Why the Integrand Reveals Local Extrema
An accumulation function records how a rate has added up as its upper endpoint moves. When the rate is positive, the accumulated quantity increases; when the rate is negative, the accumulated quantity decreases. A change from positive to negative therefore signals a possible local maximum, while a change from negative to positive signals a possible local minimum.
The key is to connect the integrand’s sign to the derivative of the accumulation function. Once that connection is established, a sign chart gives a reliable way to locate and classify local extrema.
By the Fundamental Theorem of Calculus, \(A\) is differentiable at interior points where \(f\) is continuous, and \[ A'(x)=f(x). \] Thus the sign of \(f\) is the sign of the derivative of \(A\). This is the reason the integrand can be used to classify extrema.
From Integrand Signs to Increasing and Decreasing
Because \(A'(x)=f(x)\), a positive integrand means \(A'(x)>0\), so \(A\) is increasing there. A negative integrand means \(A'(x)<0\), so \(A\) is decreasing there. The sign change at a candidate point determines whether the function changes direction.
| Sign of \(f=A'\) to the left | Sign of \(f=A'\) to the right | Behavior of \(A\) | Classification |
|---|---|---|---|
| Positive | Negative | Increasing, then decreasing | Local maximum |
| Negative | Positive | Decreasing, then increasing | Local minimum |
| Positive | Positive | Increasing on both sides | No local extremum |
| Negative | Negative | Decreasing on both sides | No local extremum |
A point where \(f(c)=0\) is a candidate for an interior extremum because \(A'(c)=0\). But being a candidate is not enough: the sign of \(f\) must change in the appropriate way. A zero where the sign stays the same does not turn increasing behavior into decreasing behavior or vice versa.
Why the Sign-Change Test Works
The connection between derivative signs and function behavior follows from the Mean Value Theorem. Suppose \(A'(x)>0\) throughout an interval. For any two points \(u<v\) in that interval, the Mean Value Theorem gives a point \(r\) between them such that \[ A(v)-A(u)=A'(r)(v-u). \] Both factors on the right are positive, so \(A(v)-A(u)>0\). Therefore \(A(v)>A(u)\), and \(A\) is increasing. If \(A'(x)<0\), the same equation gives \(A(v)-A(u)<0\), so \(A\) is decreasing.
Now suppose \(f\) is positive just to the left of \(c\) and negative just to the right. Since \(A'=f\), the preceding argument shows that \(A\) increases toward \(c\) from the left and decreases after \(c\). Values near \(c\) on either side are smaller than \(A(c)\), so \(A\) has a local maximum at \(c\). Reversing the signs proves the local-minimum case: \(A\) decreases toward \(c\), then increases away from it, so nearby values are greater than \(A(c)\).
For a continuous integrand, locate the interior points where \(f(x)=0\).
Use a sign chart or test values in the intervals separated by the candidates.
Positive-to-negative gives a local maximum; negative-to-positive gives a local minimum. No sign change gives no local extremum by this test.
The integrand’s signs classify the extrema; the initial value and accumulated integral determine the height \(A(c)\).
Worked Examples
Worked Example: A Local Maximum and a Local Minimum
Let \[ A(x)=2+\int_0^x (t-1)(t-3)\,dt,\qquad -1\leq x\leq 4. \] Find and classify the interior local extrema, and calculate their values.
Step 1: Differentiate. The integrand is continuous, so the Fundamental Theorem of Calculus gives \[ A'(x)=(x-1)(x-3). \]
Step 2: Find the candidates and signs. The derivative is zero when \(x=1\) or \(x=3\). For \(x<1\), both factors are negative, so \(A'(x)>0\). For \(1<x<3\), the first factor is positive and the second is negative, so \(A'(x)<0\). For \(x>3\), both factors are positive, so \(A'(x)>0\). Thus \(A\) increases, then decreases, then increases. It has a local maximum at \(x=1\) and a local minimum at \(x=3\).
Step 3: Calculate the accumulation values. Expanding the integrand gives \[ (t-1)(t-3)=t^2-4t+3. \] An antiderivative is \(\frac{t^3}{3}-2t^2+3t\). Therefore, \[ A(1)=2+\left(\frac{1}{3}-2+3\right)=2+\frac{4}{3}=\frac{10}{3}, \] and \[ A(3)=2+\left(\frac{27}{3}-2(9)+9\right) =2+(9-18+9)=2. \]
Check: Differentiating the accumulation function gives \(A'(x)=(x-1)(x-3)\), whose sign changes positive-to-negative at \(1\) and negative-to-positive at \(3\). This confirms the classifications. The extrema are a local maximum \(\left(1,\frac{10}{3}\right)\) and a local minimum \((3,2)\).
Worked Example: A Zero That Is Not an Extremum
Let \[ A(x)=5+\int_0^x (t+2)^2(t-1)\,dt,\qquad -3\leq x\leq 2. \] Find and classify the interior extrema. In particular, decide whether \(x=-2\) is an extremum.
Step 1: Differentiate and find candidates. The integrand is continuous, and \[ A'(x)=(x+2)^2(x-1). \] The derivative is zero at \(x=-2\) and \(x=1\).
Step 2: Inspect the signs. The factor \((x+2)^2\) is positive except at \(x=-2\), and it never changes sign. The factor \(x-1\) is negative for \(x<1\) and positive for \(x>1\). Therefore \(A'(x)<0\) on both sides of \(-2\), so \(A\) is decreasing across that point and has no extremum there. At \(x=1\), \(A'\) changes from negative to positive, so \(A\) has a local minimum.
Step 3: Calculate the values. Expanding, \[ (t+2)^2(t-1)=(t^2+4t+4)(t-1)=t^3+3t^2-4. \] An antiderivative is \(\frac{t^4}{4}+t^3-4t\). Hence \[ A(-2)=5+\left(\frac{16}{4}-8+8\right)=5+4=9 \] and \[ A(1)=5+\left(\frac{1}{4}+1-4\right) =5-\frac{11}{4}=\frac{9}{4}. \]
Check: Differentiating the displayed antiderivative gives \(t^3+3t^2-4=(t+2)^2(t-1)\). The derivative remains negative across \(-2\), but changes from negative to positive at \(1\). Thus \(x=-2\) is not an extremum, and the local minimum is \(\left(1,\frac{9}{4}\right)\).
Worked Example: Classifying Extrema from a Piecewise-Linear Integrand
Suppose \(f\) is defined on \([-2,2]\) by \[ f(x)= \begin{cases} -2x-2, & -2\leq x\leq 0,\\ 2x-2, & 0\leq x\leq 2, \end{cases} \qquad A(x)=\int_{-2}^x f(t)\,dt. \] Find and classify the interior local extrema, and calculate the values of \(A\) there.
Step 1: Identify the integrand’s zeros and signs. On the first piece, \(-2x-2=0\) when \(x=-1\). On the second piece, \(2x-2=0\) when \(x=1\). The graph described by these formulas is above the axis on \((-2,-1)\), below it on \((-1,1)\), and above it on \((1,2)\). Since \(A'(x)=f(x)\), \(A\) increases, then decreases, then increases. There is a local maximum at \(-1\) and a local minimum at \(1\).
Step 2: Evaluate \(A(-1)\). From \(-2\) to \(-1\), \(f\) forms a triangle above the axis with base \(1\) and height \(2\). Its signed area is \[ \frac{1}{2}(1)(2)=1. \] Thus \(A(-1)=1\).
Step 3: Evaluate \(A(1)\). From \(-2\) to \(-1\), the signed area is \(1\). From \(-1\) to \(1\), the graph forms a triangle below the axis with base \(2\) and height \(2\), so its signed area is \[ -\frac{1}{2}(2)(2)=-2. \] Adding the signed areas gives \(A(1)=1+(-2)=-1\).
Check: The integrand is continuous, including at \(x=0\), where both formulas give \(-2\). Therefore \(A'=f\) throughout the interior. The sign changes confirm the local maximum \((-1,1)\) and the local minimum \((1,-1)\).
Common Pitfall: Treating Every Zero as an Extremum
A common incorrect conclusion is: “The derivative is zero at \(x=-2\), so the accumulation function has a local maximum or minimum there.” The calculation \(A'(-2)=(-2+2)^2(-2-1)=0\) is correct, but the conclusion does not follow. In the second example, \(A'\) is negative immediately to both the left and right of \(-2\). The function decreases on both sides, so there is no change of direction and no local extremum there.
The correction is to check the sign on both sides of every candidate, not just substitute the candidate into the derivative. Also keep the direction straight: positive \(f\) means positive \(A'\), so \(A\) increases; negative \(f\) means \(A\) decreases. Reversing this connection reverses the maximum and minimum classifications.
What This Test Establishes
The sign-change method classifies local behavior at interior candidates under the stated continuity condition. It does not, by itself, determine whether a local extremum is also the largest or smallest value on an entire interval. That broader comparison requires considering the domain and the values of the function at other relevant points. At an endpoint, “local” behavior is relative to the portion of the domain that lies next to that endpoint; the interior sign-change test above concerns points with domain on both sides.
For accumulation problems, this method is useful whenever the rate or integrand is easier to analyze than the accumulated function itself. A sign chart can reveal where the accumulated quantity turns around without first finding a formula for its full values. If the actual height of an extremum is requested, use the initial value and the signed accumulation up to that point, as in the examples.
Check Your Understanding
Use the relationship between an accumulation function’s derivative and its integrand to answer each question.
- If \(f\) changes from positive to negative at an interior point \(c\), what kind of local extremum does \(A(x)=C+\int_a^x f(t)\,dt\) have at \(c\)?
- If \(f(c)=0\) but \(f\) is negative on both sides of \(c\), does the sign-change test identify a local extremum there? Explain.
- For \(A(x)=4+\int_0^x (t+1)(t-2)\,dt\), identify the interior candidates and classify them from the integrand’s sign.
- Why does \(f(x)>0\) on an interval imply that \(A\) is increasing there?
- What additional work is needed if a problem asks for the value, not just the location, of a local extremum?
