Behavior of Accumulation Functions Involving Area
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AP Calculus BC › Behavior of Accumulation Functions Involving Area
Given $H(x)=\int_{0}^{x}e^{-t^2},dt$, which statement about $H$ is true for all $x$?
$H$ is decreasing for $x>0$ and increasing for $x<0$
$H$ is increasing only where $e^{-x^2}$ increases
$H$ is increasing because $e^{-x^2}>0$
$H$ is decreasing because $e^{-x^2}$ decreases
$H$ is constant because $e^{-x^2}$ approaches $0$
Explanation
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.
Let $A(x)=\int_{0}^{x}(t^2+1),dt$. Which is true about the sign of $A(x)$?
$A(x)>0$ for all $x$
$A(x)=0$ whenever $t^2+1=0$
$A(x)$ has the same sign as $t^2+1-0$
$A(x)>0$ if $x>0$ and $A(x)<0$ if $x<0$
$A(x)<0$ for all $x$
Explanation
This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly analyzing the sign of the function. The integrand t² + 1 is always positive, so A(x) > 0 for x > 0 and A(x) < 0 for x < 0. This follows from the odd nature of the antiderivative x³/3 + x. The accumulation reflects the positive area for positive x and negative for negative x. A tempting distractor is A A(x)>0 for all x, but for x < 0, the integral accumulates negative values. Always evaluate the sign of the accumulation by considering the direction of integration and integrand positivity.
Given $H(x)=\int_{0}^{x}(2^{t}-1),dt$, for which $x$ is $H$ decreasing?
No real $x$
$x>0$
$x\ne0$
$x<0$
All real $x$
Explanation
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.
Given $G(x)=\int_{5}^{x}(2-\ln t),dt$, where does $G$ increase?
$x\in(0,1)$
$x>e^2$
$x<0$
$x\in(1,e)$
$0<x<e^2$
Explanation
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.
Let $A(x)=\int_{0}^{x}(t^2-4t+3),dt$. On which interval is $A$ increasing?
$(1,3)\cup(3,\infty)$
$(1,3)$
$(-\infty,1)$
$(3,\infty)$
$(-\infty,1)\cup(3,\infty)$
Explanation
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.
Let $G(x)=\int_{0}^{x}(t-1)^5,dt$. At which $x$ does $G$ have a local minimum?
$x=-1$
$x=5$
$x=1$
$x=0$
No local extrema
Explanation
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.
If $A(x)=\int_{-3}^{x}(t^2-9),dt$, on which interval is $A$ decreasing?
$(-3,0)$
$(-3,3)$
$(-\infty,-3)$
$(-\infty,-3)\cup(3,\infty)$
$(3,\infty)$
Explanation
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.
If $F(x)=\int_{-1}^{x}(1-t^2),dt$, at which $x$ does $F$ have a local maximum?
$x=-1$
$x=1$
$x=0$
No local extrema
$x=2$
Explanation
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.
Let $F(x)=\int_{0}^{x}(e^t-3),dt$. For which $x$ is $F$ decreasing?
$x>0$
$x<\ln 3$
$x<0$
$x>\ln 3$
All real $x$
Explanation
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.
If $F(x)=\int_{0}^{x}(4-2t),dt$, for which $x$ is $F$ increasing?
$x\in(0,2)$ only
$x<2$
$x<0$
$x>2$
$x\in(-\infty,0)\cup(2,\infty)$
Explanation
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.