From an Integrand to an Accumulation Function
An accumulation function records how much signed change has built up as its upper limit moves. Its defining integral gives information about its values, while the Fundamental Theorem of Calculus connects its derivative directly to the integrand. That connection lets us analyze the function without first finding a formula for it.
For example, if \(f\) is a rate, then the integral of \(f\) over an interval measures the net accumulated change across that interval. Positive values of \(f\) add to the accumulation; negative values subtract from it. The same signed-area interpretation applies when \(f\) is simply a function rather than a contextual rate.
The lower limit \(a\) is fixed, and the upper limit \(x\) varies. In particular, \(F(a)=0\), because the integral over an interval of zero length is zero.
The Fundamental Theorem Connects \(F\) and \(f\)
If \(f\) is continuous, the Fundamental Theorem of Calculus states that the derivative of its accumulation function is the integrand evaluated at the variable endpoint.
This result explains why a graph or sign chart of \(f\) can reveal how \(F\) behaves. Where \(f(x)>0\), the derivative \(F'(x)\) is positive, so \(F\) is increasing. Where \(f(x)<0\), \(F'(x)<0\), so \(F\) is decreasing. A zero of \(f\) makes \(F'(x)=0\), but that fact alone does not determine whether \(F\) has an extremum there.
Why the Derivative Equals the Integrand
The derivative measures the change in \(F\) per unit change in its input. When \(x\) changes to \(x+h\), the extra accumulated amount is exactly the integral of \(f\) over the short interval between those inputs:
For \(h\ne 0\), divide by \(h\). As \(h\) approaches zero, the average value of \(f\) over this shrinking interval approaches \(f(x)\), because \(f\) is continuous at \(x\). This works for positive and negative \(h\): the oriented integral and the denominator both reverse sign when \(h<0\).
Thus the difference quotient for \(F\) approaches \(f(x)\), which is precisely the definition of \(F'(x)\). The continuity condition matters: it ensures that values of \(f\) close to \(x\) are close to \(f(x)\), so their average over a sufficiently short interval approaches \(f(x)\).
Worked Example: Finding Values and a Minimum
Worked Example: Finding Values and a Minimum
Let \(A(x)=\int_0^x (t-2)\,dt\) for \(0\le x\le 4\). Find \(A(0)\), \(A(2)\), and \(A(4)\), and determine where \(A\) decreases, increases, and has an absolute minimum.
First, use the Fundamental Theorem of Calculus. The integrand \(t-2\) is continuous, so
The derivative is negative for \(0\le x<2\) and positive for \(2<x\le4\). Therefore \(A\) decreases on \([0,2]\), then increases on \([2,4]\). The change from negative to positive derivative identifies a minimum at \(x=2\).
To compute values, an antiderivative of \(t-2\) is \(\frac{t^2}{2}-2t\). Therefore
Substitute the requested inputs:
The endpoint values are both \(0\), while \(A(2)=-2\). Since the function decreases before \(2\) and increases afterward, \(-2\) is its absolute minimum on the stated interval. As a check, differentiating the formula gives \(A'(x)=x-2\), the integrand as required.
Sign Changes Locate Interior Extrema
An interior critical point occurs where \(F'(x)=0\) or where \(F'\) does not exist. In this setting, when \(f\) is continuous, \(F'=f\), so zeros of the integrand are the critical points to examine. The decisive evidence for a local maximum or minimum is the sign of \(f\) on either side.
On a closed interval, the endpoints must also be considered when finding absolute extrema. An interior sign change gives local behavior, not automatically the largest or smallest value on the entire interval. Compare the function values at the endpoints and at the relevant interior critical points.
Worked Example: Sign Changes and Absolute Extrema
Worked Example: Sign Changes and Absolute Extrema
Let \(B(x)=\int_0^x (t-1)(t-4)\,dt\) on \([0,6]\). Find where \(B\) increases and decreases, locate its local extrema, and determine its absolute extrema.
The integrand factors into \((t-1)(t-4)\), so its zeros are \(t=1\) and \(t=4\). Check the sign in each interval: at \(t=0\), the product is positive; at \(t=2\), it is negative; at \(t=5\), it is positive. By the Fundamental Theorem, \(B'(x)=(x-1)(x-4)\). Thus \(B\) increases on \([0,1]\), decreases on \([1,4]\), and increases on \([4,6]\). It has a local maximum at \(1\) and a local minimum at \(4\).
For exact values, expand the integrand and integrate:
Using the lower limit \(0\), evaluate at each candidate:
Compare the values \(0\), \(\frac{11}{6}\), \(-\frac{8}{3}\), and \(6\). The absolute minimum is \(-\frac{8}{3}\) at \(x=4\); the absolute maximum is \(6\) at \(x=6\). Differentiating the antiderivative gives \(x^2-5x+4=(x-1)(x-4)\), confirming the derivative and sign analysis.
Values from Signed Area
The defining integral also gives a geometric way to find values. The value \(F(x)\) is the signed area between the graph of \(f\) and the horizontal axis from \(a\) to \(x\). Regions above the axis contribute positively; regions below contribute negatively. If \(F(a)\) is not zero, as in a shifted accumulation function, add the starting value to the signed area.
When using a graph or table, keep the accumulation at each point rather than interpreting every region as positive area. A negative region reduces the running total. This matters because extrema depend on the current value as well as on whether the function is increasing or decreasing.
Worked Example: Reading Accumulation from a Piecewise-Linear Integrand
Worked Example: Reading Accumulation from a Piecewise-Linear Integrand
Suppose \(p\) is continuous and piecewise linear through the points \((0,2)\), \((2,0)\), \((4,-2)\), and \((6,0)\). Let \(P(x)=3+\int_0^x p(t)\,dt\) on \([0,6]\). Find \(P(0)\), \(P(2)\), \(P(4)\), and \(P(6)\), and identify the absolute maximum and minimum.
By the Fundamental Theorem, \(P'(x)=p(x)\). The graph of \(p\) is above the axis on \([0,2)\) and below the axis on \((2,6)\). Therefore \(P\) increases up to \(x=2\) and decreases from \(x=2\) through \(x=6\). The derivative changes from positive to negative at \(2\), so \(P\) has a local maximum there.
Compute the signed areas. From \(0\) to \(2\), the graph forms a triangle with base \(2\) and height \(2\), giving area \(\frac12(2)(2)=2\). From \(2\) to \(4\), the triangle is below the axis, so its signed area is \(-\frac12(2)(2)=-2\). From \(4\) to \(6\), it is also below the axis, with signed area \(-\frac12(2)(2)=-2\).
The function rises from \(3\) to \(5\), then decreases continuously to \(1\). Hence its absolute maximum is \(5\) at \(x=2\), and its absolute minimum is \(1\) at \(x=6\). As a check on the accumulated values, the net signed area from \(0\) to \(6\) is \(2-2-2=-2\), so \(P(6)=3-2=1\), as found.
Common Pitfall: A Zero Does Not Guarantee an Extremum
It is tempting to conclude that \(F\) has an extremum wherever \(f\) equals zero. The zero only says that the tangent slope of \(F\) is zero at that point. For an extremum, the derivative must change sign appropriately.
For a concrete example, let \(F(x)=\int_0^x (t-2)\,dt\). A mistaken argument might say: “\(F'(2)=0\), so \(F\) has a maximum at \(x=2\).” That conclusion is wrong. For \(x<2\), \(F'(x)=x-2<0\); for \(x>2\), \(F'(x)=x-2>0\). The function changes from decreasing to increasing, so \(x=2\) is a minimum, not a maximum. Indeed, \(F(2)=-2\), while \(F(0)=F(4)=0\).
The correction is to check the integrand’s sign on both sides of each zero, then compare candidate values if the question asks for absolute extrema on a closed interval.
A Reliable Analysis Process
Record its fixed lower limit, variable upper limit, and any added starting value.
For \(F(x)=\int_a^x f(t)\,dt\), write \(F'(x)=f(x)\), provided \(f\) is continuous.
Find its zeros and determine its sign between them to establish where \(F\) increases or decreases.
Evaluate the integral or add signed areas to the starting value, keeping below-axis contributions negative.
Use derivative sign changes for interior local extrema; include endpoints when comparing absolute extrema on a closed interval.
Why This Analysis Is Useful
In an AP free-response problem, an accumulation function may be defined by an integral even when no convenient explicit formula is available. You can still make precise claims about its values, monotonicity, and extrema. The integrand supplies the derivative, while the signed integral supplies the function’s accumulated value.
In a written response, connect each conclusion to the evidence. For instance, state that the accumulation function increases on an interval because its integrand is positive there. For an interior maximum, state that the integrand changes from positive to negative. For an absolute maximum on a closed interval, compare the relevant function values, including the endpoints. These explanations distinguish a justified conclusion from a guess based only on a sketch.
Check Your Understanding
Use the Fundamental Theorem of Calculus and signed accumulation to answer each question.
- If \(G(x)=\int_2^x (3-t)\,dt\), what is \(G'(x)\), and on which intervals is \(G\) increasing and decreasing?
- If the integrand changes from negative to positive at an interior point \(c\), what kind of local extremum does the accumulation function have at \(c\)?
- For \(H(x)=\int_0^x f(t)\,dt\), how should a region below the horizontal axis contribute to \(H(x)\)?
- If \(f(c)=0\) but \(f\) is negative on both sides of \(c\), does the sign-change test establish a local extremum of \(F(x)=\int_a^x f(t)\,dt\) at \(c\)? Explain.
- When finding absolute extrema of an accumulation function on a closed interval, which points must be included in the comparison?
