This question tests understanding of how an accumulation function's behavior relates to its integrand's sign. Since $A(x) = \int_0^x f(t),dt$, we have $A'(x) = f(x)$ by the Fundamental Theorem of Calculus. For $0 < x < 3$, we have $f(x) > 0$, which means $A'(x) > 0$, so $A$ is increasing on $(0,3)$. For $3 < x < 6$, we have $f(x) < 0$, which means $A'(x) < 0$, so $A$ is decreasing on $(3,6)$. Choice E incorrectly suggests that $A$ is increasing wherever $|f(x)| > 0$, but the absolute value is irrelevant—only the sign of $f$ matters. The key principle: an accumulation function increases where its integrand is positive and decreases where its integrand is negative.

This problem tests understanding of how positive integrands affect accumulation function values. Since s(x) > 0 for x > 1 and S(x) = ∫₁ˣ s(t)dt, we are accumulating positive values as x increases beyond 1. This means S(x) > 0 for all x > 1, as we're adding up positive quantities starting from S(1) = 0. Students might incorrectly choose that S is decreasing, confusing the sign of the integrand with the monotonicity of S. The fundamental principle is that integrating a positive function over a positive-length interval yields a positive result.

This question examines critical points where an integrand changes sign. Since $F(x) = \int_0^x f(t),dt$, we have $F'(x) = f(x)$ by the Fundamental Theorem of Calculus. At $x = 2$, we have $F'(2) = f(2) = 0$, making it a critical point. Since $f$ changes from positive to negative at $x = 2$, we have $F'(x) > 0$ for $x < 2$ (so $F$ is increasing) and $F'(x) < 0$ for $x > 2$ (so $F$ is decreasing). Therefore, $F$ has a local maximum at $x = 2$. Choice C incorrectly suggests $F(2) = 0$, but $F(2) = \int_0^2 f(t),dt$, which depends on the accumulated positive area from 0 to 2. Remember: when an integrand changes from positive to negative, the accumulation function reaches a local maximum at that transition point.

This question examines the relationship between an accumulation function's concavity and its integrand's monotonicity. Since $G(x) = \int_2^x g(t),dt$, we have $G'(x) = g(x)$ and $G''(x) = g'(x)$ by the Fundamental Theorem of Calculus. If $g$ is increasing on $(0,5)$, then $g'(x) > 0$ on $(0,5)$, which means $G''(x) > 0$ on $(0,5)$. Therefore, $G$ is concave up on the entire interval $(0,5)$. Choice C incorrectly focuses on where $g(x) > 0$, but the sign of $g$ determines whether $G$ is increasing or decreasing, not its concavity. Remember: an accumulation function is concave up where its integrand is increasing and concave down where its integrand is decreasing.

This problem tests basic understanding of accumulation function monotonicity based on integrand sign. Since $F'(x) = f(x)$ by the Fundamental Theorem, the function $F$ is increasing wherever its derivative is positive. Given that $f(x) > 0$ for $x > 2$, we have $F'(x) > 0$ for $x > 2$, making $F$ increasing on this interval. The other choices incorrectly relate to the sign or magnitude of $F$ itself rather than its derivative. Choice C confuses function values with derivative signs. To determine if an accumulation function is increasing, check if the integrand is positive.

This problem tests understanding of accumulation functions when the integrand is zero. Since $F'(x) = f(x) = 0$ for all $x$ in $(2,5)$, the derivative of $F$ is zero throughout this interval. This means $F$ has zero rate of change, making it constant on $(2,5)$. The function neither increases nor decreases in this region. Choice E incorrectly suggests $F(x) = 0$, but $F$ equals whatever accumulated area existed up to $x = 2$. When the integrand is zero on an interval, the accumulation function remains constant there.

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.

This problem requires identifying all local extrema in accumulation functions with multiple sign changes. Since $G'(x) = g(x)$ by the Fundamental Theorem, extrema occur where $g(x) = 0$ with sign changes. The integrand is positive on $(0,1)$, negative on $(1,3)$, and positive on $(3,4)$, creating sign changes at both $x = 1$ and $x = 3$. At $x = 1$, $g$ changes from positive to negative, creating a local maximum. At $x = 3$, $g$ changes from negative to positive, creating a local minimum. Choice C incorrectly identifies only one extremum. Accumulation functions have local extrema at every point where the integrand changes sign.

This problem tests understanding of accumulation function concavity with varying lower limits and integrand conditions. Since $F'(x) = f(x)$ and $F''(x) = f'(x)$ by differentiation, concavity depends on whether $f$ is increasing or decreasing. Given that $f$ is increasing on $(0,6)$, we have $f'(x) > 0$, making $F''(x) > 0$. This means $F$ is concave up on $(0,6)$, regardless of the lower limit value or the sign of $f(3)$. Choice B would be incorrect since concave down requires $f$ to be decreasing. The concavity of accumulation functions depends only on the integrand's monotonicity, not its sign or the integration limits.

This problem tests identification of local extrema based on sign changes in the integrand. Since $F'(x) = g(x)$ by the Fundamental Theorem, critical points occur where $g(x) = 0$. At $x = 2$, the integrand changes from positive to negative, meaning $F'(x)$ changes from positive to negative. This indicates $F$ transitions from increasing to decreasing, creating a local maximum at $x = 2$. Choice B would be incorrect since a minimum requires the opposite sign change. To find extrema in accumulation functions, locate where the integrand changes sign.