6.5 Interpreting the Behavior of Accumulation Functions Involving Area

If g(x) = ∫_a^x f(t) dt, use the Fundamental Theorem of Calculus (Part 1): - Differentiation: g′(x) = f(x). So g is an antiderivative of f and is continuous/differentiable wherever f is. - Evaluation: if F is any antiderivative of f (F′ = f), then g(x) = F(x) − F(a). In particular g(a) = 0. - Graphically: g(x) equals the net (signed) area under f from a to x. Where f > 0, g increases; where f < 0, g decreases. Where f changes sign you get local extrema of g. - Use in problems: to find g(x) explicitly, find an antiderivative F(x) of f and compute F(x) − F(a). To analyze behavior (increasing/concave/inflection), use f for slope and f′ for concavity of g. This is exactly Topic 6.5 material—see the study guide for examples (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and lots of practice problems at (https://library.fiveable.me/practice/ap-calculus).

f(x) is the original function (often a rate), while g(x) = ∫_a^x f(t) dt is the accumulation (net area) of f from a to x. Key differences and links (AP CED terms): - Role: f(t) is the integrand / rate of change; g(x) is an accumulation function (signed area, net accumulation) with initial value g(a)=0. - Derivative/Fundamental Theorem: g′(x)=f(x). So g’s slope at x equals f(x). (FUN-5.A) - Second derivative/concavity: g″(x)=f′(x). So where f′>0, g is concave up. - Increasing/decreasing: If f(x)>0 on an interval, g increases there; if f(x)<0, g decreases. Zeros of f are critical points of g (possible extrema). - Units/example: If f is “meters/second,” then g is “meters” traveled. If f(x)=2, g grows at rate 2 so g(x)=2(x−a). - Smoothness: If f is continuous, g is differentiable everywhere. This is exactly what AP asks you to interpret graphically and verbally (see Topic 6.5 study guide for examples and practice) (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr). For more review across Unit 6 and thousands of practice problems, check (https://library.fiveable.me/ap-calculus/unit-6) and (https://library.fiveable.me/practice/ap-calculus).

Use the Fundamental Theorem of Calculus Part 1 whenever your function is defined as an accumulation (definite integral) with a variable limit. Key cases: - g(x) = ∫_a^x f(t) dt. If f is continuous at x, then g′(x) = f(x). (g measures net/signed area; derivative = instantaneous accumulation rate.) - g(x) = ∫_a^{h(x)} f(t) dt. Use the chain rule: g′(x) = f(h(x))·h′(x). - g(x) = ∫_{u(x)}^{v(x)} f(t) dt. Then g′(x) = f(v(x))·v′(x) − f(u(x))·u′(x). Make sure the integrand f is continuous on the interval (this guarantees g is differentiable). Graphically: g increases where f > 0, decreases where f < 0, and g′(x) equals the height of f at x (with chain-rule adjustments if limits aren’t just x). This is exactly what Topic 6.5 / FUN-5.A in the CED tests—practice these forms on the study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and try more problems at (https://library.fiveable.me/practice/ap-calculus).

Think of g(x) = ∫_a^x f(t) dt as the net area accumulated from a to x. If you increase x by a tiny amount h, the extra area added is approximately the height f(x) times the width h (because over a very small interval f is nearly constant). So g(x + h) − g(x) ≈ f(x)·h. Divide by h and take the limit as h → 0: g′(x) = lim_{h→0} [g(x+h)−g(x)]/h = f(x). This is the Fundamental Theorem of Calculus, Part I: when f is continuous, the accumulation function g(x)=∫_a^x f(t) dt is differentiable and its derivative is the integrand f(x). Intuitively, f(x) is the instantaneous rate at which area (or any accumulated quantity) is changing—exactly what a derivative measures. This is a key AP idea in Topic 6.5 (see the study guide) (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr). For more practice on interpreting accumulation and rates, check the Unit 6 resources (https://library.fiveable.me/ap-calculus/unit-6) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Think of g(x) = ∫_a^x f(t) dt as “net area accumulated from a to x.” Use these quick rules from the FTC and the CED keywords: - g′(x) = f(x). So where f > 0, g is increasing; where f < 0, g is decreasing. - g″(x) = f′(x). So if f is increasing (f′ > 0) then g is concave up; if f is decreasing (f′ < 0) g is concave down. - Local extrema of g occur where f crosses 0 (f changes sign). If f crosses from + to −, g has a local max; from − to +, a local min. - g(x) value equals signed area: add positive areas above the axis and subtract negative areas below. Use that to compute values or compare accumulation amounts. - Remember initial value g(a) sets the baseline; everything is relative to that. On the exam you’ll be asked to read f’s graph and state intervals of increase/decrease, concavity, and compute net accumulation—practice doing that visually. For a focused walk-through, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr). For extra practice problems, check (https://library.fiveable.me/practice/ap-calculus).

If g(x) = ∫_a^x f(t) dt, the Fundamental Theorem of Calculus says g′(x) = f(x). So: - g is increasing on any interval where f(x) > 0 (the net area is growing; g has positive slope). - g is decreasing where f(x) < 0 (net area is shrinking; g has negative slope). - If f(x) = 0 at a point, g has a horizontal tangent there. Think of g(x) as “net accumulated area from a to x.” Positive signed area adds to g; negative signed area subtracts. Also useful: g″(x) = f′(x), so if f is increasing then g is concave up, and if f is decreasing g is concave down—useful for interpreting shape on the exam. This is exactly the idea tested in Topic 6.5 (FUN-5.A)—see the topic study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and the Unit 6 overview (https://library.fiveable.me/ap-calculus/unit-6). For more practice problems, check https://library.fiveable.me/practice/ap-calculus.

Use the Fundamental Theorem of Calculus: g′(x) = f(x). So to find maxima/minima of g: - Find critical points: solve f(x) = 0 or where f is undefined. Also check endpoints of the domain/interval. - Use sign changes of f to classify: - If f (i.e., g′) changes from positive to negative at c, g has a local maximum at c. - If f changes from negative to positive at c, g has a local minimum at c. - You can also use the second-derivative idea: g′′(x) = f′(x). If g′(c)=0 and f′(c)<0, g has a local max; if f′(c)>0, g has a local min (when f′ exists). - Remember g(2)=∫_2^2 f = 0, so include that endpoint if you’re looking for absolute extrema on a closed interval. Graphically: g rises where the signed area under f is increasing (f>0) and falls where f<0. For more examples and AP-style practice, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and the Unit 6 page (https://library.fiveable.me/ap-calculus/unit-6).

If g(x) = ∫_a^x f(t) dt and f(x) > 0 on an interval, then g is increasing on that interval. By the Fundamental Theorem of Calculus (Part 1), g′(x) = f(x), so a positive integrand means a positive derivative—g’s slope is positive. Intuitively, g is accumulating positive (signed) area, so for x > a you get g(x) > g(a); larger values of f make g grow faster. (If you move left of a, ∫_a^x f = negative of the positive area from x to a, so g decreases as x decreases.) This reasoning—tying g′ = f, signed area, and net accumulation—is exactly what the AP CED expects for Topic 6.5 (FUN-5.A.3). For more examples and practice, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Think of g(x) = ∫_a^x f(t) dt as “running total” of signed area under f from a to x. Use these quick rules when reading g from the graph of f: - g′(x) = f(x). So where f is positive, g is increasing; where f is negative, g is decreasing. - The steepness of g at x equals the height of f at x: larger |f(x)| → steeper slope of g. - g has a local max/min where f crosses zero (changes sign). If f goes + → − at x, g has a local max; − → + gives a local min. - g″(x) = f′(x). So g is concave up where f is increasing, concave down where f is decreasing. - Values: g(x) = g(a) + net signed area from a to x. So the numerical area between a and x tells you how much g changed. - Continuity/differentiability: if f is continuous then g is differentiable everywhere (Fundamental Theorem of Calculus, FUN-5.A). If you want practice identifying intervals and extrema from graphs like this (AP-style), check the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and try problems from the Unit 6 practice set (https://library.fiveable.me/practice/ap-calculus).

Think of g(x) = ∫_a^x f(t) dt as the running total (net accumulation) of signed area under f from a to x. By the FTC (part 1), g′(x) = f(x), so: - The value g(x) equals the net area (positive area above the t-axis minus area below) between t = a and t = x. - Where f(t) > 0, g increases (area is being added); where f(t) < 0, g decreases (area is being subtracted). - g has a horizontal tangent (local extremum) when f(x) = 0; whether it’s a max or min depends on the sign change of f. - The rate g is changing at x is exactly f(x); the concavity of g is given by g″(x) = f′(x). On the AP exam you’ll be asked to read graphs/tables of f and use these facts to describe increasing/decreasing intervals, extrema, and amounts of accumulation (definite integrals). For a focused review check the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and more unit resources (https://library.fiveable.me/ap-calculus/unit-6).

Think of g(x) = ∫_a^x f(t) dt as “net area so far.” Use the Fundamental Theorem of Calculus (Part 1): g′(x) = f(x). So: - Where f(x) > 0, g is increasing; where f(x) < 0, g is decreasing. - Critical points of g occur where f(x) = 0 (or f is undefined). Check sign changes of f to tell if g has a local max/min. - g″(x) = f′(x), so g is concave up where f′(x) > 0 and concave down where f′(x) < 0. - g(a) is the initial value (often 0); g(x) gives signed net accumulation (areas above positive, below negative). - To get values: evaluate the definite integral (antiderivative or numeric area). For sketching, integrate areas piecewise and keep signs. For AP problems, explicitly state g′ and g″, use sign charts for f and f′, and refer to net signed area when evaluating g. For extra practice and examples, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and Unit 6 resources (https://library.fiveable.me/ap-calculus/unit-6).

Because g(x) = ∫_a^x f(t) dt, the Fundamental Theorem of Calculus (Part 1) gives g′(x) = f(x). A critical point of g is any x where g′(x) = 0 or g′ is undefined. So wherever f(x) = 0 (and f is continuous so g is differentiable there), g′(x)=0 and x is a critical point of g. (This is exactly the FUN-5.A connection between an integrand and its accumulation function.) Whether that critical point is a local max, min, or neither depends on how f changes sign at x: - f: + → − (g′ goes positive → negative) ⇒ local maximum of g. - f: − → + ⇒ local minimum. - If f just touches 0 without changing sign, g has a horizontal inflection (no local extremum). For more on interpreting accumulation functions and signed area, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr). For extra practice, check the unit resources (https://library.fiveable.me/ap-calculus/unit-6) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

If g(x) is an accumulation function g(x) = ∫_a^x f(t) dt (with f continuous), use the Fundamental Theorem of Calculus (Part 1): - Step 1—g′(x): g′(x) = f(x). Intuition: the instantaneous rate of change of accumulated area equals the height of the integrand at x (signed area/net accumulation). (CED: FUN-5.A) - Step 2—g″(x): differentiate g′. If f is differentiable, g″(x) = f′(x). So concavity of g comes from f′. If the limit depends on x (variable limits): - If g(x)=∫_a^{h(x)} f(t) dt, then g′(x)=f(h(x))·h′(x) by chain rule. - Then g″(x)=d/dx[f(h(x))·h′(x)] = f′(h(x))·(h′(x))^2 + f(h(x))·h″(x). More generally, for g(x)=∫_{u(x)}^{v(x)} f(t) dt: - g′(x)=f(v(x))v′(x) − f(u(x))u′(x). Want more worked examples and AP-style practice on interpreting accumulation, continuity, and sign? Check the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and more Unit 6 resources (https://library.fiveable.me/ap-calculus/unit-6). For lots of practice problems, see (https://library.fiveable.me/practice/ap-calculus).

If g(x) = ∫_a^x f(t) dt, the Fundamental Theorem of Calculus gives g′(x) = f(x) and g″(x) = f′(x). So concavity of g tells you about f and f′ directly: - If g is concave up on an interval, then g″(x) > 0 there → f′(x) > 0 → f is increasing on that interval. Since g′ = f, g′ is increasing. - If g is concave down, then g″(x) < 0 → f′(x) < 0 → f is decreasing (so g′ is decreasing). - Combined with signs of f: g is increasing where f(x) > 0 and decreasing where f(x) < 0. In APCalc language: concavity of the accumulation function g gives the sign of the integrand’s derivative (f′) and whether the integrand f is increasing or decreasing. For more practice and examples, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and extra problems (https://library.fiveable.me/practice/ap-calculus).

Think of g(x) = ∫_a^x f(t) dt as a running total whose instantaneous rate of change is f. That gives you three quick rules to sketch g from the graph of f: - Slope rule: g′(x) = f(x). At each x, the slope of g equals the height of f there. If f(x) is large positive, g is steeply increasing; if f(x) is negative, g is decreasing. - Extrema: g has a local max or min where f crosses zero (changes sign). If f goes + to −, g has a local max; − to + gives a local min. - Concavity: g′′(x) = f′(x). Where f is increasing, g is concave up; where f is decreasing, g is concave down. Practical sketch steps: start at g(a) (usually 0), move right using slope = f(x), flatten where f = 0, steepen where f is large, and change concavity where f has turning points. Use signed areas under f to place relative vertical changes or exact values when integrals/areas are computable. These ideas are exactly what the CED calls for (FUN-5.A). For more examples and practice problems, see the Topic 6.5 study guide (https://library.fiveable.me/ap-calculus/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) and the unit practice page (https://library.fiveable.me/practice/ap-calculus).