Behavior of Accumulation Functions Involving Area - AP Calculus AB

Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f$ is increasing on $(0,2)$ and decreasing on $(2,5)$ , where is $A$ concave up?
Answer: On $(0,2)$ only This problem tests understanding of accumulation function concavity based on integrand monotonicity patterns. Since $A'(x) = f(x)$ and $A''(x) = f'(x)$ by differentiation, concavity depends on where $f'(x) > 0$ . Given that $f$ is increasing on $(0,2)$ and decreasing on $(2,5)$ , we have $f'(x) > 0$ on $(0,2)$ and $f'(x) < 0$ on $(2,5)$ . This makes $A''(x) > 0$ on $(0,2)$ only, so $A$ is concave up on $(0,2)$ only. Choice C incorrectly includes the entire domain. Accumulation functions are concave up exactly where the integrand is increasing.
Let $A(x)=\int_{2}^{x} f(t)\,dt$ , where $f(x)>0$ for $2<x<6$ and $f(x)<0$ for $x>6$ . Where is $A$ increasing?
Answer: Only on $(2,6)$ This question tests understanding of how the behavior of an integrand affects its accumulation function. Since A'(x) = f(x) by the Fundamental Theorem of Calculus, A is increasing when f(x) > 0 and decreasing when f(x) < 0. Given that f(x) > 0 only on (2,6) and f(x) < 0 for x > 6, A increases only on the interval (2,6). Students might incorrectly think A increases on the entire interval (2,∞) by forgetting that the sign of f determines whether A increases or decreases. The key strategy is to remember that an accumulation function increases when its integrand is positive and decreases when its integrand is negative.
Let $A(x)=\int_{4}^{x} f(t)\,dt$ . If $f(x)<0$ and $f'(x)<0$ on $(0,7)$ , what describes $A$ on $(0,7)$ ?
Answer: Decreasing and concave down This problem involves analyzing accumulation functions when both the integrand and its derivative are negative. Since $A'(x) = f(x) < 0$ on $(0,7)$ , the function $A$ is decreasing throughout this interval. For concavity, $A''(x) = f'(x) < 0$ , so $A$ is concave down on $(0,7)$ . Both conditions being negative creates an accumulation function that decreases at an increasing rate. Choice B would be incorrect since concave up requires $f'(x) > 0$ . When both the integrand and its derivative are negative, the accumulation function decreases with decreasing rate of change (accelerating downward).
Let $H(x)=\int_{-1}^{x} h(t)\,dt$ . If $h(x)=0$ only at $x=3$ and changes from negative to positive there, what occurs at $x=3$ for $H$ ?
Answer: $H$ has a local minimum This question examines critical points of accumulation functions and their relationship to sign changes in the integrand. Since H'(x) = h(x), when h(x) = 0 at x = 3, we have H'(3) = 0, making x = 3 a critical point of H. Because h changes from negative to positive at x = 3, H' changes from negative to positive there, indicating H transitions from decreasing to increasing. This means H has a local minimum at x = 3. Students might confuse this with a maximum by reversing the sign analysis. Remember that when the integrand changes from negative to positive, the accumulation function reaches a local minimum.
Define $G(x)=\int_{2}^{x} f(t)\,dt$ . If $f(x)=0$ at $x=7$ and $f$ changes from negative to positive there, what happens to $G$ at $x=7$ ?
Answer: $G$ has a local minimum This problem tests identification of local minima based on integrand sign changes in accumulation functions. Since $G'(x) = f(x)$ by the Fundamental Theorem, critical points occur where $f(x) = 0$ . At $x = 7$ , the integrand changes from negative to positive, meaning $G'(x)$ changes from negative to positive. This indicates $G$ changes from decreasing to increasing, creating a local minimum at $x = 7$ . Choice B would be incorrect since a maximum requires the sign change from positive to negative. When the integrand changes from negative to positive, the accumulation function transitions from decreasing to increasing, producing a local minimum.
Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)<0$ for $0<x<5$ , which statement about $A$ is correct?
Answer: $A$ is decreasing on $(0,5)$ This problem tests understanding of accumulation function monotonicity when the integrand is negative. Since $A'(x) = f(x) < 0$ for $0 < x < 5$ , the accumulation function $A$ has a negative derivative throughout this interval. This means $A$ is decreasing on $(0,5)$ . The sign of $f$ directly determines the monotonicity of $A$ , regardless of the actual values of $A(x)$ . Choice B incorrectly suggests increasing behavior, which would require $f(x) > 0$ . To determine accumulation function monotonicity, examine the sign of the integrand.
Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)>0$ but decreasing on $(0,4)$ , which describes $A$ on $(0,4)$ ?
Answer: Increasing and concave down This problem combines monotonicity and concavity analysis for accumulation functions with positive but decreasing integrands. Since $A'(x) = f(x) > 0$ on $(0,4)$ , the function $A$ is increasing throughout this interval. For concavity, $A''(x) = f'(x) < 0$ since $f$ is decreasing, making $A$ concave down on $(0,4)$ . The combination creates an increasing function with decreasing rate of increase. Choice B would be incorrect since concave up requires $f$ to be increasing. When the integrand is positive but decreasing, the accumulation function increases but is concave down.
Let $F(x)=\int_{-1}^{x} p(t)\,dt$ . If $p(2)=0$ and changes from positive to negative at $2$ , what happens to $F$ at $x=2$ ?
Answer: $F$ has a local maximum This problem tests our understanding of how sign changes in the integrand create extrema in accumulation functions. Since $F'(x) = p(x)$ by the Fundamental Theorem, critical points of $F$ occur where $p(x) = 0$ . At $x = 2$ , we have $p(2) = 0$ and $p$ changes from positive to negative, meaning $F'(x)$ changes from positive to negative. This indicates $F$ goes from increasing to decreasing, creating a local maximum at $x = 2$ . Choice A would be incorrect since a local minimum requires the integrand to change from negative to positive. To identify extrema in accumulation functions, look for sign changes in the integrand.
Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(1)=0$ and $f$ is increasing through $x=1$ , what is true about $A$ at $x=1$ ?
Answer: $A$ has an inflection point at $x=1$ This problem involves analyzing accumulation functions when the integrand crosses zero while changing monotonicity. Since $A'(x) = f(x)$ and $A''(x) = f'(x)$ by differentiation, we examine both the value and derivative of $f$ at $x = 1$ . At this point, $f(1) = 0$ means $A'(1) = 0$ , creating a critical point. Since $f$ is increasing through $x = 1$ , we have $f'(1) > 0$ , so $A''(1) > 0$ , but the sign change in $f$ creates an inflection point rather than an extremum. Choice B would be incorrect since extrema require definite sign changes in the derivative. When the integrand equals zero and is increasing, the accumulation function has an inflection point.
Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)>0$ for all $x$ and $f$ is decreasing, which describes $A$ ?
Answer: Increasing and concave down This problem requires analyzing accumulation functions with positive but decreasing integrands. Since $A'(x) = f(x) > 0$ for all $x$ , the function $A$ is increasing everywhere. For concavity, $A''(x) = f'(x) < 0$ since $f$ is decreasing, making $A$ concave down everywhere. This creates an increasing function with decreasing rate of increase. Choice B would be incorrect since concave up requires $f$ to be increasing. When the integrand is positive but decreasing, the accumulation function increases but is concave down.
Let $H(x)=\int_{-2}^{x} h(t)\,dt$ . If $h(x)>0$ on $(-2,0)$ and $h(x)=0$ on $(0,3)$ , how does $H$ behave on $(0,3)$ ?
Answer: $H$ is constant on $(0,3)$ This problem tests understanding of accumulation function behavior when the integrand becomes zero on an interval. Since $H'(x) = h(x)$ by the Fundamental Theorem, the behavior of $H$ depends on the values of $h$ . On $(0,3)$ where $h(x) = 0$ , we have $H'(x) = 0$ , making the derivative zero throughout this interval. This means $H$ has no rate of change and remains constant on $(0,3)$ . The previous behavior on $(-2,0)$ determines the constant value, but doesn't affect the constant behavior on $(0,3)$ . Choice B would be incorrect since increasing requires $h(x) > 0$ . When the integrand equals zero on an interval, the accumulation function is constant there.
Let $R(x)=\int_{3}^{x} r(t)\,dt$ . If $r$ has a minimum at $x=5$ and is differentiable, what does $R$ have at $x=5$ ?
Answer: A point of inflection This question explores the relationship between extrema of the integrand and inflection points of the accumulation function. Since R'(x) = r(x) and R''(x) = r'(x), at a minimum of r where r is differentiable, we have r'(5) = 0, which means R''(5) = 0. Because r has a minimum at x = 5, r' changes sign from negative to positive there, so R'' changes sign, indicating R has a point of inflection at x = 5. Students often expect the accumulation function to have an extremum where the integrand has an extremum, but this is incorrect. The key insight is that extrema of the integrand correspond to inflection points of the accumulation function.
Define $F(x)=\int_{-4}^{x} p(t)\,dt$ . If $p(x)<0$ for $-4<x<0$ and $p(x)>0$ for $0<x<4$ , what occurs at $x=0$ ?
Answer: $F$ has a local minimum This problem tests identification of local minima in accumulation functions based on integrand sign changes. Since $F'(x) = p(x)$ by the Fundamental Theorem, critical points occur where $p(x) = 0$ . Given the sign pattern, at $x = 0$ , the integrand changes from negative to positive, meaning $F'(x)$ changes from negative to positive. This indicates $F$ changes from decreasing to increasing, creating a local minimum at $x = 0$ . Choice B would be incorrect since a maximum requires the opposite sign change. When the integrand changes from negative to positive, the accumulation function has a local minimum.
Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f$ is negative on $(0,1)$ and positive on $(1,2)$ , which best describes $A(2)$ ?
Answer: $A(2)$ could be positive, negative, or zero depending on areas This problem requires analyzing accumulation function values based on competing positive and negative areas. Since $A(x) = \int_0^x f(t) dt$ , the value $A(2)$ represents the net area from 0 to 2. With $f$ negative on $(0,1)$ and positive on $(1,2)$ , we get $A(2) = \int_0^1 f(t) dt + \int_1^2 f(t) dt$ , where the first integral is negative and the second is positive. The sign of $A(2)$ depends on which area magnitude is larger, making it potentially positive, negative, or zero. Choice B incorrectly assumes only the final behavior matters. The value of an accumulation function depends on the net balance of all signed areas.
Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f$ is increasing and crosses the $t$ -axis at $x=4$ , what can be said about $A$ at $x=4$ ?
Answer: $A$ has an inflection point at $x=4$ This problem involves analyzing accumulation functions when the integrand crosses zero while changing monotonicity. Since $A'(x) = f(x)$ and $A''(x) = f'(x)$ by differentiation, we examine the behavior at $x = 4$ . The integrand $f$ is increasing and crosses the $t$ -axis at $x = 4$ , meaning $f(4) = 0$ and $f'(4) > 0$ . This creates $A'(4) = 0$ and $A''(4) > 0$ , but since $f$ changes sign while increasing, $A$ has an inflection point rather than an extremum. Choice B would require a definite sign change pattern in $f$ . When an increasing integrand crosses zero, the accumulation function has an inflection point.
Define $G(x)=\int_{0}^{x} g(t)\,dt$ . If $g$ is increasing on $(0,4)$ , what is true about $G$ on $(0,4)$ ?
Answer: $G$ is concave up This problem requires analyzing how properties of an integrand transfer to its accumulation function. Since G'(x) = g(x) by the Fundamental Theorem, and G''(x) = g'(x), the fact that g is increasing means g'(x) > 0, which makes G''(x) > 0. Therefore, G is concave up on (0,4). Students might mistakenly think that an increasing integrand makes the accumulation function increasing, but g being increasing tells us about G's concavity, not its monotonicity. The essential insight is that the derivative of the integrand determines the concavity of the accumulation function.
Let $G(x)=\int_{-3}^{x} p(t)\,dt$ . If $p$ is increasing on $(-3,1)$ and decreasing on $(1,5)$ , where can $G$ change concavity?
Answer: At $x=1$ only This problem requires understanding how monotonicity changes in the integrand create inflection points in accumulation functions. Since $G'(x) = p(x)$ and $G''(x) = p'(x)$ by differentiation, inflection points occur where $p'(x) = 0$ with a sign change. Given that $p$ increases on $(-3,1)$ and decreases on $(1,5)$ , we have $p'(x) > 0$ before $x = 1$ and $p'(x) < 0$ after $x = 1$ . This means $G''(x)$ changes sign at $x = 1$ , creating an inflection point. Choice D incorrectly suggests all zeros of $p$ create inflection points, but inflection points require sign changes in $p'$ . Inflection points in accumulation functions occur where the integrand changes from increasing to decreasing or vice versa.
Let $G(x)=\int_{1}^{x} g(t)\,dt$ . If $g$ is decreasing on $(1,9)$ and $g(4)=0$ with a sign change, what must occur to $G$ at $x=4$ ?
Answer: $G$ has a local extremum at $x=4$ This problem requires identifying extrema in accumulation functions when the integrand is decreasing and changes sign. Since $G'(x) = g(x)$ by the Fundamental Theorem, critical points occur where $g(x) = 0$ . At $x = 4$ , we have $g(4) = 0$ with a sign change (since $g$ is decreasing and crosses zero). This creates a critical point where $G$ has a local extremum. Since $g$ is decreasing through zero, it changes from positive to negative, making $G$ change from increasing to decreasing, creating a local maximum. Choice B would be incorrect since inflection points require changes in the integrand's monotonicity. When a decreasing integrand crosses zero with a sign change, the accumulation function has a local extremum.
Define $A(x)=\int_{2}^{x} g(t)\,dt$ . If $g$ is decreasing on $(0,5)$ , what is true about the concavity of $A$ on $(0,5)$ ?
Answer: $A$ is concave down on $(0,5)$ This problem requires understanding how the integrand's monotonicity affects the accumulation function's concavity. Since $A'(x) = g(x)$ and $A''(x) = g'(x)$ by differentiation, the concavity of $A$ depends on whether $g'(x)$ is positive or negative. Given that $g$ is decreasing on $(0,5)$ , we have $g'(x) < 0$ throughout this interval. This means $A''(x) < 0$ , so $A$ is concave down on $(0,5)$ . Choice B incorrectly suggests concave up, which would require $g$ to be increasing rather than decreasing. When analyzing accumulation function concavity, examine whether the integrand is increasing or decreasing.
Let $H(x)=\int_{0}^{x} h(t)\,dt$ . If $h'(x)>0$ for $x<0$ and $h'(x)<0$ for $x>0$ , what is true about $H$ at $x=0$ ?
Answer: $H$ can change concavity at $x=0$ This problem involves understanding how changes in the integrand's derivative affect accumulation function concavity. Since $H'(x) = h(x)$ and $H''(x) = h'(x)$ by differentiation, the concavity of $H$ depends on the sign of $h'(x)$ . Given that $h'(x) > 0$ for $x < 0$ and $h'(x) < 0$ for $x > 0$ , the second derivative $H''(x)$ changes from positive to negative at $x = 0$ . This sign change indicates that $H$ changes from concave up to concave down, creating an inflection point. Choice B would require $h(0) = 0$ with a sign change in $h$ . Changes in the integrand's monotonicity create inflection points in accumulation functions.

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