(7)\), and decide whether its sign is known.
The estimated positive contribution is \(+4.5\), and the estimated negative contribution is \(-5.5\). Their estimated total is
So the estimate predicts \(A(7)\approx-1\). But the first contribution could be anywhere from \(3.5\) to \(5.5\), and the magnitude of the second could also be anywhere from \(4.5\) to \(6.5\). The possible net totals consequently range from
This range includes both negative and positive values. The sketch-based estimates therefore do not establish the sign of the actual \(A(7)\). They support a prediction of a negative value, but a more accurate estimate or stronger information about the areas is needed to settle the sign.
Check: The estimated total is \(-1\), which lies inside the possible range \([-3,1]\). The interval includes zero, confirming that the estimate alone cannot rule out an actual value of zero or a positive value.
When Can We Conclude That an Actual Zero Exists?
A prediction that an accumulation crosses zero is not the same as a proof that it does. For a continuous accumulation, opposite signs at two endpoints do guarantee a zero between them by the Intermediate Value Theorem. In graph-based problems, continuity is appropriate, for example, when the rate is continuous or piecewise continuous on the interval; in particular, the accumulation from such a rate varies continuously with its endpoint.
In the exact-area example, \(A(5)=5>0\) and \(A(9)=-3<0\), and the accumulation is continuous on \([5,9]\). The Intermediate Value Theorem therefore guarantees a zero between those endpoints. The area calculation locates it exactly at \(x=7.5\). By contrast, an estimated endpoint value such as \(-1\), without a guaranteed actual sign, cannot be used as the negative endpoint value in this theorem.
Common Pitfall: Confusing Direction with Sign
A common error is to infer the sign of the accumulation from the sign of the rate. For instance, suppose \(A(0)=-3\) and \(f(x)=1\) on \([0,1]\). The positive rate means that \(A\) rises, but the contribution over the interval is only
Hence \(A(1)=-3+1=-2\), which is still negative. The incorrect calculation “\(f(1)>0\), so \(A(1)>0\)” confuses the rate’s sign with the accumulated value’s sign. The correction is to begin with the starting value, add the signed area, and then assess the resulting total.
The same care applies when estimating a zero. A sketch suggesting that the accumulation changes from \(2\) to \(-2\) does not prove that its actual endpoint values have opposite signs. Those estimates predict a crossing. To conclude an actual zero using continuity, first establish the actual endpoint inequalities.
A Practical Reading Routine
When given a rate graph and asked about its accumulation, use this routine:
- Mark the basepoint and record the starting value, usually \(A(a)=0\).
- Divide the rate graph into regions above and below the axis.
- Use the sign of the rate to describe where the accumulation rises and falls.
- Estimate or calculate each region’s signed area and update the running total in order.
- Use that running total—not the rate’s current sign—to decide where the accumulation is positive, negative, or zero.
- If the conclusion depends on a zero between two points, distinguish an estimate from actual endpoint signs and check the continuity condition.
This process keeps the two graphs conceptually separate: the rate graph supplies the direction and the area contributions, while the accumulation graph records the running net total.
Check Your Understanding
Use signed contributions from the rate graph, and distinguish exact conclusions from estimates.
- If \(f\) is negative throughout an interval, what does that tell you about the direction of \(A\) there? Does it determine whether \(A\) is positive or negative?
- An accumulation begins at \(A(a)=0\), and the rate graph contributes \(+6\) units of area followed by \(-4\) units. What is the final accumulation, and what is its sign?
- What does a point where \(f(x)=0\) tell you, and why does it not by itself imply \(A(x)=0\)?
- A sketch estimates \(A(u)\approx 2\) and \(A(v)\approx-2\). What additional facts are needed to conclude that the actual accumulation has a zero between \(u\) and \(v\)?
- If \(A\) is rising but is currently negative, is that possible? Explain using the meaning of a positive rate.
