AP Calculus BC Quiz: Behavior Of Accumulation Functions Involving Area

20 questions · exam conditions

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Behavior Of Accumulation Functions Involving AreaQuestion 1 of 20

Given $H(x)=\int_{0}^{x}e^{-t^2}\,dt$ , which statement about $H$ is true for all $x$ ?

H

is decreasing because

e^{-x^2}

decreases

H

is increasing because

e^{-x^2}>0

H

is constant because

e^{-x^2}

approaches

0

H

is decreasing for

x>0

and increasing for

x<0

H

is increasing only where

e^{-x^2}

increases

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AP Calculus BC Quiz

AP Calculus BC Quiz: Behavior Of Accumulation Functions Involving Area

Practice Behavior Of Accumulation Functions Involving Area in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Behavior Of Accumulation Functions Involving Area, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Given $H(x)=\int_{0}^{x}e^{-t^2}\,dt$ , which statement about $H$ is true for all $x$ ?

$H$ is decreasing because $e^{-x^2}$ decreases
$H$ is increasing because $e^{-x^2}>0$ (correct answer)
$H$ is constant because $e^{-x^2}$ approaches $0$
$H$ is decreasing for $x>0$ and increasing for $x<0$
$H$ is increasing only where $e^{-x^2}$ increases

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.

Question 2

Let $A(x)=\int_{0}^{x}(t^2+1)\,dt$ . Which is true about the sign of $A(x)$ ?

$A(x)>0$ for all $x$
$A(x)<0$ for all $x$
$A(x)>0$ if $x>0$ and $A(x)<0$ if $x<0$ (correct answer)
$A(x)$ has the same sign as $t^2+1-0$
$A(x)=0$ whenever $t^2+1=0$

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.

Question 3

Given $H(x)=\int_{0}^{x}(2^{t}-1)\,dt$ , for which $x$ is $H$ decreasing?

$x<0$ (correct answer)
$x>0$
All real $x$
No real $x$
$x\ne0$

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.

Question 4

Given $G(x)=\int_{5}^{x}(2-\ln t)\,dt$ , where does $G$ increase?

$0<x<e^2$ (correct answer)
$x>e^2$
$x<0$
$x\in(0,1)$
$x\in(1,e)$

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.

Question 5

Let $A(x)=\int_{0}^{x}(t^2-4t+3)\,dt$ . On which interval is $A$ increasing?

$(-\infty,1)$
$(1,3)$
$(3,\infty)$
$(-\infty,1)\cup(3,\infty)$ (correct answer)
$(1,3)\cup(3,\infty)$

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.

Question 6

Let $G(x)=\int_{0}^{x}(t-1)^5\,dt$ . At which $x$ does $G$ have a local minimum?

$x=1$ (correct answer)
$x=0$
$x=-1$
$x=5$
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.

Question 7

If $A(x)=\int_{-3}^{x}(t^2-9)\,dt$ , on which interval is $A$ decreasing?

$(-\infty,-3)$
$(-3,3)$ (correct answer)
$(3,\infty)$
$(-\infty,-3)\cup(3,\infty)$
$(-3,0)$

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.

Question 8

If $F(x)=\int_{-1}^{x}(1-t^2)\,dt$ , at which $x$ does $F$ have a local maximum?

$x=-1$
$x=0$
$x=1$ (correct answer)
$x=2$
No local extrema

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.

Question 9

Let $F(x)=\int_{0}^{x}(e^t-3)\,dt$ . For which $x$ is $F$ decreasing?

$x<\ln 3$ (correct answer)
$x>\ln 3$
$x<0$
$x>0$
All real $x$

Explanation: This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of 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.

Question 10

If $F(x)=\int_{0}^{x}(4-2t)\,dt$ , for which $x$ is $F$ increasing?

$x<0$
$x<2$ (correct answer)
$x>2$
$x\in(0,2)$ only
$x\in(-\infty,0)\cup(2,\infty)$

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.

Question 11

Let $A(x)=\int_{0}^{x}(t^2-6t+8)\,dt$ . Where is $A$ decreasing?

$(-\infty,2)$
$(2,4)$ (correct answer)
$(4,\infty)$
$(-\infty,2)\cup(4,\infty)$
$(2,4)\cup(4,\infty)$

Explanation: 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.

Question 12

If $A(x)=\int_{2}^{x}(t-2)e^{t}\,dt$ , where is $A$ decreasing?

$x<2$ (correct answer)
$x>2$
$x<0$
$0<x<2$
No values; $A$ always increases

Explanation: 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.

Question 13

Let $G(x)=\int_{3}^{x}\frac{1}{1+t^2}\,dt$ . Which is true about $G$ for all real $x$ ?

$G$ decreases because $\frac{1}{1+x^2}$ decreases for $x>0$
$G$ increases because $\frac{1}{1+x^2}>0$ (correct answer)
$G$ is constant because $\frac{1}{1+x^2}\to0$
$G$ increases only on $(-1,1)$
$G$ decreases on $(-\infty,0)$ and increases on $(0,\infty)$

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 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.

Question 14

Given $G(x)=\int_{1}^{x}\sqrt{t-1}\,dt$ , for which $x$ is $G$ concave up?

$x<1$
$x>1$ (correct answer)
$0<x<1$
$x\ne1$
No values of $x$

Explanation: This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining where the function is concave up. The second derivative of G(x) is (1/2)(x-1)^{-1/2}, positive for x > 1. Thus, G is concave up for x > 1, where the function is defined. The first derivative is √(x-1), increasing for x > 1. A tempting distractor is A x<1, but the integrand is not real there, and concavity is undefined. Always find where the derivative of the integrand is positive to determine concave up intervals for accumulation functions.

Question 15

If $A(x)=\int_{-2}^{x}(t+1)(t-3)\,dt$ , on which interval is $A$ decreasing?

$(-\infty,-1)$
$(-1,3)$ (correct answer)
$(3,\infty)$
$(-\infty,-1)\cup(3,\infty)$
$(-1,3)\cup(3,\infty)$

Explanation: This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly determining intervals where the function is 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.

Question 16

Let $A(x)=\int_{3}^{x}(\sqrt{t}-2)\,dt$ . On which interval is $A$ increasing?

$0<x<4$
$x>4$ (correct answer)
$x<4$
$x\in(0,3)$
$x\in(3,4)$

Explanation: This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of A(x) is √x - 2, which is positive for x > 4, indicating A increases there. For x < 4, it is negative where defined (x ≥ 0), so A decreases. The square root is defined for x ≥ 0. A tempting distractor is x < 4, but that is where A decreases. A transferable strategy for accumulation functions is to isolate variables in root expressions to find positive integrand regions.

Question 17

Let $G(x)=\int_{0}^{x}(\sin t+1)\,dt$ . Which is true about $G$ ?

$G$ decreases where $\sin x$ decreases
$G$ increases for all $x$ because $\sin x+1\ge0$ (correct answer)
$G$ is constant on intervals where $\sin x=-1$
$G$ decreases on $(0,\pi)$
$G$ increases only on $(0,\pi)$

Explanation: This problem assesses the skill of interpreting the behavior of accumulation functions defined as definite integrals, particularly analyzing monotonicity. The derivative of G(x) is sin x + 1, always non-negative. Thus, G increases for all x since the integrand is at least zero. The shift by 1 ensures no negative values. A tempting distractor is D, but on (0, π), sin x + 1 > 0, and it increases elsewhere too. Always check the sign of the integrand to confirm monotonicity of accumulation functions across the entire domain.

Question 18

Given $H(x)=\int_{-2}^{x}(t^2+4t+3)\,dt$ , on which interval is $H$ increasing?

$(-\infty,-3)$
$(-3,-1)$
$(-1,\infty)$
$(-\infty,-3)\cup(-1,\infty)$ (correct answer)
$(-3,\infty)$

Explanation: 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.

Question 19

Let $G(x)=\int_{0}^{x}\frac{t-3}{t^2+1}\,dt$ . For which $x$ is $G$ increasing?

$x<3$
$x>3$ (correct answer)
$x<0$
$0<x<3$
All real $x$

Explanation: This problem tests the skill of interpreting the behavior of accumulation functions defined as definite integrals. The derivative of G(x) is $(x-3)/(x^2 + 1)$ , which is positive for $x > 3$ since the denominator is always positive. For $x < 3$ , it is negative. The rational function has no vertical asymptotes. A tempting distractor is all real $x$ , but G decreases for $x < 3$ . A transferable strategy for accumulation functions is to analyze rational integrands by considering numerator and denominator signs separately.

Question 20

Let $G(x)=\int_{0}^{x}(t^2-2t)\,dt$ . For which $x$ does $G$ have a local minimum?

$x=0$
$x=1$
$x=2$ (correct answer)
$x=-1$
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^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.