Behavior of Accumulation Functions Involving Area - AP Calculus BC

1. If $F(x)=\int_{0}^{x}(4-2t)\,dt$, for which $x$ is $F$ increasing?
   Answer: $x<2$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals of increase. The derivative of F(x) is 4 - 2x, positive when x < 2. Thus, F increases for x < 2, including negative x where the derivative remains positive. The linear integrand changes sign at t=2, controlling the behavior. A tempting distractor is A x<0, but F also increases for 0 < x < 2. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
2. Let $G(x)=\int_{0}^{x}(t^2-2t)\,dt$. For which $x$ does $G$ have a local minimum?
   Answer: $x=2$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of G(x) is $x^2$ - 2x, which is negative between x=0 and x=2, and positive elsewhere, indicating a local minimum at x=2 where it changes from negative to positive. The quadratic integrand has roots at 0 and 2. The second derivative confirms concavity at critical points. A tempting distractor is x=0, but that is a local maximum as the sign changes from positive to negative. A transferable strategy for accumulation functions is to use the first derivative test on the integrand to classify extrema.
3. Let $G(x)=\int_{0}^{x}(t-1)^5\,dt$. At which $x$ does $G$ have a local minimum?
   Answer: $x=1$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of G(x) is $(x-1)^5$, which changes from negative to positive at x=1, indicating a local minimum there. The odd power preserves the sign of (x-1). For x < 1, G decreases; for x > 1, it increases. A tempting distractor is x=0, but the critical point is at x=1. A transferable strategy for accumulation functions is to examine sign changes in odd-powered integrands for extrema classification.
4. If $A(x)=\int_{-3}^{x}(t^2-9)\,dt$, on which interval is $A$ decreasing?
   Answer: $(-3,3)$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals where the function is decreasing. The derivative of A(x) is x² - 9, negative when |x| < 3. Thus, A decreases on (-3, 3). The quadratic integrand changes sign at ±3. A tempting distractor is D (-∞, -3) ∪ (3, ∞), but there the derivative is positive, so A increases. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
5. If $G(x)=\int_{0}^{x}(1-\sqrt{t})\,dt$, on which interval is $G$ increasing?
   Answer: $0<x<1$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals of increase. The derivative of G(x) is 1 - √x, positive for 0 < x < 1. Thus, G increases on (0, 1) where the integrand is positive. The square root ensures the sign change at x=1. A tempting distractor is C x > 1, but there the derivative is negative, so G decreases. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
6. Given $H(x)=\int_{0}^{x}(2^{t}-1)\,dt$, for which $x$ is $H$ decreasing?
   Answer: $x<0$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of H(x) is $2^x$ - 1, which is negative for x < 0, indicating H decreases there. For x > 0, it is positive, so H increases. At x=0, the derivative is zero. A tempting distractor is x > 0, but that is where H increases. A transferable strategy for accumulation functions is to compare exponential integrands to constants for sign determination.
7. If $A(x)=\int_{-2}^{x}(t+1)(t-3)\,dt$, on which interval is $A$ decreasing?
   Answer: $(-1,3)$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals where the function is decreasing. The derivative of A(x) is (x+1)(x-3), negative when -1 < x < 3. Thus, A decreases on (-1, 3). The quadratic integrand controls the sign changes at the roots. A tempting distractor is D (-∞, -1) ∪ (3, ∞), but that's where the derivative is positive, so A increases there. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
8. Let $F(x)=\int_{0}^{x}(e^t-3)\,dt$. For which $x$ is $F$ decreasing?
   Answer: $x<\ln 3$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of F(x) is $e^x$ - 3, which controls whether F is increasing or decreasing based on its sign. This integrand is negative when $e^x$ < 3, or x < ln 3, causing F to decrease in that region. For x > ln 3, the integrand is positive, so F increases there. A tempting distractor is x > ln 3, but that is where F increases, not decreases. A transferable strategy for accumulation functions is to solve inequalities for the integrand's sign to identify intervals of monotonicity.
9. If $F(x)=\int_{2}^{x}(t^4-16)\,dt$, on which interval is $F$ increasing?
   Answer: $(-\infty,-2)\cup(2,\infty)$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of F(x) is $x^4$ - 16, which is positive when $x^4$ > 16, or |x| > 2, so F increases on (-∞, -2) ∪ (2, ∞). Between -2 and 2, the integrand is negative, causing F to decrease. The fourth power ensures symmetry around zero. A tempting distractor is (-∞, -2), but that ignores the symmetric interval (2, ∞). A transferable strategy for accumulation functions is to solve for where the integrand exceeds zero to find increasing regions.
10. Given $G(x)=\int_{5}^{x}(2-\ln t)\,dt$, where does $G$ increase?
    Answer: $0<x<e^2$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining where the function increases. The derivative of G(x) is 2 - ln x, positive for x < e². Thus, G increases on (0, e²) since ln x is defined for x > 0. The logarithmic integrand changes sign at x = e². A tempting distractor is B x > e², but there the derivative is negative, so G decreases. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
11. Given $H(x)=\int_{0}^{x}e^{-t^2}\,dt$, which statement about $H$ is true for all $x$?
    Answer: $H$ is increasing because $e^{-x^2}>0$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly analyzing monotonicity for all x. The derivative of H(x) is $e^{-x²}$, which is always positive. Thus, H increases everywhere since its derivative never changes sign to negative. The integrand's positivity ensures consistent increasing behavior regardless of the lower limit. A tempting distractor is A, but while $e^{-x²}$ decreases for x > 0, the integral accumulates positively. Always check the sign of the integrand to confirm monotonicity of accumulation functions across the entire domain.
12. Let $H(x)=\int_{0}^{x}(t^2-1)^2\,dt$. Which statement about $H$ is true?
    Answer: $H$ increases for all $x$ because $(x^2-1)^2\ge0$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of H(x) is $(x^2 - 1)^2$, which is always non-negative, indicating H increases for all x. The integrand equals zero only at x=±1 but is positive elsewhere. This ensures strict increase except at isolated points. A tempting distractor is that H decreases on (-1,1), but the integrand is positive there too. A transferable strategy for accumulation functions is to recognize when the integrand is always positive for universal increasing behavior.
13. If $F(x)=\int_{-1}^{x}(1-t^2)\,dt$, at which $x$ does $F$ have a local maximum?
    Answer: $x=1$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of F(x) is 1 - $x^2$, which changes from positive to negative at x=1, indicating a local maximum there. The integrand is positive for |x| < 1 and negative for |x| > 1. At x=-1, it is a local minimum. A tempting distractor is x=-1, but the sign change there indicates a minimum. A transferable strategy for accumulation functions is to perform a sign analysis around points where the integrand is zero to find maxima.
14. If $A(x)=\int_{2}^{x}(t-2)e^{t}\,dt$, where is $A$ decreasing?
    Answer: $x<2$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of A(x) is (x-2)$e^x$, which is negative for x < 2 since $e^x$ > 0 and (x-2) < 0. For x > 2, the product is positive, so A increases. The exponential ensures no sign change independent of (x-2). A tempting distractor is x > 2, but that is where A increases, not decreases. A transferable strategy for accumulation functions is to factor the integrand and analyze each component's sign.
15. Let $G(x)=\int_{3}^{x}\frac{1}{1+t^2}\,dt$. Which is true about $G$ for all real $x$?
    Answer: $G$ increases because $\frac{1}{1+x^2}>0$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly analyzing monotonicity for all x. The derivative of G(x) is $1/(1 + x^2)$, which is always positive. Thus, G increases everywhere since its derivative is never negative or zero except asymptotically. The integrand's positivity ensures consistent behavior regardless of the lower limit. A tempting distractor is A, but while $1/(1 + x^2)$ decreases for x > 0, it remains positive, so G increases. Always check the sign of the integrand to confirm monotonicity of accumulation functions across the entire domain.
16. Let $A(x)=\int_{0}^{x}(t^2-6t+8)\,dt$. Where is $A$ decreasing?
    Answer: $(2,4)$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of A(x) is $x^2$ - 6x + 8, which is negative on (2, 4), indicating A decreases there. The quadratic is positive outside [2, 4]. Roots at 2 and 4 define the interval. A tempting distractor is (-∞, 2) ∪ (4, ∞), but that is where A increases. A transferable strategy for accumulation functions is to find roots of polynomial integrands and test sign intervals.
17. Let $H(x)=\int_{2}^{x}(t-2)^3\,dt$. On which interval is $H$ decreasing?
    Answer: $(-\infty,2)$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals where the function is decreasing. The derivative of H(x) is (x-2)³, negative for x < 2. Thus, H decreases on (-∞, 2). The odd power ensures the sign change only at x=2. A tempting distractor is B (2, ∞), but there the derivative is positive, so H increases. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.
18. Given $H(x)=\int_{-2}^{x}(t^2+4t+3)\,dt$, on which interval is $H$ increasing?
    Answer: $(-\infty,-3)\cup(-1,\infty)$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of H(x) is $x^2$ + 4x + 3, which is positive on (-∞, -3) ∪ (-1, ∞), indicating H increases there. It is negative on (-3, -1), so H decreases in that interval. The quadratic factors as (x+1)(x+3). A tempting distractor is (-3, ∞), but it includes the decreasing interval (-3, -1). A transferable strategy for accumulation functions is to factor quadratics and use test points for sign charts.
19. If $F(x)=\int_{0}^{x}(1+t^2)\sin t\,dt$, on which interval is $F$ decreasing?
    Answer: $(\pi,2\pi)$ This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of F(x) is (1 + $x^2$) sin x, which is negative where sin x < 0 since 1 + $x^2$ > 0, such as on (π, 2π). This occurs in intervals where sine is negative. The polynomial multiplier does not change sign. A tempting distractor is (0, π), but sin x > 0 there, so F increases. A transferable strategy for accumulation functions is to separate always-positive factors from oscillating ones for sign analysis.
20. Let $A(x)=\int_{0}^{x}(t^2-4t+3)\,dt$. On which interval is $A$ increasing?
    Answer: $(-\infty,1)\cup(3,\infty)$ This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals where the function is increasing. The derivative of A(x) is the integrand evaluated at x, which is x² - 4x + 3. This quadratic factors to (x-1)(x-3) and is positive when x < 1 or x > 3, indicating A(x) increases on those intervals. Since the lower limit is 0, the behavior holds for all real x as the derivative sign determines monotonicity regardless. A tempting distractor is B (1,3), but that's where the derivative is negative, so A decreases there. Always analyze the sign of the integrand to determine where an accumulation function increases or decreases.

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