Calculus 1 Quiz: Accumulation Function Behavior
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Accumulation Function BehaviorQuestion 1 of 20

Let f(t)f(t) be a continuous function. Define G(x)=xx2f(t)dtG(x) = \int_{-x}^{x^2} f(t) dt. Which of the following is an expression for G(x)G'(x)?

f(x2)f(x)f(x^2) - f(-x)
2xf(x2)f(x)2x f(x^2) - f(-x)
2xf(x2)+f(x)2x f(x^2) + f(-x)
f(x2)+f(x)f(x^2) + f(-x)
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Calculus 1 Quiz

Calculus 1 Quiz: Accumulation Function Behavior

Practice Accumulation Function Behavior in Calculus 1 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 1.

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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 function. Define G(x)=xx2f(t)dtG(x) = \int_{-x}^{x^2} f(t) dt. Which of the following is an expression for G(x)G'(x)?

  1. f(x2)f(x)f(x^2) - f(-x)
  2. 2xf(x2)f(x)2x f(x^2) - f(-x)
  3. 2xf(x2)+f(x)2x f(x^2) + f(-x) (correct answer)
  4. f(x2)+f(x)f(x^2) + f(-x)
Explanation: To differentiate an integral with variable limits in both the upper and lower bounds, we first split the integral at a constant, for example, 0: G(x)=x0f(t)dt+0x2f(t)dtG(x) = \int_{-x}^{0} f(t) dt + \int_{0}^{x^2} f(t) dt. We can rewrite the first integral as 0xf(t)dt-\int_{0}^{-x} f(t) dt. Now we differentiate term by term using the Fundamental Theorem of Calculus and the Chain Rule: G(x)=ddx(0xf(t)dt)+ddx(0x2f(t)dt)G'(x) = \frac{d}{dx} \left(-\int_{0}^{-x} f(t) dt\right) + \frac{d}{dx} \left(\int_{0}^{x^2} f(t) dt\right). This gives G(x)=f(x)(ddx(x))+f(x2)(ddx(x2))G'(x) = -f(-x) \cdot (\frac{d}{dx}(-x)) + f(x^2) \cdot (\frac{d}{dx}(x^2)). Calculating the derivatives gives G(x)=f(x)(1)+f(x2)(2x)G'(x) = -f(-x) \cdot (-1) + f(x^2) \cdot (2x). Simplifying gives G(x)=f(x)+2xf(x2)G'(x) = f(-x) + 2x f(x^2).

Question 2

Let f(t)f(t) be a continuous function and define G(x)=1xf(t)dtG(x) = \int_{1}^{x} f(t) \,dt. If the average value of ff on the interval [1,5][1, 5] is 10, what is the value of G(5)G(5)?

  1. 10
  2. 40 (correct answer)
  3. 50
  4. 2.5
Explanation: The average value of a function ff on an interval [a,b][a, b] is given by the formula 1baabf(t)dt\frac{1}{b-a} \int_{a}^{b} f(t) \,dt. In this case, the average value of ff on [1,5][1, 5] is 15115f(t)dt=1415f(t)dt\frac{1}{5-1} \int_{1}^{5} f(t) \,dt = \frac{1}{4} \int_{1}^{5} f(t) \,dt. We are given that this average value is 10. So, 1415f(t)dt=10\frac{1}{4} \int_{1}^{5} f(t) \,dt = 10. Multiplying both sides by 4 gives 15f(t)dt=40\int_{1}^{5} f(t) \,dt = 40. By the definition of G(x)G(x), G(5)=15f(t)dtG(5) = \int_{1}^{5} f(t) \,dt. Therefore, G(5)=40G(5) = 40.

Question 3

Let G(x)=0xf(t)dtG(x) = \int_{0}^{x} f(t) \,dt. If ff is a continuous and strictly decreasing function on [0,5][0, 5] and f(2)=0f(2)=0, which of the following statements must be true?

  1. G(5)>0G(5) > 0
  2. G(5)<0G(5) < 0
  3. G(5)<G(2)G(5) < G(2) (correct answer)
  4. G(5)>G(0)G(5) > G(0)
Explanation: We are given that ff is strictly decreasing and f(2)=0f(2)=0. This implies that f(t)>0f(t) > 0 for t[0,2)t \in [0, 2) and f(t)<0f(t) < 0 for t(2,5]t \in (2, 5]. The function G(x)G(x) changes its behavior based on the sign of its derivative, G(x)=f(x)G'(x) = f(x). On the interval (2,5](2, 5], f(x)<0f(x) < 0, which means G(x)<0G'(x) < 0. Therefore, G(x)G(x) is strictly decreasing on [2,5][2, 5]. Since 5>25 > 2 and GG is decreasing on this interval, it must be true that G(5)<G(2)G(5) < G(2). The sign of G(5)G(5) cannot be determined without knowing more about ff, as it depends on whether the positive area from 0 to 2 is larger or smaller than the magnitude of the negative area from 2 to 5.

Question 4

Let f(t)f(t) be a continuous and strictly positive function for all real numbers tt. Let H(x)=01xf(t)dtH(x) = \int_{0}^{1-x} f(t) \,dt. Which statement accurately describes the function H(x)H(x)?

  1. H(x)H(x) is increasing for x<1x < 1.
  2. H(x)H(x) is increasing for x>1x > 1.
  3. H(x)H(x) is always increasing.
  4. H(x)H(x) is never increasing. (correct answer)
Explanation: To determine where H(x)H(x) is increasing, we must find the sign of its derivative, H(x)H'(x). Using the Fundamental Theorem of Calculus Part 1 combined with the Chain Rule, we have H(x)=f(1x)ddx(1x)=f(1x)(1)H'(x) = f(1-x) \cdot \frac{d}{dx}(1-x) = f(1-x) \cdot (-1). Since f(t)f(t) is given to be strictly positive for all tt, f(1x)f(1-x) must be positive for all xx. Therefore, H(x)=f(1x)H'(x) = -f(1-x) is always negative. A function whose derivative is always negative is always decreasing, which means it is never increasing.

Question 5

Let G(x)=axf(t)dtG(x) = \int_{a}^{x} f(t) \,dt, where ff is a continuous function. If G(x)G(x) is always concave up, which of the following must be true about the function f(t)f(t)?

  1. f(t)f(t) is always positive.
  2. f(t)f(t) is always negative.
  3. f(t)f(t) is always increasing. (correct answer)
  4. f(t)f(t) is always decreasing.
Explanation: The concavity of G(x)G(x) is determined by its second derivative, G(x)G''(x). If G(x)G(x) is always concave up, then G(x)>0G''(x) > 0 for all xx (where it is defined). From the Fundamental Theorem of Calculus, G(x)=f(x)G'(x) = f(x). Differentiating again gives G(x)=f(x)G''(x) = f'(x). Therefore, the condition G(x)>0G''(x) > 0 implies that f(x)>0f'(x) > 0. A function whose derivative is always positive is always increasing. Thus, f(t)f(t) must be an always increasing function.

Question 6

Let f(x)f(x) be a differentiable function with f(1)=4f(1)=4. Let G(x)=x21xf(t)dtG(x) = x^2 \int_1^x f(t) dt. Find G(1)G'(1).

  1. 0
  2. 8
  3. 4 (correct answer)
  4. Cannot be determined from the given information.
Explanation: To find G(x)G'(x), we must use the Product Rule, as G(x)G(x) is the product of x2x^2 and 1xf(t)dt\int_1^x f(t) dt. The derivative is: G(x)=(ddxx2)(1xf(t)dt)+x2(ddx1xf(t)dt)G'(x) = (\frac{d}{dx}x^2) \left(\int_1^x f(t) dt\right) + x^2 \left(\frac{d}{dx}\int_1^x f(t) dt\right). Using the Power Rule and the Fundamental Theorem of Calculus, this becomes: G(x)=2x1xf(t)dt+x2f(x)G'(x) = 2x \int_1^x f(t) dt + x^2 f(x). Now we evaluate this at x=1x=1: G(1)=2(1)11f(t)dt+(1)2f(1)G'(1) = 2(1) \int_1^1 f(t) dt + (1)^2 f(1). The definite integral 11f(t)dt\int_1^1 f(t) dt is 0. So, the expression simplifies to G(1)=2(0)+1f(1)=f(1)G'(1) = 2(0) + 1 \cdot f(1) = f(1). We are given that f(1)=4f(1)=4, so G(1)=4G'(1)=4.

Question 7

Let f(t)f(t) be a differentiable function. The function ff is decreasing on (,1)(-\infty, -1), increasing on (1,4)(-1, 4), and decreasing on (4,)(4, \infty). Let G(x)=0xf(t)dtG(x) = \int_{0}^{x} f(t) \,dt. Which of the following statements describes the concavity of the graph of G(x)G(x)?

  1. G(x)G(x) is concave up on (,1)(-\infty, -1) and (4,)(4, \infty).
  2. G(x)G(x) is concave up on (1,4)(-1, 4). (correct answer)
  3. G(x)G(x) has inflection points at x=1x=-1 and x=4x=4, but the concavity is otherwise undetermined.
  4. The concavity of G(x)G(x) depends on the sign of f(t)f(t), which is not given.
Explanation: The concavity of G(x)G(x) is determined by its second derivative, G(x)G''(x). From the Fundamental Theorem of Calculus, G(x)=f(x)G'(x) = f(x). Differentiating again gives G(x)=f(x)G''(x) = f'(x). The graph of G(x)G(x) is concave up where G(x)>0G''(x) > 0, which means where f(x)>0f'(x) > 0. The derivative f(x)f'(x) is positive on intervals where f(x)f(x) is increasing. We are given that f(x)f(x) is increasing on the interval (1,4)(-1, 4). Therefore, the graph of G(x)G(x) is concave up on (1,4)(-1, 4).

Question 8

Let G(x)=2xf(t)dtG(x) = \int_{-2}^{x} f(t) \,dt and H(x)=2xf(t)dtH(x) = \int_{-2}^{x} |f(t)| \,dt. Suppose f(t)f(t) is a continuous function such that f(t)>0f(t) > 0 on [2,0)[-2, 0) and f(t)<0f(t) < 0 on (0,2](0, 2]. Which of the following statements is true for x(0,2]x \in (0, 2]?

  1. Both G(x)G(x) and H(x)H(x) are increasing.
  2. Both G(x)G(x) and H(x)H(x) are decreasing.
  3. G(x)G(x) is decreasing and H(x)H(x) is increasing. (correct answer)
  4. G(x)G(x) is increasing and H(x)H(x) is decreasing.
Explanation: We determine the behavior of the functions by examining their derivatives. For G(x)G(x), we have G(x)=f(x)G'(x) = f(x). For x(0,2]x \in (0, 2], we are given that f(x)<0f(x) < 0. Therefore, G(x)G'(x) is negative, and G(x)G(x) is decreasing on this interval. For H(x)H(x), we have H(x)=f(x)H'(x) = |f(x)|. The absolute value function always returns a non-negative value. Since f(x)f(x) is not identically zero on the interval, f(x)>0|f(x)| > 0. Therefore, H(x)H'(x) is positive, and H(x)H(x) is increasing on this interval. Thus, G(x)G(x) is decreasing and H(x)H(x) is increasing.

Question 9

The velocity of a particle is given by v(t)=t24t+3v(t) = t^2 - 4t + 3 for t0t \ge 0. Let P(x)=0xv(t)dtP(x) = \int_0^x v(t) dt be the particle's displacement. The total distance traveled by the particle on the interval [0,3][0, 3] is given by D=03v(t)dtD = \int_0^3 |v(t)| dt. Which of the following correctly relates P(3)P(3) and DD?

  1. D=P(3)D = P(3)
  2. D>P(3)D > P(3) (correct answer)
  3. D<P(3)D < P(3)
  4. D=P(3)D = -P(3)
Explanation: Displacement P(3)P(3) accounts for direction and can be negative, while total distance DD is always non-negative. They are equal only if the velocity v(t)v(t) does not change sign on the interval. Let's find the roots of v(t)=t24t+3=(t1)(t3)v(t) = t^2 - 4t + 3 = (t-1)(t-3). The roots are t=1t=1 and t=3t=3. On the interval [0,3][0, 3], the velocity is positive for t[0,1)t \in [0, 1) and negative for t(1,3)t \in (1, 3), with v(3)=0v(3) = 0. Since the velocity changes sign, we have P(3)=01v(t)dt+13v(t)dtP(3) = \int_0^1 v(t) dt + \int_1^3 v(t) dt and D=01v(t)dt+13v(t)dt=01v(t)dt13v(t)dtD = \int_0^1 v(t) dt + \int_1^3 |v(t)| dt = \int_0^1 v(t) dt - \int_1^3 v(t) dt. Since 13v(t)dt<0\int_1^3 v(t) dt < 0, we have D>P(3)D > P(3).

Question 10

Let f(t)f(t) be a continuous function such that f(t)>0f(t) > 0 on the interval (1,4)(1, 4) and f(t)<0f(t) < 0 on the intervals (,1)(-\infty, 1) and (4,)(4, \infty). Let G(x)=2xf(t)dtG(x) = \int_{2}^{x} f(t) \,dt. At which value of xx does G(x)G(x) have a local maximum?

  1. x=1x=1
  2. x=2x=2
  3. x=4x=4 (correct answer)
  4. There is not enough information to determine the location of a local maximum.
Explanation: By the Fundamental Theorem of Calculus Part 1, G(x)=f(x)G'(x) = f(x). A local maximum for G(x)G(x) occurs where its derivative, G(x)G'(x), changes from positive to negative. This is equivalent to finding where f(x)f(x) changes from positive to negative. According to the problem description, f(x)f(x) is positive on (1,4)(1, 4) and negative on (4,)(4, \infty). Therefore, f(x)f(x) changes from positive to negative at x=4x=4, indicating that G(x)G(x) has a local maximum at x=4x=4.

Question 11

Let f(t)f(t) be a differentiable function that is increasing on the interval (,3)(-\infty, 3) and decreasing on the interval (3,)(3, \infty). Let G(x)=1xf(t)dtG(x) = \int_{-1}^{x} f(t) \,dt. On which interval is the graph of G(x)G(x) concave up?

  1. (,3)(-\infty, 3) (correct answer)
  2. (3,)(3, \infty)
  3. On the interval(s) where f(t)>0f(t) > 0.
  4. On the interval(s) where G(x)>0G(x) > 0.
Explanation: The concavity of the graph of G(x)G(x) is determined by the sign of its second derivative, G(x)G''(x). By the Fundamental Theorem of Calculus, G(x)=f(x)G'(x) = f(x). Differentiating again, we get G(x)=f(x)G''(x) = f'(x). The graph of G(x)G(x) is concave up where G(x)>0G''(x) > 0, which is equivalent to f(x)>0f'(x) > 0. A function's derivative is positive on intervals where the function is increasing. We are given that f(t)f(t) is increasing on the interval (,3)(-\infty, 3). Thus, the graph of G(x)G(x) is concave up on (,3)(-\infty, 3).

Question 12

Let f(t)f(t) be a continuous function on [4,6][-4, 6]. The only zero of f(t)f(t) in the interval (4,6)(-4, 6) is at t=3t = 3. We know that f(t)>0f(t) > 0 for t[4,3)t \in [-4, 3) and f(t)<0f(t) < 0 for t(3,6]t \in (3, 6]. Let G(x)=4xf(t)dtG(x) = \int_{-4}^{x} f(t) \,dt. At what value of xx does G(x)G(x) attain its absolute maximum value on the interval [4,6][-4, 6]?

  1. x=4x=-4
  2. x=3x=3 (correct answer)
  3. x=6x=6
  4. The maximum occurs at an unknown point in the interval [4,3)[-4, 3).
Explanation: To find the absolute maximum of G(x)G(x) on the closed interval [4,6][-4, 6], we must test the endpoints and the critical points. The critical points of G(x)G(x) occur where G(x)=f(x)=0G'(x) = f(x) = 0, which is at x=3x=3. The candidates are x=4,x=3,x=-4, x=3, and x=6x=6.
  1. At x=4x=-4, G(4)=44f(t)dt=0G(-4) = \int_{-4}^{-4} f(t) \,dt = 0.
  2. For x[4,3)x \in [-4, 3), f(x)>0f(x) > 0, which means G(x)G(x) is increasing. So, G(3)>G(4)G(3) > G(-4).
  3. For x(3,6]x \in (3, 6], f(x)<0f(x) < 0, which means G(x)G(x) is decreasing. So, G(3)>G(6)G(3) > G(6). Since G(x)G(x) increases from x=4x=-4 to a maximum at x=3x=3 and then decreases until x=6x=6, the absolute maximum value must occur at x=3x=3.

Question 13

Let f(t)f(t) be a continuous function. The area of the region bounded by the graph of ff, the t-axis, and the lines t=0t=0 and t=4t=4 is 10. The area of the region bounded by the graph of ff, the t-axis, and the lines t=4t=4 and t=7t=7 is 3. Given that f(t)>0f(t) > 0 for t[0,4]t \in [0,4] and f(t)<0f(t) < 0 for t[4,7]t \in [4,7], what is the value of 07f(t)dt\int_{0}^{7} f(t) \,dt?

  1. 13
  2. 7 (correct answer)
  3. -7
  4. -13
Explanation: The definite integral abf(t)dt\int_{a}^{b} f(t) \,dt represents the signed area. We can split the integral: 07f(t)dt=04f(t)dt+47f(t)dt\int_{0}^{7} f(t) \,dt = \int_{0}^{4} f(t) \,dt + \int_{4}^{7} f(t) \,dt. On the interval [0,4][0,4], f(t)>0f(t) > 0, so the integral equals the area, which is 10. On the interval [4,7][4,7], f(t)<0f(t) < 0, so the integral is the negative of the area, which is -3. Therefore, 07f(t)dt=10+(3)=7\int_{0}^{7} f(t) \,dt = 10 + (-3) = 7.

Question 14

Let f(t)f(t) be a continuous function and define F(x)=0xf(t)dtF(x) = \int_0^x f(t) dt. Let H(x)=0xF(t)dtH(x) = \int_0^x F(t) dt. If f(t)f(t) is positive and increasing for t0t \ge 0, which of the following describes the graph of H(x)H(x) for x>0x > 0?

  1. Increasing and concave up. (correct answer)
  2. Increasing and concave down.
  3. Decreasing and concave up.
  4. Decreasing and concave down.
Explanation: We analyze H(x)H(x) using its derivatives. The first derivative is H(x)=F(x)H'(x) = F(x) by FTC. The second derivative is H(x)=F(x)H''(x) = F'(x). Using FTC on F(x)F(x), we get F(x)=f(x)F'(x) = f(x). Thus, H(x)=f(x)H''(x) = f(x). We are given that f(t)f(t) is positive for t0t \ge 0, so H(x)=f(x)>0H''(x) = f(x) > 0 for x>0x>0. A positive second derivative means the graph of H(x)H(x) is concave up. Now we check if H(x)H(x) is increasing or decreasing by looking at H(x)=F(x)=0xf(t)dtH'(x) = F(x) = \int_0^x f(t) dt. Since f(t)f(t) is positive for t0t \ge 0 and x>0x > 0, the integral 0xf(t)dt\int_0^x f(t) dt represents a positive area and is therefore positive. So, H(x)>0H'(x) > 0 for x>0x > 0, which means H(x)H(x) is increasing. Thus, H(x)H(x) is increasing and concave up for x>0x > 0.

Question 15

Let f(t)f(t) be a continuous, positive, and increasing function on the interval [0,)[0, \infty). Let A=13f(t)dtA = \int_{1}^{3} f(t) \,dt and B=24f(t)dtB = \int_{2}^{4} f(t) \,dt. Which of the following best describes the relationship between AA and BB?

  1. A<BA < B (correct answer)
  2. A>BA > B
  3. A=BA = B
  4. The relationship cannot be determined without knowing f(t)f(t).
Explanation: Both integrals are over an interval of length 2. We can compare the integrals by comparing the values of the integrand f(t)f(t) over their respective domains. Let's make a change of variables. For integral A, let u=t1u = t-1, so A=02f(u+1)duA = \int_{0}^{2} f(u+1) \,du. For integral B, let v=t2v = t-2, so B=02f(v+2)dvB = \int_{0}^{2} f(v+2) \,dv. Now we are integrating over the same interval [0,2][0, 2]. Since ff is an increasing function, for any value x(0,2]x \in (0, 2], we have x+1<x+2x+1 < x+2, which implies f(x+1)<f(x+2)f(x+1) < f(x+2). Because the integrand for BB is strictly greater than the integrand for AA over the entire interval of integration (except possibly at the lower endpoint), the value of integral BB must be greater than the value of integral AA.

Question 16

Let f(t)f(t) be a differentiable function with f(3)=0f(3)=0 and f(3)=2f'(3) = -2. Define G(x)=0xf(t)dtG(x) = \int_{0}^{x} f(t) \,dt. What can be concluded about G(x)G(x) at x=3x=3?

  1. G(x)G(x) has a local maximum at x=3x=3. (correct answer)
  2. G(x)G(x) has a local minimum at x=3x=3.
  3. G(x)G(x) has an inflection point at x=3x=3.
  4. Nothing can be concluded without more information about f(t)f(t).
Explanation: We can use the Second Derivative Test to classify the critical point of G(x)G(x). First, we find the first derivative of G(x)G(x), which is G(x)=f(x)G'(x) = f(x). At x=3x=3, G(3)=f(3)=0G'(3) = f(3) = 0. This confirms that x=3x=3 is a critical point. Next, we find the second derivative of G(x)G(x), which is G(x)=f(x)G''(x) = f'(x). At x=3x=3, G(3)=f(3)=2G''(3) = f'(3) = -2. Since G(3)=0G'(3)=0 and G(3)<0G''(3) < 0, the Second Derivative Test tells us that G(x)G(x) has a local maximum at x=3x=3.

Question 17

Let f(t)f(t) be a continuous function that is always positive. Let G(x)=x0f(t)dtG(x) = \int_{x}^{0} f(t) \,dt. Which of the following must be true about G(x)G(x) for x>0x > 0?

  1. G(x)G(x) is positive and increasing.
  2. G(x)G(x) is positive and decreasing.
  3. G(x)G(x) is negative and increasing.
  4. G(x)G(x) is negative and decreasing. (correct answer)
Explanation: First, we rewrite G(x)G(x) with standard integral bounds: G(x)=x0f(t)dt=0xf(t)dtG(x) = \int_{x}^{0} f(t) \,dt = -\int_{0}^{x} f(t) \,dt. Since f(t)f(t) is always positive and x>0x > 0, the integral 0xf(t)dt\int_{0}^{x} f(t) \,dt represents a positive area, so its value is positive. Therefore, G(x)G(x) is the negative of a positive value, which means G(x)G(x) is negative. To determine if G(x)G(x) is increasing or decreasing, we find its derivative: G(x)=ddx(0xf(t)dt)=f(x)G'(x) = \frac{d}{dx} \left( -\int_{0}^{x} f(t) \,dt \right) = -f(x). Since f(t)f(t) is always positive, f(x)-f(x) is always negative. A function with a negative derivative is always decreasing. Thus, for x>0x > 0, G(x)G(x) is negative and decreasing.

Question 18

Let f(t)f(t) be a continuous function with f(2)=5f(2) = 5. Define G(x)=1x2f(t)dtG(x) = \int_{1}^{x^2} f(t) \,dt. What is the slope of the tangent line to the graph of y=G(x)y=G(x) at x=2x=\sqrt{2}?

  1. 5
  2. 10
  3. 525\sqrt{2}
  4. 10210\sqrt{2} (correct answer)
Explanation: The slope of the tangent line is given by the derivative G(x)G'(x). To find G(x)G'(x), we use the Fundamental Theorem of Calculus Part 1 along with the Chain Rule. The derivative of the upper bound x2x^2 is 2x2x. So, G(x)=f(x2)2xG'(x) = f(x^2) \cdot 2x. To find the slope at x=2x=\sqrt{2}, we substitute this value into the expression for G(x)G'(x): G(2)=f((2)2)2(2)=f(2)22G'(\sqrt{2}) = f((\sqrt{2})^2) \cdot 2(\sqrt{2}) = f(2) \cdot 2\sqrt{2}. We are given that f(2)=5f(2) = 5, so the slope is 522=1025 \cdot 2\sqrt{2} = 10\sqrt{2}.

Question 19

Let G(x)=axf(t)dtG(x) = \int_a^x f(t) dt, where f(t)f(t) is a linear function with a negative slope. Which of the following statements provides the best description of the graph of G(x)G(x)?

  1. A line with a negative slope.
  2. A parabola opening upward.
  3. A parabola opening downward. (correct answer)
  4. A cubic function with a local maximum.
Explanation: Let f(t)=mt+bf(t) = mt + b where m<0m < 0 (negative slope). By the Fundamental Theorem of Calculus, G(x)=f(x)=mx+bG'(x) = f(x) = mx + b. The derivative of G(x)G(x) is a linear function. The function G(x)G(x) itself is the antiderivative of a linear function, which must be a quadratic function (a parabola). We can find G(x)G(x) explicitly: G(x)=ax(mt+b)dt=[12mt2+bt]ax=(12mx2+bx)(12ma2+ba)G(x) = \int_a^x (mt+b) dt = [\frac{1}{2}mt^2 + bt]_a^x = (\frac{1}{2}mx^2 + bx) - (\frac{1}{2}ma^2 + ba). This is a quadratic function of xx of the form Ax2+Bx+CAx^2+Bx+C, where the leading coefficient is A=12mA = \frac{1}{2}m. Since m<0m < 0, the coefficient AA is negative, which means the graph of G(x)G(x) is a parabola opening downward.

Question 20

Let f(t)f(t) be an odd, continuous function, meaning f(t)=f(t)f(-t) = -f(t) for all tt. Let G(x)=2xf(t)dtG(x) = \int_{-2}^{x} f(t) \,dt. If G(2)=10G(2) = 10, what is the value of G(2)G(-2)?

  1. -10
  2. 0 (correct answer)
  3. 10
  4. 20
Explanation: The value of G(2)G(-2) is found by substituting x=2x=-2 into the definition of G(x)G(x). This gives G(2)=22f(t)dtG(-2) = \int_{-2}^{-2} f(t) \,dt. A definite integral where the upper and lower limits of integration are the same is always equal to 0. The information about f(t)f(t) being an odd function and G(2)=10G(2)=10 is extra information designed to distract. The value of G(2)G(-2) is determined solely by the limits of integration.