AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/6/6-5)
Unit 6 · Topic 6.5
6.5 Interpreting the Behavior of Accumulation Functions Involving Area
If g(x) = ∫ₐˣ f(t) dt, then g′ = f and g″ = f′. That lets you read everything about g, including where it increases, where its extrema are and where it's concave up, straight from the graph of f, and find values of g by adding signed areas.
Key terms
- accumulation function
- increasing and decreasing
- relative extrema
- concavity
- signed area
The key translation: f is g's derivative
When g(x) = ∫ₐˣ f(t) dt and f is continuous, the Fundamental Theorem gives g′(x) = f(x). So the graph you're handed (f) is the derivative of the function you're asked about (g). Every derivative test from Unit 5 now applies, with f playing the role of g′.
| On the graph of f you see… | So g is… |
|---|---|
| f above the t-axis (f > 0) | increasing |
| f below the t-axis (f < 0) | decreasing |
| f crosses from + to − | at a relative maximum |
| f crosses from − to + | at a relative minimum |
| f increasing (graph of f rising) | concave up |
| f decreasing (graph of f falling) | concave down |
| f has a local max or min (f changes direction) | at a point of inflection |
Finding values of g with areas
g(x) is the signed area from a to x. For x to the right of a, add areas above the axis and subtract areas below it. For x to the left of a, the integral runs backward, so the signs flip: g(x) = ∫ₐˣ f(t) dt = −∫ₓᵃ f(t) dt.
A quick habit: write each region's signed area on your sketch before answering anything. Most parts of the question are then just adding those numbers.
Absolute extrema of g
To find the absolute maximum or minimum of g on a closed interval, use the Candidates Test (5.5): check g at each endpoint and at each critical point where f changes sign. The biggest value is the absolute max, and the smallest is the absolute min. A sign change of f gives a local extremum, but only comparing values tells you which one is absolute.
Justifications that earn points
AP scorers (the teachers who grade free-response answers) want the reason tied to f, the derivative of g. Say “g has a relative maximum at x = 2 because g′(x) = f(x) changes from positive to negative there.” Saying only “the graph goes up then down” isn't enough.
For concavity, say “g is concave up on (4, 8) because g′ = f is increasing there.” For inflection points, say “g′ = f changes from decreasing to increasing at x = 4.” Corners or vertical tangents on the graph of f don't create inflection points of g unless f actually changes from increasing to decreasing (or the reverse) there.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Reading g from a graph of f
The graph of f on 0 ≤ t ≤ 8 has three pieces: a segment from (0, 2) to (2, 0); the lower half of the circle with center (4, 0) and radius 2, running from (2, 0) down to (4, −2) and back up to (6, 0); and a segment from (6, 0) to (8, 2). Let g(x) = ∫₀ˣ f(t) dt. (a) Find g(2), g(4), g(6) and g(8). (b) Where does g have relative extrema? (c) Where is g concave up? (d) Find the absolute maximum and minimum of g on [0, 8].
Show the solutionHide the solution
- Step 1: Signed areas: triangle on [0, 2] = ½(2)(2) = +2; quarter circle on [2, 4] = −¼π(2²) = −π; quarter circle on [4, 6] = −π; triangle on [6, 8] = +2.
- Step 2: (a) g(2) = 2, g(4) = 2 − π, g(6) = 2 − 2π, g(8) = 2 − 2π + 2 = 4 − 2π.
- Step 3: (b) g′ = f. f changes from positive to negative at x = 2, so g has a relative maximum at x = 2. f changes from negative to positive at x = 6, so g has a relative minimum at x = 6.
- Step 4: (c) g″ = f′. f is decreasing on (0, 4) and increasing on (4, 8), so g is concave down on (0, 4) and concave up on (4, 8). (x = 4 is a point of inflection.)
- Step 5: (d) Candidates: g(0) = 0, g(2) = 2, g(6) = 2 − 2π ≈ −4.28, g(8) = 4 − 2π ≈ −2.28.
- Step 6: Largest is 2 and smallest is 2 − 2π.
Answer: (a) 2, 2 − π, 2 − 2π, 4 − 2π. (b) Relative max at x = 2, relative min at x = 6. (c) Concave up on (4, 8). (d) Absolute max 2 at x = 2; absolute min 2 − 2π at x = 6.
- Example 2
Trap: a lower limit that isn't the left end
Using the same f, let h(x) = ∫₂ˣ f(t) dt. Find h(0) and h(6).
Show the solutionHide the solution
- Step 1: h(6) = ∫₂⁶ f(t) dt = −π − π = −2π.
- Step 2: h(0) = ∫₂⁰ f(t) dt. The limits run backward, so h(0) = −∫₀² f(t) dt = −2.
- Step 3: Common wrong answer: h(0) = 2, from using the area without flipping the sign.
Answer: h(0) = −2 and h(6) = −2π.
Common mistakes
- Treating the graph of f as the graph of g. Ask yourself which function you're looking at before you read anything off.
- Saying g has an extremum wherever f has a maximum or minimum. g's extrema come from f's sign changes; f's own peaks and valleys give g's inflection points.
- Forgetting that areas left of the lower limit get a negative sign.
- Picking an absolute extremum from a sign chart alone, without comparing values at the endpoints.
On the exam
- This is one of the most common free-response setups. Expect to find g at a few points, g′ and g″ at a point, intervals of increase or concavity, and an absolute extremum.
- Always write g′(x) = f(x) somewhere in your answer and refer to f in every justification.
Connected topics
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Check yourself
5 questions on 6.5 Interpreting the Behavior of Accumulation Functions Involving Area. Pick an answer to see if you got it, and why.
Let g be the function defined by g(x) = ∫₀ˣ (e^(sin t) − 1.5) dt for 0 < x < 3. At what value of x does g have a relative maximum?
The function f is continuous on the closed interval [0, 8]. On [0, 4], the graph of f is the lower half of the circle of radius 2 centered at (2, 0). On [4, 8], the graph of f consists of two line segments, one from (4, 0) to (6, 2) and one from (6, 2) to (8, −2).
Let g be the function defined by g(x) = ∫₀ˣ f(t) dt for 0 ≤ x ≤ 8.
Described function
On the open interval 0 < x < 8, at what value of x does g have a relative maximum?
On which of the following intervals is the graph of g concave down?
The function f is continuous on the closed interval [−2, 6]. The graph of f consists of four line segments connecting the points (−2, 0), (0, 4), (4, −4), (5, 0) and (6, 2), in that order.
Let g be the function defined by g(x) = ∫₀ˣ f(t) dt for −2 ≤ x ≤ 6.
Described graph of a piecewise linear function
On the open interval −2 < x < 6, at what value of x does g have a relative minimum?
What is the absolute minimum value of g on the closed interval [−2, 6]?
0 of 5 answered