Accumulation Function Behavior - Calculus 1

1. Let $f(t)$ be a differentiable function with $f(3)=0$ and $f'(3) = -2$. Define $G(x) = \int_{0}^{x} f(t) \,dt$. What can be concluded about $G(x)$ at $x=3$?
   Answer: $G(x)$ has a local maximum at $x=3$. We can use the Second Derivative Test to classify the critical point of $G(x)$. First, we find the first derivative of $G(x)$, which is $G'(x) = f(x)$. At $x=3$, $G'(3) = f(3) = 0$. This confirms that $x=3$ is a critical point. Next, we find the second derivative of $G(x)$, which is $G''(x) = f'(x)$. At $x=3$, $G''(3) = f'(3) = -2$. Since $G'(3)=0$ and $G''(3) < 0$, the Second Derivative Test tells us that $G(x)$ has a local maximum at $x=3$.
2. Let $G(x) = \int_a^x f(t) dt$, where $f(t)$ is a linear function with a negative slope. Which of the following statements provides the best description of the graph of $G(x)$?
   Answer: A parabola opening downward. Let $f(t) = mt + b$ where $m < 0$ (negative slope). By the Fundamental Theorem of Calculus, $G'(x) = f(x) = mx + b$. The derivative of $G(x)$ is a linear function. The function $G(x)$ itself is the antiderivative of a linear function, which must be a quadratic function (a parabola). We can find $G(x)$ explicitly: $G(x) = \int_a^x (mt+b) dt = [\frac{1}{2}mt^2 + bt]_a^x = (\frac{1}{2}mx^2 + bx) - (\frac{1}{2}ma^2 + ba)$. This is a quadratic function of $x$ of the form $Ax^2+Bx+C$, where the leading coefficient is $A = \frac{1}{2}m$. Since $m < 0$, the coefficient $A$ is negative, which means the graph of $G(x)$ is a parabola opening downward.
3. Let $f(t)$ be a continuous function and define $F(x) = \int_0^x f(t) dt$. Let $H(x) = \int_0^x F(t) dt$. If $f(t)$ is positive and increasing for $t \ge 0$, which of the following describes the graph of $H(x)$ for $x > 0$?
   Answer: Increasing and concave up. We analyze $H(x)$ using its derivatives. The first derivative is $H'(x) = F(x)$ by FTC. The second derivative is $H''(x) = F'(x)$. Using FTC on $F(x)$, we get $F'(x) = f(x)$. Thus, $H''(x) = f(x)$. We are given that $f(t)$ is positive for $t \ge 0$, so $H''(x) = f(x) > 0$ for $x>0$. A positive second derivative means the graph of $H(x)$ is concave up. Now we check if $H(x)$ is increasing or decreasing by looking at $H'(x) = F(x) = \int_0^x f(t) dt$. Since $f(t)$ is positive for $t \ge 0$ and $x > 0$, the integral $\int_0^x f(t) dt$ represents a positive area and is therefore positive. So, $H'(x) > 0$ for $x > 0$, which means $H(x)$ is increasing. Thus, $H(x)$ is increasing and concave up for $x > 0$.
4. Let $G(x) = \int_{0}^{x} f(t) \,dt$. If $f$ is a continuous and strictly decreasing function on $[0, 5]$ and $f(2)=0$, which of the following statements must be true?
   Answer: $G(5) < G(2)$ We are given that $f$ is strictly decreasing and $f(2)=0$. This implies that $f(t) > 0$ for $t \in [0, 2)$ and $f(t) < 0$ for $t \in (2, 5]$. The function $G(x)$ changes its behavior based on the sign of its derivative, $G'(x) = f(x)$. On the interval $(2, 5]$, $f(x) < 0$, which means $G'(x) < 0$. Therefore, $G(x)$ is strictly decreasing on $[2, 5]$. Since $5 > 2$ and $G$ is decreasing on this interval, it must be true that $G(5) < G(2)$. The sign of $G(5)$ cannot be determined without knowing more about $f$, 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.
5. Let $G(x) = \int_{a}^{x} f(t) \,dt$, where $f$ is a continuous function. If $G(x)$ is always concave up, which of the following must be true about the function $f(t)$?
   Answer: $f(t)$ is always increasing. The concavity of $G(x)$ is determined by its second derivative, $G''(x)$. If $G(x)$ is always concave up, then $G''(x) > 0$ for all $x$ (where it is defined). From the Fundamental Theorem of Calculus, $G'(x) = f(x)$. Differentiating again gives $G''(x) = f'(x)$. Therefore, the condition $G''(x) > 0$ implies that $f'(x) > 0$. A function whose derivative is always positive is always increasing. Thus, $f(t)$ must be an always increasing function.
6. Let $f(t)$ be a continuous function on $[-4, 6]$. The only zero of $f(t)$ in the interval $(-4, 6)$ is at $t = 3$. We know that $f(t) > 0$ for $t \in [-4, 3)$ and $f(t) < 0$ for $t \in (3, 6]$. Let $G(x) = \int_{-4}^{x} f(t) \,dt$. At what value of $x$ does $G(x)$ attain its absolute maximum value on the interval $[-4, 6]$?
   Answer: $x=3$ To find the absolute maximum of $G(x)$ on the closed interval $[-4, 6]$, we must test the endpoints and the critical points. The critical points of $G(x)$ occur where $G'(x) = f(x) = 0$, which is at $x=3$. The candidates are $x=-4, x=3,$ and $x=6$. 1. At $x=-4$, $G(-4) = \int_{-4}^{-4} f(t) \,dt = 0$. 2. For $x \in [-4, 3)$, $f(x) > 0$, which means $G(x)$ is increasing. So, $G(3) > G(-4)$. 3. For $x \in (3, 6]$, $f(x) < 0$, which means $G(x)$ is decreasing. So, $G(3) > G(6)$. Since $G(x)$ increases from $x=-4$ to a maximum at $x=3$ and then decreases until $x=6$, the absolute maximum value must occur at $x=3$.
7. Let $f(t)$ be a continuous function. The area of the region bounded by the graph of $f$, the t-axis, and the lines $t=0$ and $t=4$ is 10. The area of the region bounded by the graph of $f$, the t-axis, and the lines $t=4$ and $t=7$ is 3. Given that $f(t) > 0$ for $t \in [0,4]$ and $f(t) < 0$ for $t \in [4,7]$, what is the value of $\int_{0}^{7} f(t) \,dt$?
   Answer: 7 The definite integral $\int_{a}^{b} f(t) \,dt$ represents the signed area. We can split the integral: $\int_{0}^{7} f(t) \,dt = \int_{0}^{4} f(t) \,dt + \int_{4}^{7} f(t) \,dt$. On the interval $[0,4]$, $f(t) > 0$, so the integral equals the area, which is 10. On the interval $[4,7]$, $f(t) < 0$, so the integral is the negative of the area, which is -3. Therefore, $\int_{0}^{7} f(t) \,dt = 10 + (-3) = 7$.
8. Let $f(t)$ be a continuous function such that $f(t) < 0$ for all $t$ in the interval $[2, 6]$. Let $G(x) = \int_{2}^{x} f(t) \,dt$. Which of the following statements must be true?
   Answer: $G(3) > G(5)$ The function $G(x)$ represents the net signed area under the curve of $f(t)$ from $t=2$ to $t=x$. We can analyze the behavior of $G(x)$ by examining its derivative, $G'(x) = f(x)$. Since we are given that $f(t) < 0$ on $[2, 6]$, $G'(x)$ is negative on this interval. A function with a negative derivative is decreasing. Therefore, for any $x_1 < x_2$ in the interval, $G(x_1) > G(x_2)$. Since $3 < 5$, it must be that $G(3) > G(5)$.
9. The velocity of a particle is given by $v(t) = t^2 - 4t + 3$ for $t \ge 0$. Let $P(x) = \int_0^x v(t) dt$ be the particle's displacement. The total distance traveled by the particle on the interval $[0, 3]$ is given by $D = \int_0^3 |v(t)| dt$. Which of the following correctly relates $P(3)$ and $D$?
   Answer: $D > P(3)$ Displacement $P(3)$ accounts for direction and can be negative, while total distance $D$ is always non-negative. They are equal only if the velocity $v(t)$ does not change sign on the interval. Let's find the roots of $v(t) = t^2 - 4t + 3 = (t-1)(t-3)$. The roots are $t=1$ and $t=3$. On the interval $[0, 3]$, the velocity is positive for $t \in [0, 1)$ and negative for $t \in (1, 3)$, with $v(3) = 0$. Since the velocity changes sign, we have $P(3) = \int_0^1 v(t) dt + \int_1^3 v(t) dt$ and $D = \int_0^1 v(t) dt + \int_1^3 |v(t)| dt = \int_0^1 v(t) dt - \int_1^3 v(t) dt$. Since $\int_1^3 v(t) dt < 0$, we have $D > P(3)$.
10. Let $G(x) = \int_{-2}^{x} f(t) \,dt$ and $H(x) = \int_{-2}^{x} |f(t)| \,dt$. Suppose $f(t)$ is a continuous function such that $f(t) > 0$ on $[-2, 0)$ and $f(t) < 0$ on $(0, 2]$. Which of the following statements is true for $x \in (0, 2]$?
    Answer: $G(x)$ is decreasing and $H(x)$ is increasing. We determine the behavior of the functions by examining their derivatives. For $G(x)$, we have $G'(x) = f(x)$. For $x \in (0, 2]$, we are given that $f(x) < 0$. Therefore, $G'(x)$ is negative, and $G(x)$ is decreasing on this interval. For $H(x)$, we have $H'(x) = |f(x)|$. The absolute value function always returns a non-negative value. Since $f(x)$ is not identically zero on the interval, $|f(x)| > 0$. Therefore, $H'(x)$ is positive, and $H(x)$ is increasing on this interval. Thus, $G(x)$ is decreasing and $H(x)$ is increasing.
11. Let $G(x) = \int_{2}^{x} f(t) \,dt$. If $G(x)$ is a linear function of $x$ with a slope of 3, which of the following must be true about $f(t)$?
    Answer: $f(t) = 3$ for all $t$. If $G(x)$ is a linear function with a slope of 3, then $G'(x) = 3$ for all $x$. By the Fundamental Theorem of Calculus, we know that $G'(x) = f(x)$. Therefore, we must have $f(x) = 3$ for all $x$. This means that $f(t)$ is a constant function with a value of 3.
12. Let $f(t)$ be a continuous function that is always positive. Let $G(x) = \int_{x}^{0} f(t) \,dt$. Which of the following must be true about $G(x)$ for $x > 0$?
    Answer: $G(x)$ is negative and decreasing. First, we rewrite $G(x)$ with standard integral bounds: $G(x) = \int_{x}^{0} f(t) \,dt = -\int_{0}^{x} f(t) \,dt$. Since $f(t)$ is always positive and $x > 0$, the integral $\int_{0}^{x} f(t) \,dt$ represents a positive area, so its value is positive. Therefore, $G(x)$ is the negative of a positive value, which means $G(x)$ is negative. To determine if $G(x)$ is increasing or decreasing, we find its derivative: $G'(x) = \frac{d}{dx} \left( -\int_{0}^{x} f(t) \,dt \right) = -f(x)$. Since $f(t)$ is always positive, $-f(x)$ is always negative. A function with a negative derivative is always decreasing. Thus, for $x > 0$, $G(x)$ is negative and decreasing.
13. Let $f(t)$ be a continuous, positive, and increasing function on the interval $[0, \infty)$. Let $A = \int_{1}^{3} f(t) \,dt$ and $B = \int_{2}^{4} f(t) \,dt$. Which of the following best describes the relationship between $A$ and $B$?
    Answer: $A < B$ Both integrals are over an interval of length 2. We can compare the integrals by comparing the values of the integrand $f(t)$ over their respective domains. Let's make a change of variables. For integral A, let $u = t-1$, so $A = \int_{0}^{2} f(u+1) \,du$. For integral B, let $v = t-2$, so $B = \int_{0}^{2} f(v+2) \,dv$. Now we are integrating over the same interval $[0, 2]$. Since $f$ is an increasing function, for any value $x \in (0, 2]$, we have $x+1 < x+2$, which implies $f(x+1) < f(x+2)$. Because the integrand for $B$ is strictly greater than the integrand for $A$ over the entire interval of integration (except possibly at the lower endpoint), the value of integral $B$ must be greater than the value of integral $A$.
14. Let $f(t)$ be a differentiable function that is increasing on the interval $(-\infty, 3)$ and decreasing on the interval $(3, \infty)$. Let $G(x) = \int_{-1}^{x} f(t) \,dt$. On which interval is the graph of $G(x)$ concave up?
    Answer: $(-\infty, 3)$ The concavity of the graph of $G(x)$ is determined by the sign of its second derivative, $G''(x)$. By the Fundamental Theorem of Calculus, $G'(x) = f(x)$. Differentiating again, we get $G''(x) = f'(x)$. The graph of $G(x)$ is concave up where $G''(x) > 0$, which is equivalent to $f'(x) > 0$. A function's derivative is positive on intervals where the function is increasing. We are given that $f(t)$ is increasing on the interval $(-\infty, 3)$. Thus, the graph of $G(x)$ is concave up on $(-\infty, 3)$.
15. Let $f(t)$ be a continuous function such that $f(t) > 0$ on the interval $(1, 4)$ and $f(t) < 0$ on the intervals $(-\infty, 1)$ and $(4, \infty)$. Let $G(x) = \int_{2}^{x} f(t) \,dt$. At which value of $x$ does $G(x)$ have a local maximum?
    Answer: $x=4$ By the Fundamental Theorem of Calculus Part 1, $G'(x) = f(x)$. A local maximum for $G(x)$ occurs where its derivative, $G'(x)$, changes from positive to negative. This is equivalent to finding where $f(x)$ changes from positive to negative. According to the problem description, $f(x)$ is positive on $(1, 4)$ and negative on $(4, \infty)$. Therefore, $f(x)$ changes from positive to negative at $x=4$, indicating that $G(x)$ has a local maximum at $x=4$.
16. Let $f(t)$ be a differentiable function. The function $f$ is decreasing on $(-\infty, -1)$, increasing on $(-1, 4)$, and decreasing on $(4, \infty)$. Let $G(x) = \int_{0}^{x} f(t) \,dt$. Which of the following statements describes the concavity of the graph of $G(x)$?
    Answer: $G(x)$ is concave up on $(-1, 4)$. The concavity of $G(x)$ is determined by its second derivative, $G''(x)$. From the Fundamental Theorem of Calculus, $G'(x) = f(x)$. Differentiating again gives $G''(x) = f'(x)$. The graph of $G(x)$ is concave up where $G''(x) > 0$, which means where $f'(x) > 0$. The derivative $f'(x)$ is positive on intervals where $f(x)$ is increasing. We are given that $f(x)$ is increasing on the interval $(-1, 4)$. Therefore, the graph of $G(x)$ is concave up on $(-1, 4)$.
17. Let $f(x)$ be a differentiable function with $f(1)=4$. Let $G(x) = x^2 \int_1^x f(t) dt$. Find $G'(1)$.
    Answer: 4 To find $G'(x)$, we must use the Product Rule, as $G(x)$ is the product of $x^2$ and $\int_1^x f(t) dt$. The derivative is: $G'(x) = (\frac{d}{dx}x^2) \left(\int_1^x f(t) dt\right) + x^2 \left(\frac{d}{dx}\int_1^x f(t) dt\right)$. Using the Power Rule and the Fundamental Theorem of Calculus, this becomes: $G'(x) = 2x \int_1^x f(t) dt + x^2 f(x)$. Now we evaluate this at $x=1$: $G'(1) = 2(1) \int_1^1 f(t) dt + (1)^2 f(1)$. The definite integral $\int_1^1 f(t) dt$ is 0. So, the expression simplifies to $G'(1) = 2(0) + 1 \cdot f(1) = f(1)$. We are given that $f(1)=4$, so $G'(1)=4$.
18. Let $f(t)$ be a continuous function and define $G(x) = \int_{1}^{x} f(t) \,dt$. If the average value of $f$ on the interval $[1, 5]$ is 10, what is the value of $G(5)$?
    Answer: 40 The average value of a function $f$ on an interval $[a, b]$ is given by the formula $\frac{1}{b-a} \int_{a}^{b} f(t) \,dt$. In this case, the average value of $f$ on $[1, 5]$ is $\frac{1}{5-1} \int_{1}^{5} f(t) \,dt = \frac{1}{4} \int_{1}^{5} f(t) \,dt$. We are given that this average value is 10. So, $\frac{1}{4} \int_{1}^{5} f(t) \,dt = 10$. Multiplying both sides by 4 gives $\int_{1}^{5} f(t) \,dt = 40$. By the definition of $G(x)$, $G(5) = \int_{1}^{5} f(t) \,dt$. Therefore, $G(5) = 40$.
19. Let $f(t)$ be a continuous function with $f(2) = 5$. Define $G(x) = \int_{1}^{x^2} f(t) \,dt$. What is the slope of the tangent line to the graph of $y=G(x)$ at $x=\sqrt{2}$?
    Answer: $10\sqrt{2}$ The slope of the tangent line is given by the derivative $G'(x)$. To find $G'(x)$, we use the Fundamental Theorem of Calculus Part 1 along with the Chain Rule. The derivative of the upper bound $x^2$ is $2x$. So, $G'(x) = f(x^2) \cdot 2x$. To find the slope at $x=\sqrt{2}$, we substitute this value into the expression for $G'(x)$: $G'(\sqrt{2}) = f((\sqrt{2})^2) \cdot 2(\sqrt{2}) = f(2) \cdot 2\sqrt{2}$. We are given that $f(2) = 5$, so the slope is $5 \cdot 2\sqrt{2} = 10\sqrt{2}$.
20. Let $f(t)$ be a continuous and strictly positive function for all real numbers $t$. Let $H(x) = \int_{0}^{1-x} f(t) \,dt$. Which statement accurately describes the function $H(x)$?
    Answer: $H(x)$ is never increasing. To determine where $H(x)$ is increasing, we must find the sign of its derivative, $H'(x)$. Using the Fundamental Theorem of Calculus Part 1 combined with the Chain Rule, we have $H'(x) = f(1-x) \cdot \frac{d}{dx}(1-x) = f(1-x) \cdot (-1)$. Since $f(t)$ is given to be strictly positive for all $t$, $f(1-x)$ must be positive for all $x$. Therefore, $H'(x) = -f(1-x)$ is always negative. A function whose derivative is always negative is always decreasing, which means it is never increasing.

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