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How do you analyze the increasing, decreasing and concavity behavior of an accumulation function from the graph of its integrand?

Topic 6.5 Interpreting the Behavior of Accumulation Functions Involving Area: analyze extrema and concavity of an accumulation function using the graph of the integrand.

A focused answer to AP Calculus AB Topic 6.5, analyzing the increasing/decreasing, extrema, and concavity behavior of an accumulation function from the graph of its integrand, with worked area-based reasoning.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. The translation from f to g
  3. A worked area analysis
  4. Computing values of g as areas
  5. Why g lags one level behind f
  6. Absolute extrema of an accumulation function
  7. A common multi-part exam structure

What this topic is asking

The College Board (Topic 6.5) applies the analytical toolkit of Unit 5 to an accumulation function g(x)=axf(t)dtg(x) = \int_a^x f(t)\,dt, using the graph of the integrand ff. Since g=fg' = f and g=fg'' = f', the sign and slope of ff control the behavior of gg.

The translation from f to g

A worked area analysis

Computing values of g as areas

Beyond behavior, the exam asks for values of gg at specific points, which you read as signed areas under ff. For example g(4)=04f(t)dtg(4) = \int_0^4 f(t)\,dt is the net signed area of the region between ff and the axis from 00 to 44, computed by adding areas above the axis and subtracting areas below. When ff is made of line segments and simple shapes, this is straightforward geometry. Combining value computations (signed area) with behavior analysis (sign and slope of ff) is the full version of this topic, and it appears as a multi-part free-response question almost every year.

Why g lags one level behind f

The key mental model is that gg sits one level above ff: ff plays the role of gg's first derivative, and ff' plays the role of gg's second derivative. So features you would normally read from ff' and ff'' when analyzing a function ff are instead read from ff and ff' when analyzing gg. The most common error is treating the graph of ff as if it were the graph of gg itself, for instance saying gg has a maximum where ff has a maximum. It does not: gg has a maximum where ff crosses zero from positive to negative. Keeping the level shift straight is the whole skill.

Absolute extrema of an accumulation function

To find the absolute maximum or minimum of g(x)=axf(t)dtg(x) = \int_a^x f(t)\,dt on a closed interval, treat it as a candidates-test problem on gg. The critical points of gg are where g(x)=f(x)=0g'(x) = f(x) = 0, that is, where the graph of ff crosses the axis. Evaluate gg (as a signed area) at those critical points and at the endpoints of the interval, then compare. The largest value is the absolute maximum and the smallest the absolute minimum. This ties the accumulation-function analysis directly to the Unit 5 optimization framework, with the integrand ff playing the role of gg' throughout. The only extra work is computing the candidate values of gg as signed areas rather than from a formula.

A common multi-part exam structure

These accumulation questions usually come as a multi-part free response built on one graph of ff. A typical sequence asks for a value of gg at a point (signed area), then where gg is increasing or decreasing (sign of ff), then the relative extrema of gg (sign changes of ff), and finally the concavity or a point of inflection of gg (slope of ff). Working the parts in this order, and justifying each with the correct feature of ff or ff', earns the marks systematically. The recurring pitfall across all parts is the level confusion, so before answering each part, restate which feature of ff controls the requested feature of gg.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). Let g(x)=0xf(t)dtg(x) = \int_0^x f(t)\,dt, where ff is continuous. If f(x)>0f(x) > 0 on an interval, then on that interval gg is (A) increasing (B) decreasing (C) constant (D) concave down
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The correct answer is (A), increasing.

Since g(x)=f(x)g'(x) = f(x) by the FTC, f(x)>0f(x) > 0 means g(x)>0g'(x) > 0, so gg is increasing on that interval.

AP 2024 (style)4 marksSection II (free response). Let g(x)=0xf(t)dtg(x) = \int_0^x f(t)\,dt, where the graph of ff consists of line segments: ff is positive on (0,3)(0, 3), crosses zero at x=3x = 3 (going positive to negative), and is negative on (3,5)(3, 5). (a) On what interval is gg increasing? Justify. (b) Does gg have a relative maximum or minimum at x=3x = 3? Justify.
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A 4-point accumulation-behavior question.

(a) (2 points) g(x)=f(x)g'(x) = f(x). Since f>0f > 0 on (0,3)(0, 3), g>0g' > 0 there, so gg is increasing on (0,3)(0, 3).
(b) (2 points) At x=3x = 3, g(x)=f(x)g'(x) = f(x) changes from positive to negative, so by the First Derivative Test gg has a relative maximum at x=3x = 3.

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