Accumulation Function Behavior - Calculus 2

1. Let $f(t)$ and $g(t)$ be continuous positive functions for all $t \ge 0$, and suppose $f(t) > g(t)$ for all $t \ge 0$. Let $F(x) = \int_0^x f(t) \, dt$ and $G(x) = \int_0^x g(t) \, dt$. Which of the following statements is NOT necessarily true for $x > 0$?
   Answer: The graphs of both $F(x)$ and $G(x)$ are concave up. A is true because $f(t) > g(t) \ge 0$, the accumulated area for F will be greater than for G. B is true because $F'(x) = f(x)$ and $G'(x) = g(x)$, and we are given $f(x) > g(x)$. D is true because the derivative of $H(x)$ is $H'(x) = F'(x) - G'(x) = f(x) - g(x)$, which is given to be positive. A function with a positive derivative is increasing. C is not necessarily true. The concavity of $F(x)$ is given by $F''(x) = f'(x)$. We do not have any information about whether $f(t)$ or $g(t)$ are increasing or decreasing, so we cannot determine the concavity of $F(x)$ or $G(x)$.
2. Let $A(x) = \int_1^x |\sin(\pi t)| \, dt$. The function $A(x)$ represents the total area between $y = |\sin(\pi t)|$ and the $t$-axis from $t = 1$ to $t = x$. What is the behavior of $A'(x)$ at $x = 3$?
   Answer: $A'(3) = 0$ but $A'(x)$ does not change sign at $x = 3$ By the Fundamental Theorem of Calculus, $A'(x) = |\sin(\pi x)|$. At $x = 3$, we have $\sin(3\pi) = 0$, so $A'(3) = |\sin(3\pi)| = 0$. However, since $|\sin(\pi x)| \geq 0$ for all $x$, and $|\sin(\pi x)| > 0$ except at integer values of $x$, we have $A'(x) \geq 0$ always. Just before and after $x = 3$, $A'(x) > 0$, so there's no sign change. The function $A(x)$ is always increasing except at isolated points where $A'(x) = 0$. Choice D is incorrect because $A(x)$ is differentiable everywhere since we're integrating $|\sin(\pi t)|$ which is continuous.
3. Let $F(x) = \int_1^x f(t) \, dt$ where $f$ is a continuous function on $[0, 5]$. If $f(t) > 0$ for $t \in [1, 3)$, $f(3) = 0$, and $f(t) < 0$ for $t \in (3, 5]$, which statement about $F(x)$ is necessarily true?
   Answer: $F(x)$ is increasing on $[1, 3)$ and decreasing on $(3, 5]$ By the Fundamental Theorem of Calculus, $F'(x) = f(x)$. Since $f(t) > 0$ for $t \in [1, 3)$, we have $F'(x) > 0$ on this interval, so $F$ is increasing. Since $f(t) < 0$ for $t \in (3, 5]$, we have $F'(x) < 0$ on this interval, so $F$ is decreasing. Choice A is incorrect because we don't know $F(1)$ vs $F(5)$. Choice C is wrong because $F$ has a local maximum, not minimum, at $x = 3$. Choice D is incorrect because concavity depends on $F''(x) = f'(x)$, which we don't know.
4. Let $f(t)$ be a continuous function such that $f(t) < 0$ on $(-\infty, -1)$ and $f(t) > 0$ on $(-1, \infty)$. Let the function $H(x)$ be defined as $H(x) = \int_x^2 f(t) \, dt$. Which of the following statements about $H(x)$ is correct?
   Answer: H(x) has a local maximum at $x=-1$. First, rewrite $H(x) = -\int_2^x f(t) \, dt$. By the FTC, $H'(x) = -f(x)$. Critical points occur when $H'(x)=0$, so $f(x)=0$ at $x=-1$. To classify this point, examine the sign of $H'(x)$. For $x < -1$, $f(x) < 0$, so $H'(x) = -f(x) > 0$. For $x > -1$, $f(x) > 0$, so $H'(x) = -f(x) < 0$. Since $H'(x)$ changes from positive to negative at $x=-1$, $H(x)$ has a local maximum at $x=-1$.
5. Let $f(t)$ be a continuous function, and define $F(x) = \int_0^x f(t) \, dt$. If $F(x)$ has a point of inflection at $x=3$ and a critical point at $x=5$, what can be concluded about the function $f$?
   Answer: $f$ has a local extremum at $x=3$ and $f(5)=0$. $F'(x) = f(x)$ and $F''(x) = f'(x)$. A point of inflection for $F(x)$ at $x=3$ means $F''(3)=0$ and $F''(x)$ changes sign around $x=3$. This is equivalent to $f'(3)=0$ and $f'(x)$ changing sign, which means $f$ has a local extremum at $x=3$. A critical point for $F(x)$ at $x=5$ means $F'(5)=0$, which is equivalent to $f(5)=0$.
6. Let $f(t)$ be a continuous function such that $\int_1^x f(t) \, dt = x^2 e^{x-1} - 1$. What is the value of $f(1)$?
   Answer: 3 Let $F(x) = \int_1^x f(t) \, dt$. According to the FTC, $F'(x) = f(x)$. We are given $F(x) = x^2 e^{x-1} - 1$. We need to find the derivative of this expression using the product rule. $f(x) = F'(x) = (2x)(e^{x-1}) + (x^2)(e^{x-1} \cdot 1)$. Now, we evaluate $f(x)$ at $x=1$: $f(1) = (2\cdot 1)(e^{1-1}) + (1^2)(e^{1-1}) = 2e^0 + 1e^0 = 2(1) + 1(1) = 3$.
7. Let $f(t)$ be a continuous function such that $f(t) < 0$ on $(-\infty, 2)$ and $f(t) > 0$ on $(2, \infty)$, with $f(2)=0$. Define $F(x) = \int_4^x f(t) \, dt$. Which of the following statements about the value of the local extremum of $F(x)$ is true?
   Answer: F(x) has a local minimum at $x=2$ with a value of $F(2) < 0$. By the FTC, $F'(x) = f(x)$. The critical point is at $x=2$ since $f(2)=0$. For $x<2$, $f(x)<0$, so $F(x)$ is decreasing. For $x>2$, $f(x)>0$, so $F(x)$ is increasing. Thus, $F(x)$ has a local minimum at $x=2$. The value is $F(2) = \int_4^2 f(t) \, dt = -\int_2^4 f(t) \, dt$. On the interval $(2,4)$, $f(t) > 0$, so the integral $\int_2^4 f(t) \, dt$ is positive. Therefore, $F(2)$ must be negative.
8. Let $f(t)$ be a continuous function. The accumulation function $G(x) = \int_c^x f(t) \, dt$ has a local maximum at $x=a$ and a point of inflection at $x=b$. Which of the following conditions on $f$ must be met?
   Answer: $f(a)=0$ with $f$ changing from positive to negative, and $f'(b)$ changes sign. For $G(x)$ to have a local maximum at $x=a$, its derivative $G'(x)=f(x)$ must change from positive to negative at $x=a$. This implies $f(a)=0$ and the sign change condition. For $G(x)$ to have a point of inflection at $x=b$, its second derivative $G''(x)=f'(x)$ must change sign at $x=b$. This means that $f(x)$ must have a local extremum at $x=b$. Choice C correctly states both conditions.
9. Let $v(t)$ be a continuous velocity function for a particle moving along a line. The particle's velocity is positive on $(0, 2)$ and negative on $(2, 5)$. Let $D(T) = \int_0^T v(t) \, dt$ represent the particle's displacement at time $T$, and let $S(T) = \int_0^T |v(t)| \, dt$ represent the total distance traveled. Which of the following statements must be true for any time $T$ in the interval $(2, 5)$?
   Answer: $D(T) < S(T)$ and $D(T)$ is decreasing. For any $T > 2$, the integral for $S(T)$ will include the interval $(2, T)$ where $v(t)$ is negative. On this interval, $|v(t)| = -v(t)$. Thus, $S(T) = \int_0^2 v(t) \, dt + \int_2^T -v(t) \, dt$, while $D(T) = \int_0^2 v(t) \, dt + \int_2^T v(t) \, dt$. Since $\int_2^T -v(t) \, dt > \int_2^T v(t) \, dt$, it must be that $S(T) > D(T)$. The rate of change of displacement is $D'(T) = v(T)$. For $T$ in $(2, 5)$, $v(T) < 0$, so $D(T)$ is decreasing.
10. Let $f(t)$ be an odd continuous function, i.e., $f(-t) = -f(t)$ for all $t$. Let $F(x) = \int_{-3}^x f(t) \, dt$. If $\int_0^3 f(t) \, dt = 5$, what is the value of $F(3)$?
    Answer: 0 We want to find $F(3) = \int_{-3}^3 f(t) \, dt$. Since $f(t)$ is an odd function, its integral over a symmetric interval $[-a, a]$ is zero. Therefore, $\int_{-3}^3 f(t) \, dt = 0$. The information that $\int_0^3 f(t) \, dt = 5$ implies that $\int_{-3}^0 f(t) \, dt = -5$, and their sum is 0.
11. Let $f(t)$ be a differentiable function that is decreasing on the interval $(-\infty, 0)$ and increasing on the interval $(0, \infty)$. It is also known that $f(0) < 0$. Let $H(x) = \int_5^x f(t) \, dt$. Which statement accurately describes the behavior of $H(x)$ at $x=0$?
    Answer: H(x) has a point of inflection and is decreasing at $x=0$. We analyze the derivatives of $H(x)$. $H'(x) = f(x)$ and $H''(x) = f'(x)$. The behavior of $H(x)$ at $x=0$ depends on $H'(0)$ and $H''(0)$. We are given that $f(0) < 0$, so $H'(0) < 0$, which means $H(x)$ is decreasing at $x=0$. For a point of inflection, $H''(x)$ must change sign. $H''(x) = f'(x)$. We are told $f(t)$ changes from decreasing to increasing at $t=0$. This means $f'(t)$ changes from negative to positive at $t=0$. Therefore, $H''(x)$ changes sign at $x=0$, indicating a point of inflection.
12. The accumulation function $F(x) = \int_0^x f(t) \, dt$ has the property that $F(2) = 8$, $F(4) = 8$, and $F(6) = 12$. If $f$ is continuous and $f(t) > 0$ for all $t > 4$, what can be concluded about $f$ on the interval $[0, 4]$?
    Answer: The net area under $f$ from $t = 2$ to $t = 4$ is zero, but $f$ need not be zero on $[2, 4]$ Since $F(2) = F(4) = 8$, we have $\int_2^4 f(t) \, dt = F(4) - F(2) = 0$. This means the net area is zero, but $f$ could have both positive and negative values on $[2,4]$ that cancel out, or $f$ could be identically zero on $[2,4]$. Choice A is too restrictive - the net area being zero doesn't require $f \equiv 0$. Choice B is incorrect because $f$ could change sign multiple times. Choice D makes an unsupported claim about the signs on specific intervals. Only choice C correctly states that the net area is zero without making additional unwarranted assumptions about $f$.
13. Let $f(t)$ be a function such that $\int_2^5 f(t) \, dt = 8$ and $f(5) = 3$. Let $G(x) = \int_2^x f(t) \, dt$. What is the equation of the line tangent to the graph of $y=G(x)$ at $x=5$?
    Answer: y - 8 = 3(x - 5) The equation of a tangent line is $y - y_0 = m(x - x_0)$. Here, $x_0 = 5$. The y-coordinate is $y_0 = G(5) = \int_2^5 f(t) \, dt$, which is given as 8. The slope is $m = G'(5)$. By the FTC, $G'(x) = f(x)$, so the slope is $m = f(5)$, which is given as 3. Substituting these values into the point-slope formula gives $y - 8 = 3(x - 5)$.
14. Let $f(t)$ be a continuous function that is positive on the interval $(-\infty, 3)$ and negative on the interval $(3, \infty)$. Define an accumulation function $G(x) = \int_1^x f(t) \, dt$. Which of the following statements correctly describes $G(x)$ at $x=3$?
    Answer: G(x) has a local maximum at $x=3$. According to the Fundamental Theorem of Calculus Part 2, $G'(x) = f(x)$. To find local extrema of $G(x)$, we can use the First Derivative Test on $G'(x) = f(x)$. A critical point occurs where $f(x) = 0$, which is at $x=3$. For $x < 3$, $f(x) > 0$, so $G(x)$ is increasing. For $x > 3$, $f(x) < 0$, so $G(x)$ is decreasing. Since $G'(x)$ changes from positive to negative at $x=3$, $G(x)$ has a local maximum at this point.
15. Let $f(t)$ be a continuous function. If $G(x) = \int_0^{x^3} f(t) \, dt$, what is $G'(x)$?
    Answer: $3x^2 f(x^3)$ This is a direct application of the Fundamental Theorem of Calculus combined with the Chain Rule. Let $u(x) = x^3$. Then $G(x) = \int_0^{u(x)} f(t) \, dt$. The derivative is $G'(x) = f(u(x)) \cdot u'(x)$. Here, $u'(x) = 3x^2$. So, $G'(x) = f(x^3) \cdot 3x^2$.
16. Let $f(t)$ be a continuous function and $F(x) = \int_{-1}^x f(t) \, dt$. Given that $F(-1)=0$ and $F(3)=0$, and $F(x) > 0$ on $(-1, 3)$, which of the following statements must be true about the function $f$?
    Answer: $\int_{-1}^3 f(t) \, dt = 0$ By the definition of $F(x)$, we have $\int_{-1}^3 f(t) \, dt = F(3) = 0$. This is a direct consequence of the given information. Choice B is incorrect because $f$ 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 $f(3) < 0$ (since $F(x) > 0$ just to the left of 3 and $F(3) = 0$, so $F$ must be decreasing at $x = 3$), we cannot determine the sign of $f(-1)$ from the given information. Choice D has no basis in the given conditions.
17. Let $f(t)$ be a continuous function such that $f(t) < 0$ for all $t$. Define $F(x) = \int_1^{x^2} f(t) \, dt$. For what values of $x$ is $F(x)$ increasing?
    Answer: x < 0 By the FTC and Chain Rule, $F'(x) = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x$. We want to find where $F(x)$ is increasing, which is where $F'(x) > 0$. So we need $f(x^2) \cdot 2x > 0$. We are given that $f(t) < 0$ for all $t$. Since $x^2$ is always non-negative, $f(x^2)$ will be negative (assuming $x \neq 0$ such that $x^2$ falls in the domain of $f$). So we have $(negative) \cdot 2x > 0$. For this inequality to hold, the term $2x$ must be negative. This occurs when $x < 0$.
18. Let $f(t)$ be a continuous function. Define $G(x) = \int_{-2}^x f(t) \, dt$. It is known that $f(t)$ is decreasing on $(-\infty, 3)$ and increasing on $(3, \infty)$. Which of the following statements must be true about $G(x)$?
    Answer: G(x) has a point of inflection at $x=3$. We examine the derivatives of $G(x)$. $G'(x) = f(x)$ and $G''(x) = f'(x)$. A point of inflection occurs where the concavity, determined by the sign of $G''(x)$, changes. This means we need to find where $f'(x)$ changes sign. The problem states that $f(t)$ changes from decreasing to increasing at $t=3$. This means its derivative, $f'(t)$, changes from negative to positive at $t=3$. Therefore, $G''(x)$ changes sign at $x=3$, indicating that $G(x)$ has a point of inflection at $x=3$.
19. Let $F(x) = \int_x^{x+3} t^2 \, dt$. At what value of $x$ is the instantaneous rate of change of $F(x)$ equal to 21?
    Answer: x = 2 The instantaneous rate of change of $F(x)$ is $F'(x)$. Using the general form of the FTC, $F'(x) = (x+3)^2 \cdot 1 - x^2 \cdot 1$. Simplifying, $F'(x) = (x^2 + 6x + 9) - x^2 = 6x + 9$. We want to find $x$ such that $F'(x) = 21$. So, we solve the equation $6x + 9 = 21$. This gives $6x = 12$, so $x=2$.
20. The function $f(t)$ is continuous and has a single root at $t=c$. The accumulation function $G(x) = \int_a^x f(t) \, dt$ is decreasing for $x<c$ and increasing for $x>c$. Which of the following must be true?
    Answer: $f(t) < 0$ for $t<c$ and $f(t) > 0$ for $t>c$. The behavior of $G(x)$ is determined by its derivative, $G'(x) = f(x)$. If $G(x)$ is decreasing, its derivative must be negative. Thus, for $x<c$, $G'(x) = f(x) < 0$. If $G(x)$ is increasing, its derivative must be positive. Thus, for $x>c$, $G'(x) = f(x) > 0$. This directly corresponds to the conditions in choice B.

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