Difficulty: Intermediate | Prerequisites: FTC Part 1, basic geometry (areas of triangles, rectangles, circles)
An accumulation function g(x) = integral from a to x of f(t) dt collects the signed area under f from a fixed starting point. Rather than computing antiderivatives algebraically, this topic asks you to read a graph of f and use geometry (triangles, rectangles, semicircles) to evaluate g at specific points, determine where g is increasing or decreasing, and locate its absolute extrema. This bridges the abstract FTC with visual, geometric reasoning and is a staple of Calc II exams.
When g is defined as the integral of f from a constant to x, the value of g at any point is the net signed area under f's graph. g increases wherever f is positive, decreases wherever f is negative, and its critical points occur where f crosses zero.
Accumulation function
A function defined by g(x) = integral from a to x of f(t) dt. It accumulates the signed area under f starting from a fixed point a. Think of it as a running total of area as you slide x along the number line.
Signed area
Area above the x-axis counts as positive; area below the x-axis counts as negative. When you compute g(x) geometrically, regions below the axis subtract from the total.
Critical point of g
A value of x where g'(x) = f(x) = 0 or where f is undefined. Since g'(x) = f(x) by FTC1, the critical points of g are exactly the zeros of f.
Absolute maximum on a closed interval
The largest value of g(x) on a closed interval [a, b]. By the Extreme Value Theorem, you find it by checking all critical points inside the interval plus the two endpoints, then comparing values.
When f is given as a graph of line segments and circular arcs, compute g(x) = integral from a to x of f(t) dt by finding the area of each geometric region between a and x.
Triangles: area = (1/2)(base)(height)
Rectangles: area = (base)(height)
Semicircles: area = (1/2)(pi)(r^2)
Regions below the x-axis contribute negative area
Worked Example (PS1, 2a): g(x) = integral from 2 to x of f(t) dt
To find g(3): integrate f from 2 to 3. The region forms a triangle with base 1 and height 3 sitting atop a rectangle of base 1 and height 1/2. Reading from the graph: g(3) = (1)(3)(1/2) + 1/2 = 7/2.
To find g(0): integrate from 2 to 0, which reverses direction. g(0) = -integral from 0 to 2 of f(t) dt. The region from 0 to 2 forms a rectangle of base 2 and height 4 times 1/2, giving g(0) = -(2 times 4)(1/2) = -4.
To find g(-4): integrate from 2 to -4. Reverse and split: g(-4) = -integral from -4 to 2 of f(t) dt = -(integral from -4 to 0 + integral from 0 to 2). Compute each piece using the graph's geometry. The problem set solution gives g(-4) = 12 - 2pi (accounting for semicircular arcs below the axis).
Since g'(x) = f(x):
g is increasing wherever f(x) > 0 (the graph of f is above the x-axis)
g is decreasing wherever f(x) < 0 (the graph of f is below the x-axis)
Read the intervals directly off the graph of f. In the problem set, f is positive on (-6, -17/4) union (0, 5), so g is increasing on those intervals.
On a closed interval [a, b], the absolute maximum of g occurs at one of:
The endpoints a and b
The critical points where f(x) = 0 (i.e. where f changes sign)
Procedure:
Identify all x-values in [a, b] where f crosses zero
Compute g at each of those points and at both endpoints using geometric area
Compare all values; the largest is the absolute maximum
In the problem set, on [-6, 7], the candidates are the endpoints and the zeros of f. After computing, g(5) = 7.5 is the largest, so x = 5 is the location of the absolute maximum.
Students confuse the graph of f with the graph of g. The graph you are given is f, and g is its integral. g's value at a point is the area under f, not the height of f.
When the lower limit of integration is not zero, students sometimes compute area starting from zero anyway. The lower limit a is the starting point; g(a) = 0, and you measure area from there.
Students forget to negate the integral when the upper limit is less than the lower limit. Integral from 2 to 0 of f(t) dt = negative integral from 0 to 2 of f(t) dt.
When finding absolute extrema, students check only where f = 0 and forget the endpoints. On a closed interval, the endpoints must always be checked.
Graph-based accumulation problems are a favourite on Calc II exams because they test conceptual understanding, not just computation.
You will almost certainly be asked to identify where g is increasing or decreasing by reading the sign of f from a graph. Have the logic g'(x) = f(x) thoroughly internalised.
The absolute-maximum question tests whether you remember the Extreme Value Theorem procedure: critical points plus endpoints.
Geometric area calculations (especially with semicircles and piecewise-linear graphs) require careful attention to signs and to the direction of integration.
True or False: g(a) = 0 when g(x) = integral from a to x of f(t) dt. (True)
Fill in the blank: g is increasing on intervals where f is ____. (positive)
True or False: If f(x) < 0 on [3, 5], then g(5) < g(3). (True)
Fill in the blank: To find the absolute max of g on [a, b], check all ____ and both ____. (critical points; endpoints)
Q: If f is positive on (0, 4) and negative on (4, 7), where does g (defined from 0) have a local maximum?
A: At x = 4, because g'(x) = f(x) changes from positive to negative there.
Q: Suppose g(x) = integral from 1 to x of f(t) dt and the graph of f shows a semicircle of radius 2 below the x-axis from x = 1 to x = 5. What is g(5)?
A: The area of a semicircle of radius 2 is (1/2)(pi)(4) = 2 pi. Since it sits below the axis, g(5) = -2 pi.
Q: Why is g(a) = 0 always true for g(x) = integral from a to x of f(t) dt?
A: Because the integral from a to a of any function is zero; there is no width to accumulate area over.
Q: Can g have an absolute maximum at a point where f is negative?
A: Yes, at an endpoint. f being negative means g is decreasing there, but the endpoint could still be the highest value on the interval if g decreased from that endpoint onward.
This topic is a direct application of the FTC differentiation covered in Problem 1. The relationship g'(x) = f(x) is the same formula, used here in reverse: instead of differentiating, you reason about g's behaviour from f's graph.
Accumulation functions connect forward to the study of area between curves and to applications of integration (volumes, work, average value), all of which require computing definite integrals and reasoning about signed area.
Accumulation function, signed area, net area, integral from graph, geometric integration, FTC applications, increasing decreasing from derivative sign, absolute maximum closed interval, Extreme Value Theorem, Calc 2 NYU, area under piecewise graph, integral of graph