Reading a Derivative From an Integrand
An accumulation function records the signed accumulation of an integrand from a fixed starting point. Once the integrand is shown as a graph or table, the Fundamental Theorem of Calculus can often give the derivative of the accumulation function without requiring us to calculate the accumulated amount itself.
The essential reading rule is simple: at a point where the integrand is continuous, the derivative of the accumulation function is the integrand’s value there. But interpreting a graph or table carefully also means checking continuity and distinguishing what the value at one point tells us from what the sign over a whole interval tells us.
The lower limit \(a\) determines where accumulation starts, but it does not affect this derivative rule. The derivative at \(c\) is determined by the integrand’s value at \(c\), provided the continuity condition is met.
What a Graph or Table Tells You
Suppose the graph is the graph of \(y=f(t)\), not the graph of \(A\). At \(t=c\), find the graph’s vertical coordinate. If \(f\) is continuous at \(c\), that coordinate is \(A'(c)\). The horizontal coordinate identifies the input; it is the vertical height that gives the derivative.
Once the derivative’s sign is known, it describes the accumulation function’s direction of change. A positive derivative means \(A\) is rising at that point, and a negative derivative means it is falling at that point. To conclude that \(A\) is increasing or decreasing throughout a neighborhood or interval, use the sign of \(f\) throughout that region—not just its value at one isolated point.
| Information about \(f\) | Conclusion about \(A\) |
|---|---|
| \(f(c)>0\), with \(f\) continuous at \(c\) | \(A'(c)>0\); continuity also makes \(f\) positive near \(c\), so \(A\) increases nearby. |
| \(f(c)<0\), with \(f\) continuous at \(c\) | \(A'(c)<0\); continuity also makes \(f\) negative near \(c\), so \(A\) decreases nearby. |
| \(f(c)=0\), with \(f\) continuous at \(c\) | \(A'(c)=0\); this alone does not determine whether \(A\) increases, decreases, or has a local extremum. |
A graph that shows the integrand across an interval can provide the sign information needed to describe increasing or decreasing behavior there. A table of a few point values usually gives only the integrand values at those points. Unless the table also supplies information about continuity and signs between them, it does not by itself establish how \(A\) behaves throughout the intervening intervals.
Why the Derivative Rule Works
The derivative of \(A\) at \(c\) is defined by a difference quotient. The change in accumulation between \(c\) and a nearby point \(c+h\) is the integral of \(f\) over that short interval. Dividing by \(h\) gives the average value of \(f\) over the interval; continuity ensures that this average approaches \(f(c)\).
For \(h\ne 0\), subtract the two accumulation values. Additivity of integrals gives \(A(c+h)-A(c)=\int_c^{c+h}f(t)\,dt\).
The difference between the difference quotient and \(f(c)\) is the average of \(f(t)-f(c)\) over the short interval.
As \(h\) approaches zero, continuity at \(c\) makes every value \(f(t)\) on the shrinking interval as close to \(f(c)\) as desired. Thus the average difference tends to zero.
More explicitly, continuity at \(c\) says that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(|f(t)-f(c)|<\varepsilon\) whenever \(|t-c|<\delta\). For \(0<|h|<\delta\), the absolute difference between the quotient and \(f(c)\) is bounded as follows:
This proves that the difference quotient tends to \(f(c)\), so \(A'(c)=f(c)\). The argument works for positive and negative \(h\); the absolute value handles either orientation.
Worked Example: Reading a Continuous Graph
Worked Example: Reading a Continuous Graph
Let \(A(x)=\int_0^x f(t)\,dt\). A graph shows that \(f\) is continuous at \(x=2\), that \(f(2)=3\), and that \(f(t)>0\) throughout the interval \(1.5<t<2.5\). Find \(A'(2)\) and describe the behavior of \(A\) near \(2\).
Because \(f\) is continuous at \(2\), the accumulation derivative rule applies:
The derivative is positive. In addition, the graph gives the stronger interval information \(f(t)>0\) for \(1.5<t<2.5\). For any \(x_1,x_2\) with \(1.5<x_1<x_2<2.5\),
Therefore \(A\) is increasing on that interval, in particular near \(2\). Check: the derivative value read from the graph is \(3\), which is positive and agrees with the increasing behavior indicated by the positive graph heights around \(2\).
Worked Example: Using a Table With Continuity Information
Worked Example: Using a Table With Continuity Information
Let \(A(x)=\int_0^x f(t)\,dt\). The table gives values of \(f\), and the accompanying information states that \(f\) is continuous at each listed input.
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(f(x)\) | 4 | 0 | \(-3\) |
At \(x=1\), continuity allows the rule to be applied, so \(A'(1)=f(1)=4\). The derivative is positive; since \(f\) is continuous at \(1\), it remains positive in some neighborhood of \(1\), and \(A\) increases there.
At \(x=2\), \(A'(2)=f(2)=0\). This gives a horizontal tangent, but the table entry alone does not establish whether \(A\) has a local maximum or minimum. Suppose the graph or additional table information also shows that \(f\) is positive just to the left of \(2\) and negative just to the right. Then \(A\) increases on the left and decreases on the right, so \(A\) has a local maximum at \(2\).
At \(x=3\), continuity and the table give \(A'(3)=f(3)=-3\). The derivative is negative, and \(A\) decreases in a neighborhood of \(3\).
Check: substituting each table value into \(A'(x)=f(x)\) gives \(A'(1)=4\), \(A'(2)=0\), and \(A'(3)=-3\). The positive, zero, and negative signs agree respectively with increasing behavior near \(1\), a stationary point at \(2\), and decreasing behavior near \(3\). The local maximum conclusion at \(2\) uses the stated sign change, not merely the zero in the table.
Worked Example: A Zero and a Sign Change
Worked Example: A Zero and a Sign Change
Let \(B(x)=\int_{-1}^x g(t)\,dt\). A graph shows that \(g\) is continuous at \(0\), that \(g(0)=0\), that \(g(t)<0\) for \(-0.5<t<0\), and that \(g(t)>0\) for \(0<t<0.5\). Determine \(B'(0)\) and the local behavior of \(B\).
Since \(g\) is continuous at \(0\),
For \(-0.5<x_1<x_2<0\), the integrand is negative throughout the interval, so \(B(x_2)-B(x_1)=\int_{x_1}^{x_2}g(t)\,dt<0\). Thus \(B\) decreases to the left of \(0\). For \(0<x_1<x_2<0.5\), the integrand is positive, so \(B(x_2)-B(x_1)>0\). Thus \(B\) increases to the right of \(0\).
Consequently, \(B\) has a local minimum at \(0\). Check: the derivative at \(0\) is zero, and the sign pattern changes from negative to positive, matching decrease followed by increase. The zero derivative alone would not have been enough to reach this conclusion.
What Point Values Cannot Establish
A crucial condition in the derivative rule is continuity of the integrand at the point where the derivative is sought. A table that lists \(f(c)\) does not, by itself, establish continuity there. Nor does a statement that the graph lies below the axis establish that the listed point value is the derivative unless continuity at that point is also known.
Worked Example: Why Continuity at the Point Matters
Define \(f(t)=-1\) for every \(t\), except that \(f(1.5)=-2\), and let \(A(x)=\int_0^x f(t)\,dt\). The integrand is integrable, but it is not continuous at \(1.5\).
Changing a bounded integrand at one point does not change its integral. Therefore, for every \(x\), \(A(x)=-x\). Differentiating this explicit expression gives \(A'(x)=-1\), so \(A'(1.5)=-1\), not \(-2\).
The tempting but incorrect calculation is \(A'(1.5)=f(1.5)=-2\). It misapplies the rule because continuity at \(1.5\) was not established. The correction is to check the hypothesis before reading the derivative from the point value. In this example, direct differentiation of \(A(x)=-x\) verifies that the derivative is \(-1\).
Using Signs to Describe Change
For \(x_1<x_2\), additivity of the definite integral gives \(A(x_2)-A(x_1)=\int_{x_1}^{x_2}f(t)\,dt\). This identity explains why the sign of the integrand over an interval controls the change in accumulation. If \(f\) is continuous and positive throughout an interval, every subinterval has positive integral, so \(A\) is increasing there. If \(f\) is continuous and negative throughout, \(A\) is decreasing there. If \(f\) is zero at one point, inspect the signs on either side before claiming a local maximum or minimum.
At an endpoint of the domain of \(A\), interpret a derivative using a one-sided limit from within the domain. For example, at the left endpoint \(a\), the relevant difference quotient uses \(h>0\). If \(f\) is continuous from within the interval at \(a\), the right-hand derivative is \(f(a)\). An ordinary two-sided derivative is not defined at a domain endpoint unless the function is also defined on a surrounding open interval.
Check Your Understanding
Use the accumulation derivative rule carefully, and distinguish pointwise information from sign information over an interval.
- If \(A(x)=\int_2^x f(t)\,dt\), \(f\) is continuous at \(4\), and \(f(4)=-5\), what is \(A'(4)\), and what does its sign say?
- A table gives \(f(1)=0\) and says \(f\) is continuous at \(1\). Does this alone prove that the accumulation function has a local maximum or minimum at \(1\)? Explain.
- If a continuous integrand is positive throughout \((3,5)\), what can be concluded about the accumulation function on that interval?
- Why is it not valid to infer \(A'(c)=f(c)\) from a table entry when continuity at \(c\) is not known?
- If the integrand changes from negative to positive across a point where it is continuous and zero, what local behavior does the accumulation function have there?
