Interpreting the Behavior of Accumulation Functions Involving Area

3. Visual Connection

5. Concept Development

Begin with the accumulation definition

Let \(G(x)=\int_a^x f(t)\,dt\). The input \(x\) determines how far accumulation extends from the fixed base point \(a\). Every conclusion about \(G\) should be traced to the values or behavior of \(f\).

The value of G is signed area

To find \(G(c)\), add signed areas from \(a\) to \(c\). Regions above the axis add and regions below subtract. This accumulated value is not the graph height \(f(c)\).

The base point anchors the graph

Every basic accumulation function satisfies \(G(a)=0\), giving a known point \((a,0)\). Changing the base point changes the vertical placement of the graph but not its derivative.

The integrand is the derivative graph

The height of \(f\) is the slope of \(G\), while the slope of \(f\) controls the concavity of \(G\).

The sign of f controls direction

The accumulation function increases where \(f>0\) and decreases where \(f<0\). Whether \(f\) itself is rising or falling does not determine this direction.

Critical points come from zeros and breaks of f

Candidates for critical numbers of \(G\) occur where \(f=0\) or where \(f\), and hence \(G'\), is undefined while \(G\) exists. Partition the domain at these inputs.

Classify local extrema with sign changes

A positive-to-negative change in \(f\) gives a local maximum of \(G\). A negative-to-positive change gives a local minimum. Touching the axis without changing sign creates no extremum.

Absolute extrema require accumulated values

On a closed interval, compare \(G\) at endpoints and every critical number. The greatest and least signed-area totals determine absolute extrema; the tallest point of \(f\) does not.

Monotonicity of f controls concavity

The function \(G\) is concave up where \(f\) increases and concave down where \(f\) decreases. This follows from \(G''=f'\).

Inflection points come from behavior changes in f

A change from increasing to decreasing or decreasing to increasing in \(f\) creates a concavity change in \(G\), provided the required continuity holds. A zero of \(f\) usually concerns an extremum of \(G\), not an inflection point.

Four sign-and-shape combinations

• \(f>0\) and increasing: \(G\) increases and is concave up.
• \(f>0\) and decreasing: \(G\) increases and is concave down.
• \(f<0\) and increasing: \(G\) decreases and is concave up.
• \(f<0\) and decreasing: \(G\) decreases and is concave down.

Value, slope, and concavity are different

The sign of \(G(x)\) describes accumulated net area. The sign of \(f(x)=G'(x)\) describes current direction. The monotonicity of \(f\) describes how the slope of \(G\) changes.

Sketch G systematically

Anchor \(G(a)=0\), mark zeros and breaks of \(f\), accumulate signed areas for key values, then connect them using slopes from \(f\) and concavity from the monotonicity of \(f\).

Different base points create vertical translations

If \(A(x)=\int_a^x f(t)\,dt\) and \(B(x)=C+\int_b^x f(t)\,dt\), then \(A'=B'=f\). Their difference is constant, so the graphs have identical shape.

Discontinuities require care

A jump in \(f\) can create a corner in \(G\). The accumulation function may remain continuous while failing to be differentiable at that input, so inspect one-sided slopes.

A reliable AP reasoning routine

Use signed area for values, the sign of \(f\) for direction and extrema, and the monotonicity of \(f\) for concavity and inflection. State every conclusion in terms of \(G\).

Common errors

Frequent errors include setting \(G=f\), claiming \(G\) increases because \(f\) increases, classifying every zero of \(f\) as an extremum, and comparing heights of \(f\) instead of accumulated values for absolute extrema.

6. Detailed Worked Example and Error Check

Example 1: Complete analytical analysis. Let \(G(x)=\int_0^x(t^2-4)\,dt\). Since \(G'(x)=x^2-4\), \(G\) increases on \(( -\infty,-2)\) and \((2,\infty)\), and decreases on \((-2,2)\). It has a local maximum at \(x=-2\) and a local minimum at \(x=2\). Since \(G''(x)=2x\), it is concave down for \(x<0\), concave up for \(x>0\), and has an inflection point at \((0,0)\).

Example 2: Read behavior from a graph description. Suppose \(f>0\) on \((-3,1)\), \(f<0\) on \((1,4)\), \(f\) increases on \((-3,-1)\), decreases on \((-1,3)\), and increases on \((3,4)\). Then \(G(x)=\int_{-3}^x f(t)\,dt\) has a local maximum at \(x=1\), is concave up on \((-3,-1)\) and \((3,4)\), and concave down on \((-1,3)\), with inflection points at \(x=-1,3\).

Example 3: Compare accumulated values. Suppose \(f\) is positive on \((-2,0)\), negative on \((0,2)\), and positive on \((2,4)\), with signed areas \(3,-5,4\). For \(G(x)=\int_{-2}^x f(t)\,dt\),

The absolute maximum on \([-2,4]\) is \(3\) at \(x=0\), and the absolute minimum is \(-2\) at \(x=2\).

Example 4: Base-point changes. Suppose \(\int_0^2 f(t)\,dt=3\), \(A(x)=\int_0^x f(t)\,dt\), and \(B(x)=7+\int_2^x f(t)\,dt\). Then

Both have derivative \(f\), and \(B\) is a four-unit upward translation of \(A\).

Example 5: Positive but decreasing integrand. If \(f>0\) and \(f'<0\), then \(G'=f>0\) and \(G''=f'<0\). Thus \(G\) increases and is concave down: its values rise at a decreasing rate.

Example 6: Negative but increasing integrand. If \(f<0\) and \(f'>0\), then \(G\) decreases and is concave up. Its negative slopes become less steep.

Example 7: A zero without an extremum. If \(f(c)=0\) but \(f\) is positive on both sides, then \(G'(c)=0\) while \(G\) increases through \(c\), so no extremum occurs. If \(f\) changes from decreasing to increasing there, \(G\) instead has an inflection point.