FTC & Accumulation Functions - Calculus 2

1. Let $f$ be a function such that $f'(x) = \cos(x^2)$. If $f(1) = 5$, which of the following expressions represents $f(2)$?
   Answer: $5 + \int_1^2 \cos(x^2) dx$ This question is a direct application of the accumulation concept derived from the Fundamental Theorem of Calculus Part 2. The theorem states that $\int_a^b F'(x) dx = F(b) - F(a)$. We can rewrite this as $F(b) = F(a) + \int_a^b F'(x) dx$. In this problem, we are given $f'(x) = \cos(x^2)$ and we want to find $f(2)$ given $f(1)$. Using the formula with $F=f$, $a=1$, and $b=2$, we get $f(2) = f(1) + \int_1^2 f'(x) dx$. Substituting the given values, we have $f(2) = 5 + \int_1^2 \cos(x^2) dx$.
2. Let $F(x) = \int_0^{\sin(x)} t^2 dt$. Find $F'(\pi/2)$.
   Answer: $0$ Using the Fundamental Theorem with chain rule: $F'(x) = (\sin(x))^2 \cdot \frac{d}{dx}[\sin(x)] = \sin^2(x) \cdot \cos(x)$. At $x = \pi/2$: $F'(\pi/2) = \sin^2(\pi/2) \cdot \cos(\pi/2) = 1^2 \cdot 0 = 0$. The key insight is that $\cos(\pi/2) = 0$, making the entire expression equal to zero. Choice A shows the correct formula but doesn't evaluate it. Choice B forgets the chain rule factor $\cos(x)$. Choice C gives only the chain rule factor without the integrand evaluation.
3. Let $F(x) = \int_1^x (t^2 - 4t + 3) \ln(t) dt$ for $x > 0$. At which of the following values of $x$ does $F(x)$ have a local minimum?
   Answer: $x=3$ To find local extrema of $F(x)$, we must find its critical points by taking the derivative and setting it to zero. By the Fundamental Theorem of Calculus Part 1, $F'(x) = (x^2 - 4x + 3)\ln(x)$. We set $F'(x) = 0$. This occurs if $x^2 - 4x + 3 = 0$ or $\ln(x) = 0$. Factoring gives $(x-1)(x-3) = 0$, so $x=1$ or $x=3$. Also, $\ln(x)=0$ gives $x=1$. The critical points are $x=1$ and $x=3$. We use the first derivative test. For $0 < x < 1$, $F'(x) = (+)(-) = -$. For $1 < x < 3$, $F'(x) = (-)(+) = -$. For $x > 3$, $F'(x) = (+)(+) = +$. Since $F'(x)$ changes from negative to positive at $x=3$, $F(x)$ has a local minimum at $x=3$. At $x=1$, the derivative does not change sign, so it is not a local extremum.
4. Let $F(x) = \int_x^{x+2} (4t+1) dt$. What is the value of $F'(1)$?
   Answer: 8 We can solve this problem in two ways. Method 1: Use FTC Part 1 with variable limits. We can write $F(x) = \int_x^c (4t+1) dt + \int_c^{x+2} (4t+1) dt = -\int_c^x (4t+1) dt + \int_c^{x+2} (4t+1) dt$. Taking the derivative, $F'(x) = -(4x+1) + (4(x+2)+1) \cdot (1) = -4x-1 + 4x+8+1 = 8$. Since $F'(x)=8$ for all $x$, $F'(1)=8$. Method 2: Evaluate the integral first. $F(x) = [2t^2+t]_x^{x+2} = (2(x+2)^2 + (x+2)) - (2x^2+x) = (2(x^2+4x+4) + x+2) - 2x^2-x = (2x^2+8x+8+x+2) - 2x^2-x = 8x+10$. Then, $F'(x) = 8$. So, $F'(1)=8$.
5. Let $H(x) = \int_{x^2}^{4} \cos(t) dt$. Which of the following correctly expresses $H'(x)$?
   Answer: $-\cos(x^2) \cdot 2x$ Since the variable $x$ appears in the lower limit, we can rewrite $H(x) = \int_{x^2}^{4} \cos(t) dt = -\int_{4}^{x^2} \cos(t) dt$. Now applying the Fundamental Theorem with chain rule: $H'(x) = -\cos(x^2) \cdot \frac{d}{dx}(x^2) = -\cos(x^2) \cdot 2x$. The negative sign is crucial because the variable is in the lower limit. Choice B misses the negative sign, while choices C and D incorrectly use $\sin$ instead of $\cos$.
6. Let $G(x) = \int_{\sin(x)}^{\cos(x)} e^{t^2} dt$. What is the value of $G'(\pi/4)$?
   Answer: $-\sqrt{2e}$ Using the Fundamental Theorem of Calculus with the Chain Rule for an integral of the form $\int_{h(x)}^{g(x)} f(t) dt$, the derivative is $f(g(x))g'(x) - f(h(x))h'(x)$. Here, $f(t) = e^{t^2}$, $g(x) = \cos(x)$, and $h(x) = \sin(x)$. The derivatives are $g'(x) = -\sin(x)$ and $h'(x) = \cos(x)$. So, $G'(x) = e^{(\cos x)^2} \cdot (-\sin x) - e^{(\sin x)^2} \cdot (\cos x)$. Evaluating at $x = \pi/4$, we have $\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}$. Thus, $G'(\pi/4) = e^{(\sqrt{2}/2)^2} \cdot (-\frac{\sqrt{2}}{2}) - e^{(\sqrt{2}/2)^2} \cdot (\frac{\sqrt{2}}{2}) = e^{1/2} \cdot (-\frac{\sqrt{2}}{2}) - e^{1/2} \cdot (\frac{\sqrt{2}}{2}) = -2 e^{1/2} \frac{\sqrt{2}}{2} = -\sqrt{2}e^{1/2} = -\sqrt{2e}$.
7. What is the value of the limit $\lim_{x \to 0} \frac{1}{x^3} \int_0^x \sin(t^2) dt$?
   Answer: $1/3$ The limit is of the form $\frac{0}{0}$ because $\int_0^0 \sin(t^2) dt = 0$ and $0^3 = 0$. We can apply L'Hôpital's Rule. Let the numerator be $N(x) = \int_0^x \sin(t^2) dt$ and the denominator be $D(x) = x^3$. By FTC Part 1, $N'(x) = \sin(x^2)$. The derivative of the denominator is $D'(x) = 3x^2$. Applying L'Hôpital's Rule, the limit becomes $\lim_{x \to 0} \frac{\sin(x^2)}{3x^2}$. This limit can be rewritten as $\frac{1}{3} \lim_{x \to 0} \frac{\sin(x^2)}{x^2}$. Using the known limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$, with $u = x^2$, the expression becomes $\frac{1}{3} \cdot 1 = \frac{1}{3}$.
8. Let $H(x) = \int_x^e \frac{1}{t} dt$ for $x>0$. What is the value of $H'(1/e)$?
   Answer: $-e$ This question tests the Fundamental Theorem of Calculus, specifically how to differentiate an integral where the variable appears in the limits of integration rather than inside the integrand. When you have $H(x) = \int_x^e \frac{1}{t} dt$, you need to apply the FTC carefully because $x$ is the lower limit. The general rule states that if $F(x) = \int_{g(x)}^{h(x)} f(t) dt$, then $F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$. For our function, $g(x) = x$, $h(x) = e$ (constant), and $f(t) = \frac{1}{t}$. Therefore: $H'(x) = \frac{1}{e} \cdot 0 - \frac{1}{x} \cdot 1 = -\frac{1}{x}$ At $x = \frac{1}{e}$: $H'(1/e) = -\frac{1}{1/e} = -e$ Choice D is correct. Choice A ($e$) represents forgetting the negative sign that comes from differentiating the lower limit. Choice B ($-1$) likely comes from incorrectly substituting $x = \frac{1}{e}$ before taking the derivative, or confusing this with the derivative of $\ln(x)$. Choice C ($1$) combines both errors—forgetting the negative sign and making a substitution mistake. Study tip: Remember that when the variable is in the lower limit of integration, you get a negative sign when applying the FTC. A quick way to remember: $\frac{d}{dx}\int_x^a f(t)dt = -f(x)$ and $\frac{d}{dx}\int_a^x f(t)dt = f(x)$.
9. Let $g(x) = \int_{-2}^x |t| dt$. Which of the following is the value of $g(1)$?
   Answer: $5/2$ To evaluate $g(1) = \int_{-2}^1 |t| dt$, we must split the integral at $t=0$ because the definition of $|t|$ changes there. For $t<0$, $|t| = -t$, and for $t \ge 0$, $|t| = t$. The integral becomes $\int_{-2}^0 -t dt + \int_0^1 t dt$. Evaluating the first integral: $[-\frac{t^2}{2}]_{-2}^0 = (0) - (-\frac{(-2)^2}{2}) = -(-\frac{4}{2}) = 2$. Evaluating the second integral: $[\frac{t^2}{2}]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2}$. The total value is $2 + \frac{1}{2} = \frac{5}{2}$.
10. Let $f(x) = \int_{x^2}^{x^3} \frac{1}{\ln t} dt$ for $x>1$. Find $f'(x)$.
    Answer: $\frac{3x^2}{\ln(x^3)} - \frac{2x}{\ln(x^2)}$ We use the Fundamental Theorem of Calculus for variable limits of integration: $\frac{d}{dx}\int_{h(x)}^{g(x)} F(t) dt = F(g(x))g'(x) - F(h(x))h'(x)$. In this problem, $F(t) = \frac{1}{\ln t}$, $g(x) = x^3$, and $h(x) = x^2$. The derivatives are $g'(x) = 3x^2$ and $h'(x) = 2x$. Applying the formula: $f'(x) = \frac{1}{\ln(x^3)} \cdot (3x^2) - \frac{1}{\ln(x^2)} \cdot (2x) = \frac{3x^2}{\ln(x^3)} - \frac{2x}{\ln(x^2)}$.
11. The rate at which a contaminant is leaking into a pond is given by $r(t) = 2 - \sqrt{t}$ gallons per hour, where $t$ is time in hours. The leak is stopped at $t=4$ hours. Let $A(t) = \int_0^t r(x) dx$. Which statement correctly describes the amount of contaminant in the pond?
    Answer: The amount of contaminant is at its maximum at $t=4$, because the rate of leakage is non-negative on the interval $[0, 4]$. The function $A(t) = \int_0^t r(x) dx$ represents the total accumulated amount of contaminant at time $t$. The rate of change of this amount is $A'(t) = r(t)$. The amount of contaminant increases as long as the rate $r(t)$ is positive. We set $r(t) = 2 - \sqrt{t} > 0$, which implies $2 > \sqrt{t}$, or $t < 4$. The rate is zero at $t=4$. Since the rate is positive for $0 \le t < 4$ and zero at $t=4$, the accumulated amount $A(t)$ is increasing on the entire interval $[0, 4]$. Therefore, the maximum amount of contaminant occurs at the end of the interval, at $t=4$.
12. Consider the function $G(x) = \int_0^x (t^2 + 1) dt$. If $G(a) = 12$ for some positive value $a$, what is $G'(a)$?
    Answer: $a^2 + 1$ By the Fundamental Theorem of Calculus, if $G(x) = \int_0^x f(t) dt$, then $G'(x) = f(x)$. Here, $f(t) = t^2 + 1$, so $G'(x) = x^2 + 1$ for all $x$. Therefore, $G'(a) = a^2 + 1$, regardless of the specific value of $a$. The condition $G(a) = 12$ tells us which value of $a$ we're considering, but doesn't affect the derivative formula. Choice B gives the antiderivative (which would equal 12), choice C confuses the function value with its derivative, and choice D gives the reciprocal of the correct answer.
13. Let $F(x) = \int_3^x (c-t)e^{-t^2} dt$. If $F(x)$ has a local maximum at $x=5$, what is the value of the constant $c$?
    Answer: 5 This problem tests your understanding of the Fundamental Theorem of Calculus and how to find critical points of functions defined by integrals. When you see a function defined as an integral with a variable upper limit, you'll need to differentiate it to find where extrema occur. To find where $F(x)$ has a local maximum, you need to find where $F'(x) = 0$. Using the Fundamental Theorem of Calculus, if $F(x) = \int_3^x (c-t)e^{-t^2} dt$, then $F'(x) = (c-x)e^{-x^2}$. Since $F(x)$ has a local maximum at $x = 5$, we know $F'(5) = 0$. Substituting: $F'(5) = (c-5)e^{-25} = 0$. Since $e^{-25} > 0$ (exponential functions are always positive), the only way this product equals zero is if $c - 5 = 0$, which means $c = 5$. Looking at the wrong answers: Choice (A) gives $c = 3$, which would make $F'(5) = -2e^{-25} \neq 0$. Choice (B) represents $(5-c)e^{-25}$, which is actually the negative of $F'(5)$ and represents a common algebra error. Choice (C) gives $c = e^{-25}$, which would make $F'(5) = (e^{-25} - 5)e^{-25} \neq 0$ since $e^{-25} \neq 5$. Remember: when finding critical points of integral functions, differentiate using the Fundamental Theorem of Calculus first, then set the derivative equal to zero. The exponential factor will never be zero, so focus on when the polynomial factor equals zero.
14. Let $f$ be a continuous function. The value of $\int_0^2 f(x) dx = 5$ and $\int_0^4 f(x) dx = 2$. What is the value of $\int_2^4 (3f(x) + 2x) dx$?
    Answer: 3 This problem tests your understanding of definite integral properties, particularly how to split integrals over adjacent intervals and handle linear combinations of functions. You need to find $\int_2^4 f(x) dx$ first. Using the fundamental property that integrals over adjacent intervals add up: $\int_0^4 f(x) dx = \int_0^2 f(x) dx + \int_2^4 f(x) dx$. Substituting the given values: $2 = 5 + \int_2^4 f(x) dx$, so $\int_2^4 f(x) dx = -3$. Now you can evaluate the target integral using linearity properties: $\int_2^4 (3f(x) + 2x) dx = 3\int_2^4 f(x) dx + 2\int_2^4 x dx$ The first term gives $3(-3) = -9$. For the second term: $\int_2^4 x dx = \frac{x^2}{2}\Big|_2^4 = \frac{16}{2} - \frac{4}{2} = 6$. So $2\int_2^4 x dx = 12$. Therefore: $\int_2^4 (3f(x) + 2x) dx = -9 + 12 = 3$. The answer is D. Choice A (-9) represents forgetting the $2x$ term entirely. Choice B (21) likely comes from incorrectly calculating $\int_2^4 f(x) dx = 3$ instead of $-3$, then getting $9 + 12 = 21$. Choice C (-3) is just the value of $\int_2^4 f(x) dx$ without including the polynomial term. Strategy tip: Always break these problems into steps: first find any missing definite integrals using interval addition properties, then apply linearity to separate the integral of sums and constant multiples.
15. The velocity of a particle moving along the x-axis is given by $v(t) = 3t^2 - 12$ for $t \ge 0$. The particle starts at position $x=5$ at time $t=0$. What is the position of the particle at $t=3$?
    Answer: -4 The position function $x(t)$ is the antiderivative of the velocity function $v(t)$. The position at time $t=3$ can be found using the initial position and the net displacement from $t=0$ to $t=3$. The displacement is given by the definite integral of the velocity function: $\int_0^3 v(t) dt$. The position is $x(3) = x(0) + \int_0^3 (3t^2 - 12) dt$. First, calculate the integral: $\int_0^3 (3t^2 - 12) dt = [t^3 - 12t]_0^3 = (3^3 - 12(3)) - (0) = 27 - 36 = -9$. This is the displacement. The final position is the initial position plus the displacement: $x(3) = 5 + (-9) = -4$.
16. Consider $g(x) = \int_{x}^{2x} e^t dt$. Which expression represents $g'(x)$?
    Answer: $2e^{2x} - e^x$ Since both limits of integration contain $x$, we use: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)$. Here, $u(x) = x$, $v(x) = 2x$, and $f(t) = e^t$. So $u'(x) = 1$, $v'(x) = 2$, $f(v(x)) = e^{2x}$, and $f(u(x)) = e^x$. Therefore: $g'(x) = e^{2x} \cdot 2 - e^x \cdot 1 = 2e^{2x} - e^x$. Choice A forgets to multiply $e^{2x}$ by the derivative of $2x$. Choice C treats both limits as if their derivatives were 1. Choice D incorrectly adds instead of subtracting.
17. Let $g(x) = \frac{d}{dx} \int_0^x (t^2 - 3t) dt$. What is the instantaneous rate of change of $g(x)$ at $x=2$?
    Answer: 1 This is a multi-step derivative problem. First, we must find the function $g(x)$. By the Fundamental Theorem of Calculus Part 1, $g(x) = \frac{d}{dx} \int_0^x (t^2 - 3t) dt = x^2 - 3x$. The question asks for the instantaneous rate of change of $g(x)$ at $x=2$, which means we need to find $g'(2)$. First, we find the derivative of $g(x)$: $g'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3$. Then, we evaluate $g'(x)$ at $x=2$: $g'(2) = 2(2) - 3 = 4 - 3 = 1$.
18. If $G(x) = \int_0^x f(t) dt$ where $f$ is continuous, and $G(2) = 8$, $G(5) = 20$, what is the average value of $f$ on the interval $[2, 5]$? Give your answer in simplest form.
    Answer: $4$ This question tests your understanding of the Fundamental Theorem of Calculus and the formula for average value of a function. When you see a function defined as an integral with a variable upper limit, think about how changes in that integral relate to the original function. The key insight is recognizing what $G(5) - G(2)$ represents. Since $G(x) = \int_0^x f(t) dt$, we have: $G(5) - G(2) = \int_0^5 f(t) dt - \int_0^2 f(t) dt = \int_2^5 f(t) dt$ This gives us $\int_2^5 f(t) dt = 20 - 8 = 12$. The average value of a continuous function $f$ on interval $[a,b]$ is $\frac{1}{b-a}\int_a^b f(t) dt$. Therefore, the average value on $[2,5]$ is: $\frac{1}{5-2}\int_2^5 f(t) dt = \frac{1}{3} \cdot 12 = 4$ Choice A ($\frac{12}{3}$) shows the correct setup but leaves the fraction unreduced. While mathematically equivalent to 4, this suggests incomplete work. Choice C (6) likely comes from incorrectly using the interval length as 2 instead of 3, giving $\frac{12}{2} = 6$. Choice D ($\frac{20}{5}$) represents a common error: dividing $G(5)$ by the right endpoint rather than using the difference $G(5) - G(2)$ and the correct interval length. Remember: when finding average value over $[a,b]$, you need the integral specifically over that interval, not from 0. Use properties of definite integrals to extract the piece you need.
19. Let $f$ be a continuous function. If $G(x) = \int_1^x f(t) dt$, $G(1)=0$ and $G(3)=4$, what is the average value of $f$ on the interval $[1, 3]$?
    Answer: 2 The average value of a function $f(t)$ on an interval $[a, b]$ is given by the formula $\frac{1}{b-a} \int_a^b f(t) dt$. In this case, the interval is $[1, 3]$, so the average value is $\frac{1}{3-1} \int_1^3 f(t) dt$. We are given that $G(x) = \int_1^x f(t) dt$. Therefore, the integral $\int_1^3 f(t) dt$ is equal to $G(3)$. We are given that $G(3)=4$. Substituting this into the average value formula, we get $\frac{1}{2} \cdot G(3) = \frac{1}{2} \cdot 4 = 2$. The information $G(1)=0$ is consistent with the definition of $G(x)$ but is not needed for the final calculation.
20. Let $g(x) = \int_0^x \frac{e^t}{t^2+1} dt$. On which interval is the graph of $y=g(x)$ concave up?
    Answer: For all $x$ except $x=1$ The concavity of $g(x)$ is determined by the sign of its second derivative, $g''(x)$. First, we find the first derivative using the Fundamental Theorem of Calculus Part 1: $g'(x) = \frac{e^x}{x^2+1}$. Next, we find the second derivative using the quotient rule: $g''(x) = \frac{(e^x)'(x^2+1) - e^x(x^2+1)'}{(x^2+1)^2} = \frac{e^x(x^2+1) - e^x(2x)}{(x^2+1)^2} = \frac{e^x(x^2-2x+1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$. The graph of $g(x)$ is concave up where $g''(x) > 0$. The denominator $(x^2+1)^2$ is always positive. The term $e^x$ is always positive. The term $(x-1)^2$ is positive for all $x \neq 1$ and is zero at $x=1$. Therefore, $g''(x) > 0$ for all $x$ except $x=1$.

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