Calculus 2 Quiz: Accumulation Function Behavior
20 questions · exam conditions
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Accumulation Function BehaviorQuestion 1 of 20

Let f(t)f(t) be a continuous, positive, and decreasing function for all t0t \ge 0. Let A(x)=0xf(t)dtA(x) = \int_0^x f(t) \, dt. Which of the following inequalities must be true for x>0x > 0?

A(x)>xf(x)A(x) > x \cdot f(x)
A(x)<xf(x)A(x) < x \cdot f(x)
A(x)=xf(x)A(x) = x \cdot f(x)
A(x)<xf(0)A(x) < x \cdot f(0)
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Calculus 2 Quiz

Calculus 2 Quiz: Accumulation Function Behavior

Practice Accumulation Function Behavior in Calculus 2 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 Accumulation Function Behavior, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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.

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Question 1

Let f(t)f(t) be a continuous, positive, and decreasing function for all t0t \ge 0. Let A(x)=0xf(t)dtA(x) = \int_0^x f(t) \, dt. Which of the following inequalities must be true for x>0x > 0?

  1. A(x)>xf(x)A(x) > x \cdot f(x) (correct answer)
  2. A(x)<xf(x)A(x) < x \cdot f(x)
  3. A(x)=xf(x)A(x) = x \cdot f(x)
  4. A(x)<xf(0)A(x) < x \cdot f(0)
Explanation: Geometrically, A(x)A(x) represents the area under the curve of f(t)f(t) from t=0t=0 to t=xt=x. The expression xf(x)x \cdot f(x) represents the area of a rectangle with width xx and height f(x)f(x). Since f(t)f(t) is a decreasing function, for any tt in (0,x)(0, x), f(t)>f(x)f(t) > f(x). The rectangle of area xf(x)x \cdot f(x) is therefore inscribed under the curve of f(t)f(t) on [0,x][0, x], and its area must be less than the area under the curve. Thus, A(x)>xf(x)A(x) > x \cdot f(x).

Question 2

The function f(t)f(t) is continuous and has a single root at t=ct=c. The accumulation function G(x)=axf(t)dtG(x) = \int_a^x f(t) \, dt is decreasing for x<cx<c and increasing for x>cx>c. Which of the following must be true?

  1. f(t)>0f(t) > 0 for t<ct<c and f(t)<0f(t) < 0 for t>ct>c.
  2. f(t)<0f(t) < 0 for t<ct<c and f(t)>0f(t) > 0 for t>ct>c. (correct answer)
  3. The value of aa must be equal to cc.
  4. f(t)f(t) must be a linear function.
Explanation: The behavior of G(x)G(x) is determined by its derivative, G(x)=f(x)G'(x) = f(x). If G(x)G(x) is decreasing, its derivative must be negative. Thus, for x<cx<c, G(x)=f(x)<0G'(x) = f(x) < 0. If G(x)G(x) is increasing, its derivative must be positive. Thus, for x>cx>c, G(x)=f(x)>0G'(x) = f(x) > 0. This directly corresponds to the conditions in choice B.

Question 3

Let f(t)f(t) be a continuous function that is positive on the interval (,3)(-\infty, 3) and negative on the interval (3,)(3, \infty). Define an accumulation function G(x)=1xf(t)dtG(x) = \int_1^x f(t) \, dt. Which of the following statements correctly describes G(x)G(x) at x=3x=3?

  1. G(x) has a local minimum at x=3x=3.
  2. G(x) has a local maximum at x=3x=3. (correct answer)
  3. G(x) has a point of inflection at x=3x=3.
  4. G(x) has a root at x=3x=3.
Explanation: According to the Fundamental Theorem of Calculus Part 2, G(x)=f(x)G'(x) = f(x). To find local extrema of G(x)G(x), we can use the First Derivative Test on G(x)=f(x)G'(x) = f(x). A critical point occurs where f(x)=0f(x) = 0, which is at x=3x=3. For x<3x < 3, f(x)>0f(x) > 0, so G(x)G(x) is increasing. For x>3x > 3, f(x)<0f(x) < 0, so G(x)G(x) is decreasing. Since G(x)G'(x) changes from positive to negative at x=3x=3, G(x)G(x) has a local maximum at this point.

Question 4

Let f(t)f(t) be a continuous function such that f(t)<0f(t) < 0 on (,2)(-\infty, 2) and f(t)>0f(t) > 0 on (2,)(2, \infty), with f(2)=0f(2)=0. Define F(x)=4xf(t)dtF(x) = \int_4^x f(t) \, dt. Which of the following statements about the value of the local extremum of F(x)F(x) is true?

  1. F(x) has a local minimum at x=2x=2 with a value of F(2)=0F(2) = 0.
  2. F(x) has a local maximum at x=2x=2 with a value of F(2)<0F(2) < 0.
  3. F(x) has a local minimum at x=2x=2 with a value of F(2)<0F(2) < 0. (correct answer)
  4. F(x) has a local maximum at x=2x=2 with a value of F(2)>0F(2) > 0.
Explanation: By the FTC, F(x)=f(x)F'(x) = f(x). The critical point is at x=2x=2 since f(2)=0f(2)=0. For x<2x<2, f(x)<0f(x)<0, so F(x)F(x) is decreasing. For x>2x>2, f(x)>0f(x)>0, so F(x)F(x) is increasing. Thus, F(x)F(x) has a local minimum at x=2x=2. The value is F(2)=42f(t)dt=24f(t)dtF(2) = \int_4^2 f(t) \, dt = -\int_2^4 f(t) \, dt. On the interval (2,4)(2,4), f(t)>0f(t) > 0, so the integral 24f(t)dt\int_2^4 f(t) \, dt is positive. Therefore, F(2)F(2) must be negative.

Question 5

Let f(t)f(t) be a differentiable function that is decreasing on the interval (,0)(-\infty, 0) and increasing on the interval (0,)(0, \infty). It is also known that f(0)<0f(0) < 0. Let H(x)=5xf(t)dtH(x) = \int_5^x f(t) \, dt. Which statement accurately describes the behavior of H(x)H(x) at x=0x=0?

  1. H(x) has a local minimum at x=0x=0.
  2. H(x) has a local maximum at x=0x=0.
  3. H(x) has a point of inflection and is decreasing at x=0x=0. (correct answer)
  4. H(x) has a point of inflection and is increasing at x=0x=0.
Explanation: We analyze the derivatives of H(x)H(x). H(x)=f(x)H'(x) = f(x) and H(x)=f(x)H''(x) = f'(x). The behavior of H(x)H(x) at x=0x=0 depends on H(0)H'(0) and H(0)H''(0). We are given that f(0)<0f(0) < 0, so H(0)<0H'(0) < 0, which means H(x)H(x) is decreasing at x=0x=0. For a point of inflection, H(x)H''(x) must change sign. H(x)=f(x)H''(x) = f'(x). We are told f(t)f(t) changes from decreasing to increasing at t=0t=0. This means f(t)f'(t) changes from negative to positive at t=0t=0. Therefore, H(x)H''(x) changes sign at x=0x=0, indicating a point of inflection.

Question 6

Let f(t)f(t) be a twice-differentiable function. The function f(t)f(t) has a local maximum at t=4t=4 and a single root at t=1t=1. Let F(x)=0xf(t)dtF(x) = \int_0^x f(t) \, dt. On what interval must the graph of F(x)F(x) be concave down?

  1. The interval where f(t)<0f(t) < 0.
  2. The interval (,1)(-\infty, 1).
  3. The interval where f(t)f(t) is decreasing. (correct answer)
  4. The interval where f(t)f(t) is increasing.
Explanation: The concavity of F(x)F(x) is determined by the sign of its second derivative, F(x)F''(x). We have F(x)=f(x)F'(x) = f(x) and F(x)=f(x)F''(x) = f'(x). The graph of F(x)F(x) is concave down where F(x)<0F''(x) < 0, which means f(x)<0f'(x) < 0. The condition f(x)<0f'(x) < 0 means that the function f(x)f(x) is decreasing. Since f(t)f(t) has a local maximum at t=4t=4, we know f(4)=0f'(4) = 0, and f(t)f(t) is decreasing in some neighborhood to the right of 4 where f(t)<0f'(t) < 0.

Question 7

Let f(t)f(t) be an odd continuous function, i.e., f(t)=f(t)f(-t) = -f(t) for all tt. Let F(x)=3xf(t)dtF(x) = \int_{-3}^x f(t) \, dt. If 03f(t)dt=5\int_0^3 f(t) \, dt = 5, what is the value of F(3)F(3)?

  1. -5
  2. 0 (correct answer)
  3. 5
  4. 10
Explanation: We want to find F(3)=33f(t)dtF(3) = \int_{-3}^3 f(t) \, dt. Since f(t)f(t) is an odd function, its integral over a symmetric interval [a,a][-a, a] is zero. Therefore, 33f(t)dt=0\int_{-3}^3 f(t) \, dt = 0. The information that 03f(t)dt=5\int_0^3 f(t) \, dt = 5 implies that 30f(t)dt=5\int_{-3}^0 f(t) \, dt = -5, and their sum is 0.

Question 8

Let f(t)f(t) be a continuous function. Define G(x)=0xf(t)dtG(x) = \int_0^x f(t) \, dt. If G(x)G(x) is always increasing for x>0x>0 and is concave down for x>0x>0, which of the following must be true about f(t)f(t) for t>0t>0?

  1. f(t)>0f(t) > 0 and f(t)f(t) is increasing.
  2. f(t)>0f(t) > 0 and f(t)f(t) is decreasing. (correct answer)
  3. f(t)<0f(t) < 0 and f(t)f(t) is increasing.
  4. f(t)<0f(t) < 0 and f(t)f(t) is decreasing.
Explanation: The derivatives of G(x)G(x) are G(x)=f(x)G'(x) = f(x) and G(x)=f(x)G''(x) = f'(x). The statement that G(x)G(x) is always increasing for x>0x>0 means that G(x)>0G'(x) > 0 for x>0x>0, which implies f(x)>0f(x) > 0 for x>0x>0. The statement that G(x)G(x) is concave down for x>0x>0 means that G(x)<0G''(x) < 0 for x>0x>0, which implies f(x)<0f'(x) < 0 for x>0x>0. If f(x)<0f'(x) < 0, then the function f(x)f(x) is decreasing. Thus, for t>0t>0, f(t)f(t) must be positive and decreasing.

Question 9

Let f(t)f(t) be a continuous function such that f(t)<0f(t) < 0 for all tt. Define F(x)=1x2f(t)dtF(x) = \int_1^{x^2} f(t) \, dt. For what values of xx is F(x)F(x) increasing?

  1. x < 0 (correct answer)
  2. x > 0
  3. x > 1
  4. F(x) is never increasing.
Explanation: By the FTC and Chain Rule, F(x)=f(x2)ddx(x2)=f(x2)2xF'(x) = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x. We want to find where F(x)F(x) is increasing, which is where F(x)>0F'(x) > 0. So we need f(x2)2x>0f(x^2) \cdot 2x > 0. We are given that f(t)<0f(t) < 0 for all tt. Since x2x^2 is always non-negative, f(x2)f(x^2) will be negative (assuming x0x \neq 0 such that x2x^2 falls in the domain of ff). So we have (negative)2x>0(negative) \cdot 2x > 0. For this inequality to hold, the term 2x2x must be negative. This occurs when x<0x < 0.

Question 10

Let F(x)=xx+3t2dtF(x) = \int_x^{x+3} t^2 \, dt. At what value of xx is the instantaneous rate of change of F(x)F(x) equal to 21?

  1. x = -4
  2. x = -1
  3. x = 1
  4. x = 2 (correct answer)
Explanation: The instantaneous rate of change of F(x)F(x) is F(x)F'(x). Using the general form of the FTC, F(x)=(x+3)21x21F'(x) = (x+3)^2 \cdot 1 - x^2 \cdot 1. Simplifying, F(x)=(x2+6x+9)x2=6x+9F'(x) = (x^2 + 6x + 9) - x^2 = 6x + 9. We want to find xx such that F(x)=21F'(x) = 21. So, we solve the equation 6x+9=216x + 9 = 21. This gives 6x=126x = 12, so x=2x=2.

Question 11

Let f(t)f(t) be a continuous function, and define F(x)=0xf(t)dtF(x) = \int_0^x f(t) \, dt. If F(x)F(x) has a point of inflection at x=3x=3 and a critical point at x=5x=5, what can be concluded about the function ff?

  1. f(3)=0f(3)=0 and f(5)f(5) is a local extremum.
  2. ff has a local extremum at x=3x=3 and f(5)=0f(5)=0. (correct answer)
  3. f(3)=0f(3)=0 and f(5)=0f'(5)=0.
  4. ff has a local extremum at x=3x=3 and ff has a local extremum at x=5x=5.
Explanation: F(x)=f(x)F'(x) = f(x) and F(x)=f(x)F''(x) = f'(x). A point of inflection for F(x)F(x) at x=3x=3 means F(3)=0F''(3)=0 and F(x)F''(x) changes sign around x=3x=3. This is equivalent to f(3)=0f'(3)=0 and f(x)f'(x) changing sign, which means ff has a local extremum at x=3x=3. A critical point for F(x)F(x) at x=5x=5 means F(5)=0F'(5)=0, which is equivalent to f(5)=0f(5)=0.

Question 12

Let f(t)f(t) be a continuous function and F(x)=1xf(t)dtF(x) = \int_{-1}^x f(t) \, dt. Given that F(1)=0F(-1)=0 and F(3)=0F(3)=0, and F(x)>0F(x) > 0 on (1,3)(-1, 3), which of the following statements must be true about the function ff?

  1. 13f(t)dt=0\int_{-1}^3 f(t) \, dt = 0 (correct answer)
  2. f(t)>0f(t) > 0 for all tt in (1,3)(-1, 3)
  3. f(1)>0f(-1) > 0 and f(3)<0f(3) < 0
  4. f(t)f(t) must be an even function
Explanation: By the definition of F(x)F(x), we have 13f(t)dt=F(3)=0\int_{-1}^3 f(t) \, dt = F(3) = 0. This is a direct consequence of the given information. Choice B is incorrect because ff could have both positive and negative values while still producing a net area of zero. Choice C is incorrect because while we can determine that f(3)<0f(3) < 0 (since F(x)>0F(x) > 0 just to the left of 3 and F(3)=0F(3) = 0, so FF must be decreasing at x=3x = 3), we cannot determine the sign of f(1)f(-1) from the given information. Choice D has no basis in the given conditions.

Question 13

Let f(t)f(t) be a continuous function where f(t)>0f(t) > 0 on (1,3)(1, 3) and f(t)<0f(t) < 0 on (0,1)(3,4)(0, 1) \cup (3, 4). It is known that 13f(t)dt=6\int_1^3 f(t) \, dt = 6 and 34f(t)dt=6\int_3^4 f(t) \, dt = -6. Define G(x)=1xf(t)dtG(x) = \int_1^x f(t) \, dt. On the interval [1,4][1, 4], the absolute maximum value of G(x)G(x) occurs at which value of xx?

  1. x = 1
  2. x = 3 (correct answer)
  3. x = 4
  4. The maximum value occurs at both x=1 and x=4.
Explanation: To find the absolute maximum of G(x)G(x) on [1,4][1, 4], we must test the critical points and the endpoints. The derivative is G(x)=f(x)G'(x) = f(x). Critical points are where f(x)=0f(x) = 0, which occurs at x=1x=1 and x=3x=3. The candidates for the location of the absolute maximum are the endpoints x=1,x=4x=1, x=4 and the critical point x=3x=3. We evaluate G(x)G(x) at these points: G(1)=11f(t)dt=0G(1) = \int_1^1 f(t) \, dt = 0. G(3)=13f(t)dt=6G(3) = \int_1^3 f(t) \, dt = 6. G(4)=14f(t)dt=13f(t)dt+34f(t)dt=6+(6)=0G(4) = \int_1^4 f(t) \, dt = \int_1^3 f(t) \, dt + \int_3^4 f(t) \, dt = 6 + (-6) = 0. Comparing the values {0, 6, 0}, the maximum value is 6, which occurs at x=3x=3.

Question 14

Let f(t)f(t) be a continuous function. Define G(x)=2xf(t)dtG(x) = \int_{-2}^x f(t) \, dt. It is known that f(t)f(t) is decreasing on (,3)(-\infty, 3) and increasing on (3,)(3, \infty). Which of the following statements must be true about G(x)G(x)?

  1. G(x) has a local minimum at x=3x=3.
  2. G(x) has a local maximum at x=3x=3.
  3. G(x) has a point of inflection at x=3x=3. (correct answer)
  4. G(x) is concave up for all xx.
Explanation: We examine the derivatives of G(x)G(x). G(x)=f(x)G'(x) = f(x) and G(x)=f(x)G''(x) = f'(x). A point of inflection occurs where the concavity, determined by the sign of G(x)G''(x), changes. This means we need to find where f(x)f'(x) changes sign. The problem states that f(t)f(t) changes from decreasing to increasing at t=3t=3. This means its derivative, f(t)f'(t), changes from negative to positive at t=3t=3. Therefore, G(x)G''(x) changes sign at x=3x=3, indicating that G(x)G(x) has a point of inflection at x=3x=3.

Question 15

Let f(t)f(t) be a continuous function. Given 03f(t)dt=6\int_0^3 f(t) \, dt = 6, 37f(t)dt=10\int_3^7 f(t) \, dt = -10, and 78f(t)dt=3\int_7^8 f(t) \, dt = 3. Let F(x)=0xf(t)dtF(x) = \int_0^x f(t) \, dt. Which of the following correctly orders the values of F(0),F(3),F(7),F(8)F(0), F(3), F(7), F(8) from least to greatest?

  1. F(7) < F(8) < F(0) < F(3) (correct answer)
  2. F(0) < F(8) < F(3) < F(7)
  3. F(7) < F(0) < F(8) < F(3)
  4. F(8) < F(7) < F(0) < F(3)
Explanation: We calculate each value: F(0)=00f(t)dt=0F(0) = \int_0^0 f(t) \, dt = 0. F(3)=03f(t)dt=6F(3) = \int_0^3 f(t) \, dt = 6. F(7)=07f(t)dt=03f(t)dt+37f(t)dt=6+(10)=4F(7) = \int_0^7 f(t) \, dt = \int_0^3 f(t) \, dt + \int_3^7 f(t) \, dt = 6 + (-10) = -4. F(8)=08f(t)dt=F(7)+78f(t)dt=4+3=1F(8) = \int_0^8 f(t) \, dt = F(7) + \int_7^8 f(t) \, dt = -4 + 3 = -1. The values are F(0)=0,F(3)=6,F(7)=4,F(8)=1F(0)=0, F(3)=6, F(7)=-4, F(8)=-1. Ordering them gives: 4<1<0<6-4 < -1 < 0 < 6, which corresponds to F(7)<F(8)<F(0)<F(3)F(7) < F(8) < F(0) < F(3).

Question 16

Let f(t)f(t) be a continuous function. The accumulation function G(x)=cxf(t)dtG(x) = \int_c^x f(t) \, dt has a local maximum at x=ax=a and a point of inflection at x=bx=b. Which of the following conditions on ff must be met?

  1. f(a)=0f(a)=0 and f(b)=0f(b)=0.
  2. f(a)<0f'(a) < 0 and f(b)f'(b) changes sign.
  3. f(a)=0f(a)=0 with ff changing from positive to negative, and f(b)f'(b) changes sign. (correct answer)
  4. f(a)=0f(a)=0 with ff changing from negative to positive, and f(b)=0f(b)=0.
Explanation: For G(x)G(x) to have a local maximum at x=ax=a, its derivative G(x)=f(x)G'(x)=f(x) must change from positive to negative at x=ax=a. This implies f(a)=0f(a)=0 and the sign change condition. For G(x)G(x) to have a point of inflection at x=bx=b, its second derivative G(x)=f(x)G''(x)=f'(x) must change sign at x=bx=b. This means that f(x)f(x) must have a local extremum at x=bx=b. Choice C correctly states both conditions.

Question 17

Let f(t)f(t) and g(t)g(t) be continuous positive functions for all t0t \ge 0, and suppose f(t)>g(t)f(t) > g(t) for all t0t \ge 0. Let F(x)=0xf(t)dtF(x) = \int_0^x f(t) \, dt and G(x)=0xg(t)dtG(x) = \int_0^x g(t) \, dt. Which of the following statements is NOT necessarily true for x>0x > 0?

  1. F(x)>G(x)F(x) > G(x)
  2. F(x)>G(x)F'(x) > G'(x)
  3. The graphs of both F(x)F(x) and G(x)G(x) are concave up. (correct answer)
  4. The function H(x)=F(x)G(x)H(x) = F(x) - G(x) is strictly increasing.
Explanation: A is true because f(t)>g(t)0f(t) > g(t) \ge 0, the accumulated area for F will be greater than for G. B is true because F(x)=f(x)F'(x) = f(x) and G(x)=g(x)G'(x) = g(x), and we are given f(x)>g(x)f(x) > g(x). D is true because the derivative of H(x)H(x) is H(x)=F(x)G(x)=f(x)g(x)H'(x) = F'(x) - G'(x) = f(x) - g(x), which is given to be positive. A function with a positive derivative is increasing. C is not necessarily true. The concavity of F(x)F(x) is given by F(x)=f(x)F''(x) = f'(x). We do not have any information about whether f(t)f(t) or g(t)g(t) are increasing or decreasing, so we cannot determine the concavity of F(x)F(x) or G(x)G(x).

Question 18

Let v(t)v(t) be a continuous velocity function for a particle moving along a line. The particle's velocity is positive on (0,2)(0, 2) and negative on (2,5)(2, 5). Let D(T)=0Tv(t)dtD(T) = \int_0^T v(t) \, dt represent the particle's displacement at time TT, and let S(T)=0Tv(t)dtS(T) = \int_0^T |v(t)| \, dt represent the total distance traveled. Which of the following statements must be true for any time TT in the interval (2,5)(2, 5)?

  1. D(T)=S(T)D(T) = S(T) and D(T)D(T) is increasing.
  2. D(T)<S(T)D(T) < S(T) and D(T)D(T) is decreasing. (correct answer)
  3. D(T)<S(T)D(T) < S(T) and D(T)D(T) is increasing.
  4. D(T)>S(T)D(T) > S(T) and D(T)D(T) is decreasing.
Explanation: For any T>2T > 2, the integral for S(T)S(T) will include the interval (2,T)(2, T) where v(t)v(t) is negative. On this interval, v(t)=v(t)|v(t)| = -v(t). Thus, S(T)=02v(t)dt+2Tv(t)dtS(T) = \int_0^2 v(t) \, dt + \int_2^T -v(t) \, dt, while D(T)=02v(t)dt+2Tv(t)dtD(T) = \int_0^2 v(t) \, dt + \int_2^T v(t) \, dt. Since 2Tv(t)dt>2Tv(t)dt\int_2^T -v(t) \, dt > \int_2^T v(t) \, dt, it must be that S(T)>D(T)S(T) > D(T). The rate of change of displacement is D(T)=v(T)D'(T) = v(T). For TT in (2,5)(2, 5), v(T)<0v(T) < 0, so D(T)D(T) is decreasing.

Question 19

Let f(t)f(t) be a continuous function such that f(t)<0f(t) < 0 on (,1)(-\infty, -1) and f(t)>0f(t) > 0 on (1,)(-1, \infty). Let the function H(x)H(x) be defined as H(x)=x2f(t)dtH(x) = \int_x^2 f(t) \, dt. Which of the following statements about H(x)H(x) is correct?

  1. H(x) has a local maximum at x=1x=-1. (correct answer)
  2. H(x) has a local minimum at x=1x=-1.
  3. H(x) is always an increasing function.
  4. H(x) has a point of inflection at x=1x=-1.
Explanation: First, rewrite H(x)=2xf(t)dtH(x) = -\int_2^x f(t) \, dt. By the FTC, H(x)=f(x)H'(x) = -f(x). Critical points occur when H(x)=0H'(x)=0, so f(x)=0f(x)=0 at x=1x=-1. To classify this point, examine the sign of H(x)H'(x). For x<1x < -1, f(x)<0f(x) < 0, so H(x)=f(x)>0H'(x) = -f(x) > 0. For x>1x > -1, f(x)>0f(x) > 0, so H(x)=f(x)<0H'(x) = -f(x) < 0. Since H(x)H'(x) changes from positive to negative at x=1x=-1, H(x)H(x) has a local maximum at x=1x=-1.

Question 20

Let A(x)=1xsin(πt)dtA(x) = \int_1^x |\sin(\pi t)| \, dt. The function A(x)A(x) represents the total area between y=sin(πt)y = |\sin(\pi t)| and the tt-axis from t=1t = 1 to t=xt = x. What is the behavior of A(x)A'(x) at x=3x = 3?

  1. A(3)=0A'(3) = 0 and A(x)A'(x) changes from positive to negative as xx passes through 33
  2. A(3)=0A'(3) = 0 and A(x)A'(x) changes from negative to positive as xx passes through 33
  3. A(3)=0A'(3) = 0 but A(x)A'(x) does not change sign at x=3x = 3 (correct answer)
  4. A(3)A'(3) does not exist because A(x)A(x) has a corner at x=3x = 3
Explanation: By the Fundamental Theorem of Calculus, A(x)=sin(πx)A'(x) = |\sin(\pi x)|. At x=3x = 3, we have sin(3π)=0\sin(3\pi) = 0, so A(3)=sin(3π)=0A'(3) = |\sin(3\pi)| = 0. However, since sin(πx)0|\sin(\pi x)| \geq 0 for all xx, and sin(πx)>0|\sin(\pi x)| > 0 except at integer values of xx, we have A(x)0A'(x) \geq 0 always. Just before and after x=3x = 3, A(x)>0A'(x) > 0, so there's no sign change. The function A(x)A(x) is always increasing except at isolated points where A(x)=0A'(x) = 0. Choice D is incorrect because A(x)A(x) is differentiable everywhere since we're integrating sin(πt)|\sin(\pi t)| which is continuous.