FTC & Definite Integrals - Calculus 2

1. Let $g(x) = \int_0^x \frac{1}{t^2 + 1} dt$. If $g(a) = \frac{\pi}{6}$ for some positive value $a$, then $g'(a)$ equals:
   Answer: $\frac{3}{4}$ By FTC, $g'(x) = \frac{1}{x^2 + 1}$. Since $g(x) = \arctan(x)$ and $g(a) = \frac{\pi}{6}$, we have $\arctan(a) = \frac{\pi}{6}$, so $a = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Therefore, $g'(a) = \frac{1}{a^2 + 1} = \frac{1}{(\frac{1}{\sqrt{3}})^2 + 1} = \frac{1}{\frac{1}{3} + 1} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$. Choices A and B are common trigonometric values that students might confuse. Choice C inverts the correct answer.
2. If $h(x)$ is continuous and $\int_1^5 h(x) dx = 8$, then $\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} h\left(1 + \frac{4k}{n}\right)$ equals:
   Answer: $8$ When you see a limit involving a sum with terms like $\frac{1}{n}\sum_{k=1}^{n} f\left(a + \frac{(b-a)k}{n}\right)$, you're looking at a Riemann sum that approaches a definite integral as $n \to \infty$. Let's identify the components in $\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} h\left(1 + \frac{4k}{n}\right)$. Here, $\Delta x = \frac{4}{n}$, the lower bound is $a = 1$, and the upper bound is $b = 1 + 4 = 5$. The expression $1 + \frac{4k}{n}$ represents sample points in the interval $[1,5]$ using right endpoints. By the definition of definite integrals as limits of Riemann sums, this limit equals: $\int_1^5 h(x) dx$ Since we're given that $\int_1^5 h(x) dx = 8$, the answer is A) 8. Let's examine why the other choices are wrong: - B) 2 might result from incorrectly dividing the integral by the width of the interval ($8 ÷ 4 = 2$), but Riemann sums converge to the integral itself, not some scaled version. - C) 16 could come from mistakenly multiplying the given integral by 2, perhaps confusing the coefficient 4 in the sum. - D) 32 likely results from multiplying by 4, incorrectly thinking the factor $\frac{4}{n}$ somehow multiplies the final result. Study tip: When you see Riemann sum limits, identify the interval endpoints and recognize that the limit always equals the definite integral over that interval—no additional scaling needed.
3. Let $F(x) = \int_{x^2}^{x^3} e^{t^2} dt$. What is the value of $F'(1)$? Give your answer in simplest form.
   Answer: $e$ This question requires applying the Fundamental Theorem of Calculus, Part 1, with the chain rule. The formula for the derivative of an integral with variable limits is $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)$. Here, $f(t) = e^{t^2}$, $b(x) = x^3$, and $a(x) = x^2$. Thus, $b'(x) = 3x^2$ and $a'(x) = 2x$. Applying the formula, we get $F'(x) = e^{(x^3)^2} \cdot (3x^2) - e^{(x^2)^2} \cdot (2x) = 3x^2 e^{x^6} - 2x e^{x^4}$. Now, we evaluate at $x=1$: $F'(1) = 3(1)^2 e^{1^6} - 2(1) e^{1^4} = 3e^1 - 2e^1 = e$.
4. What is the value of the definite integral $\int_{-3}^3 |x^2 - 4| dx$?
   Answer: $\frac{46}{3}$ The integrand $|x^2 - 4|$ changes its definition based on the sign of $x^2 - 4$. The roots of $x^2 - 4 = 0$ are $x = -2$ and $x = 2$. For $x \in [-3, -2] \cup [2, 3]$, $x^2 - 4 \ge 0$. For $x \in (-2, 2)$, $x^2 - 4 < 0$. We must split the integral: $\int_{-3}^{-2} (x^2 - 4) dx + \int_{-2}^2 -(x^2 - 4) dx + \int_2^3 (x^2 - 4) dx$. Evaluating each part: $[\frac{x^3}{3} - 4x]_{-3}^{-2} = (-\frac{8}{3}+8) - (-9+12) = \frac{16}{3} - 3 = \frac{7}{3}$. $[4x - \frac{x^3}{3}]_{-2}^2 = (8-\frac{8}{3}) - (-8+\frac{8}{3}) = 2(8-\frac{8}{3}) = \frac{32}{3}$. $[\frac{x^3}{3} - 4x]_2^3 = (9-12) - (\frac{8}{3}-8) = -3 - (-\frac{16}{3}) = \frac{7}{3}$. The total is $\frac{7}{3} + \frac{32}{3} + \frac{7}{3} = \frac{46}{3}$.
5. Let $G(x) = \int_1^x (t^2-3t+2) dt$. For what values of $x$ is $G(x)$ decreasing?
   Answer: $1 < x < 2$ When you encounter a function defined as a definite integral with a variable upper limit, you're dealing with the Fundamental Theorem of Calculus. To determine where such a function is increasing or decreasing, you need to find where its derivative is positive or negative. By the Fundamental Theorem of Calculus, $G'(x) = t^2-3t+2$ evaluated at $t = x$, so $G'(x) = x^2-3x+2$. The function $G(x)$ is decreasing when $G'(x) < 0$. Let's factor the derivative: $x^2-3x+2 = (x-1)(x-2)$. To find where this is negative, analyze the sign changes. The expression equals zero when $x = 1$ or $x = 2$. For $x < 1$, both factors are negative, so the product is positive. For $1 < x < 2$, the first factor is positive and the second is negative, making the product negative. For $x > 2$, both factors are positive, so the product is positive again. Therefore, $G'(x) < 0$ only when $1 < x < 2$, making D correct. Choice A ($x < 3/2$) incorrectly identifies where the derivative changes sign—this might come from confusing the vertex of the parabola with the roots. Choice B ($x > 2$) gives where the function is increasing, not decreasing. Choice C ($x < 1$ or $x > 2$) also identifies where $G(x)$ is increasing. Study tip: Always remember that for integral functions $F(x) = \int_a^x f(t)dt$, the derivative is simply $F'(x) = f(x)$. Then use sign analysis to determine intervals of increase/decrease.
6. If $f$ is an odd function continuous on $[-3, 3]$ and $\int_0^3 f(x) dx = 7$, then $\int_{-3}^3 [f(x) + x^2] dx$ equals:
   Answer: $18$ When you encounter integrals involving odd functions, immediately think about symmetry properties. An odd function satisfies $f(-x) = -f(x)$, which creates powerful simplifications when integrating over symmetric intervals. Let's break down $\int_{-3}^3 [f(x) + x^2] dx$ by splitting it: $\int_{-3}^3 f(x) dx + \int_{-3}^3 x^2 dx$. For the first integral, since $f(x)$ is odd and we're integrating over the symmetric interval $[-3,3]$, we have $\int_{-3}^3 f(x) dx = 0$. This happens because the area from $-3$ to $0$ exactly cancels the area from $0$ to $3$. For the second integral, $x^2$ is an even function, so $\int_{-3}^3 x^2 dx = 2\int_0^3 x^2 dx = 2 \cdot \frac{x^3}{3}\Big|_0^3 = 2 \cdot 9 = 18$. Therefore, the total is $0 + 18 = 18$, which is choice A. The given information $\int_0^3 f(x) dx = 7$ is a distractor. Choice B (14) results from doubling this value, mistakenly thinking $\int_{-3}^3 f(x) dx = 2 \cdot 7 = 14$. Choice C (25) comes from adding $7 + 18$, incorrectly using the half-interval value. Choice D (32) likely stems from errors in calculating $\int_{-3}^3 x^2 dx$ or misapplying the given information. Remember: for odd functions on symmetric intervals, $\int_{-a}^a f(x) dx = 0$ always. Don't let extra given information distract you from applying fundamental properties.
7. The average value of the function $g(x) = e^x$ on the interval $[0, k]$ is $e-1$. What is the value of $k$?
   Answer: $1$ The formula for the average value of a function $g(x)$ on an interval $[a, b]$ is $\frac{1}{b-a} \int_a^b g(x) dx$. We are given that the average value of $g(x) = e^x$ on $[0, k]$ is $e-1$. Setting up the equation: $\frac{1}{k-0} \int_0^k e^x dx = e-1$. The integral of $e^x$ is $e^x$, so we have $\frac{1}{k} [e^x]_0^k = e-1$. Evaluating the antiderivative: $\frac{1}{k} (e^k - e^0) = e-1$, which simplifies to $\frac{e^k - 1}{k} = e-1$. Therefore, $e^k - 1 = k(e-1)$. Testing $k=1$: $e^1 - 1 = 1(e-1)$, which gives $e-1 = e-1$. This is true, so $k=1$.
8. Given the equation $\int_y^x \sin(t^3) dt = 2$, find the value of $\frac{dy}{dx}$ at a point $(x,y)$ on the curve.
   Answer: $\frac{\sin(x^3)}{\sin(y^3)}$ We need to use implicit differentiation with respect to $x$. Differentiating both sides of the equation $\int_y^x \sin(t^3) dt = 2$ gives: $\frac{d}{dx} \int_y^x \sin(t^3) dt = \frac{d}{dx}(2)$. Using the Fundamental Theorem of Calculus, Part 1, with the chain rule for the lower limit $y$, which is a function of $x$: $\sin(x^3) \cdot \frac{d}{dx}(x) - \sin(y^3) \cdot \frac{dy}{dx} = 0$. This simplifies to $\sin(x^3) - \sin(y^3) \frac{dy}{dx} = 0$. Now, we solve for $\frac{dy}{dx}$: $\sin(y^3) \frac{dy}{dx} = \sin(x^3)$, so $\frac{dy}{dx} = \frac{\sin(x^3)}{\sin(y^3)}$.
9. Let $G(x) = \int_0^x (\int_1^{\cos t} u^2 du) dt$. What is the value of $G''(\pi/2)$?
   Answer: $0$ This problem involves nested applications of the Fundamental Theorem of Calculus, Part 1. Let $H(t) = \int_1^{\cos t} u^2 du$. Then $G(x) = \int_0^x H(t) dt$. The first derivative is $G'(x) = H(x) = \int_1^{\cos x} u^2 du$. To find the second derivative, we differentiate $G'(x)$ with respect to $x$, again using FTC1 with the chain rule: $G''(x) = \frac{d}{dx} \int_1^{\cos x} u^2 du = ((\cos x)^2) \cdot (-\sin x) = -\cos^2(x) \sin(x)$. Now, we evaluate at $x = \pi/2$: $G''(\pi/2) = -\cos^2(\pi/2) \sin(\pi/2) = -(0)^2(1) = 0$.
10. Let $f$ be a continuous function. If $\int_0^3 f(x) dx = 6$ and the average value of $f$ on $[3, 8]$ is 4, what is the average value of $f$ on $[0, 8]$?
    Answer: $3.25$ When you encounter problems involving average values over different intervals, you need to connect the definition of average value with properties of definite integrals. The average value of a function $f$ on interval $[a,b]$ is $\frac{1}{b-a}\int_a^b f(x)dx$. You're given that $\int_0^3 f(x)dx = 6$ and the average value on $[3,8]$ is 4. Since the average value on $[3,8]$ is 4, you can find: $\frac{1}{8-3}\int_3^8 f(x)dx = 4$, so $\int_3^8 f(x)dx = 4 \times 5 = 20$. Now you can find the average value on $[0,8]$. Using the additive property of integrals: $\int_0^8 f(x)dx = \int_0^3 f(x)dx + \int_3^8 f(x)dx = 6 + 20 = 26$. Therefore, the average value on $[0,8]$ is $\frac{26}{8-0} = \frac{26}{8} = 3.25$, which is answer D. Looking at the wrong answers: A (5) might come from averaging the two given average values, but this ignores that the intervals have different lengths. B (10) could result from incorrectly calculating $\frac{20}{2}$. C (4.5) might come from adding the average values and dividing by 2, again not accounting for interval lengths. Study tip: Remember that average values over combined intervals depend on both the individual averages AND the lengths of each interval. You can't simply average the average values unless the intervals are equal in length.
11. If $\int_a^b f(x) dx = 5$ and $\int_a^b g(x) dx = -2$, which of the following statements is NOT necessarily true?
    Answer: $\int_a^b (f(x) \cdot g(x)) dx = -10$ Let's analyze each statement using the properties of definite integrals. A: $\int_a^b (f(x) - g(x)) dx = \int_a^b f(x) dx - \int_a^b g(x) dx = 5 - (-2) = 7$. This is true by the linearity property. B: $\int_b^a 3f(x) dx = -3 \int_a^b f(x) dx = -3(5) = -15$. This is true by the constant multiple rule and the property of reversing limits. C: In general, $\int_a^b (f(x) \cdot g(x)) dx \neq (\int_a^b f(x) dx) \cdot (\int_a^b g(x) dx)$. The integral of a product is not the product of the integrals. This statement is not necessarily true. D: Let $h(x) = f(x) - g(x)$. From A, we know $\int_a^b h(x) dx = 7$. If $h(x) \le 0$ for all $x \in [a,b]$, then its integral would be non-positive. Since the integral is positive (7), there must be at least one point $c$ where $h(c) > 0$, which means $f(c) > g(c)$. So this statement is necessarily true.
12. Let $F(x) = \int_{x^2}^{2x} \sin(t^3) dt$. Using the Fundamental Theorem of Calculus, $F'(x)$ equals:
    Answer: $2\sin(8x^3) - 2x\sin(x^6)$ Using the chain rule with FTC: $F'(x) = \sin((2x)^3) \cdot \frac{d}{dx}(2x) - \sin((x^2)^3) \cdot \frac{d}{dx}(x^2) = \sin(8x^3) \cdot 2 - \sin(x^6) \cdot 2x = 2\sin(8x^3) - 2x\sin(x^6)$. Choice A omits the chain rule derivatives. Choice B has the correct derivatives but wrong simplification of the cubic terms. Choice C has correct simplification but omits the chain rule derivatives.
13. The equation of the tangent line to the curve $y = \int_x^4 \frac{dt}{1+\sqrt{t}}$ at $x=4$ is:
    Answer: $y = -\frac{1}{3}(x-4)$ To find the equation of a tangent line, we need a point and a slope. The point is at $x=4$. The y-coordinate is $y(4) = \int_4^4 \frac{dt}{1+\sqrt{t}} = 0$. So the point is $(4, 0)$. The slope is the derivative $y'$ evaluated at $x=4$. First, rewrite the integral as $y = -\int_4^x \frac{dt}{1+\sqrt{t}}$. Using the Fundamental Theorem of Calculus, Part 1, $y' = -\frac{1}{1+\sqrt{x}}$. Now, evaluate the slope at $x=4$: $y'(4) = -\frac{1}{1+\sqrt{4}} = -\frac{1}{1+2} = -\frac{1}{3}$. Using the point-slope form $y - y_1 = m(x - x_1)$, we get $y - 0 = -\frac{1}{3}(x - 4)$, which is $y = -\frac{1}{3}(x-4)$.
14. Let $G(x) = \int_2^x f(t) dt$, where the function $f(t)$ is always positive and decreasing. Which statement accurately describes the function $G(x)$ for $x > 2$?
    Answer: $G(x)$ is increasing and concave down. According to the Fundamental Theorem of Calculus, Part 1, $G'(x) = f(x)$. Since we are given that $f(t)$ is always positive, $G'(x) > 0$, which means $G(x)$ is an increasing function. To determine concavity, we examine the second derivative, $G''(x)$. Differentiating $G'(x)$ gives $G''(x) = f'(x)$. We are given that $f(t)$ is a decreasing function, which means its derivative, $f'(x)$, is negative. Therefore, $G''(x) < 0$, which implies that $G(x)$ is concave down. Thus, $G(x)$ is increasing and concave down.
15. If $f(x)$ is a continuous odd function and $\int_{-2}^5 f(x) dx = 8$, what is the value of $\int_2^5 f(x) dx$?
    Answer: $8$ We can split the given integral using the additivity property: $\int_{-2}^5 f(x) dx = \int_{-2}^2 f(x) dx + \int_2^5 f(x) dx$. Since $f(x)$ is an odd function, its integral over a symmetric interval $[-a, a]$ is zero. Therefore, $\int_{-2}^2 f(x) dx = 0$. Substituting this into our equation gives $8 = 0 + \int_2^5 f(x) dx$. Thus, $\int_2^5 f(x) dx = 8$.
16. Let $f$ be a continuous function. Given $\int_1^7 f(x) dx = 10$ and $\int_5^7 (f(x) - 3) dx = 2$, what is the value of $\int_1^5 f(x) dx$?
    Answer: $2$ This problem requires using the properties of definite integrals. First, simplify the second given integral: $\int_5^7 (f(x) - 3) dx = \int_5^7 f(x) dx - \int_5^7 3 dx = 2$. The second term is $3(7-5) = 3(2) = 6$. So, $\int_5^7 f(x) dx - 6 = 2$, which implies $\int_5^7 f(x) dx = 8$. Now use the additivity property of integrals: $\int_1^7 f(x) dx = \int_1^5 f(x) dx + \int_5^7 f(x) dx$. Substituting the known values: $10 = \int_1^5 f(x) dx + 8$. Solving for the unknown integral gives $\int_1^5 f(x) dx = 10 - 8 = 2$.
17. Let $f(x)$ be a continuous function. If $\int_0^4 f(x) dx = 10$, what is the value of $\int_0^2 f(2x) dx$?
    Answer: $5$ This problem requires using u-substitution on a definite integral. Let's evaluate $\int_0^2 f(2x) dx$. Let $u = 2x$. Then $du = 2 dx$, or $dx = \frac{1}{2} du$. We also need to change the limits of integration. When $x=0$, $u = 2(0) = 0$. When $x=2$, $u = 2(2) = 4$. Substituting into the integral, we get $\int_0^4 f(u) \cdot \frac{1}{2} du = \frac{1}{2} \int_0^4 f(u) du$. Since we are given $\int_0^4 f(x) dx = 10$, the value of $\int_0^4 f(u) du$ is also 10 (the variable of integration does not change the value of the definite integral). Therefore, the result is $\frac{1}{2} (10) = 5$.
18. What is the value of $\lim_{x \to 0} \frac{1}{x} \int_0^x \cos(t^2 + t) dt$?
    Answer: $1$ This limit is in the form of the definition of a derivative. Let $G(x) = \int_0^x \cos(t^2 + t) dt$. The expression can be rewritten as $\lim_{x \to 0} \frac{G(x) - G(0)}{x - 0}$, since $G(0) = \int_0^0 \cos(t^2 + t) dt = 0$. This is the definition of $G'(0)$. By the Fundamental Theorem of Calculus, Part 1, $G'(x) = \cos(x^2 + x)$. Therefore, the value of the limit is $G'(0) = \cos(0^2 + 0) = \cos(0) = 1$.
19. Which of the following inequalities is correct without explicit calculation?
    Answer: $\int_0^1 e^{-x} dx < \int_0^1 e^{-x^2} dx$ This question uses the comparison property of integrals: if $f(x) \ge g(x)$ on $[a, b]$, then $\int_a^b f(x) dx \ge \int_a^b g(x) dx$. For option C: On the interval $(0, 1)$, we have $0 < x^2 < x < 1$, which implies $-x^2 > -x$. Since the exponential function is strictly increasing, $e^{-x^2} > e^{-x}$ for all $x \in (0, 1)$. By the comparison property, $\int_0^1 e^{-x^2} dx > \int_0^1 e^{-x} dx$, making option C correct. The other options can be shown to be false by similar comparison arguments.
20. A student attempts to evaluate $\int_{-2}^2 \frac{1}{x^4} dx$ and finds $[-\frac{1}{3x^3}]_{-2}^2 = (-\frac{1}{24}) - (-\frac{1}{-24}) = -\frac{1}{12}$. Which of the following best explains the student's error?
    Answer: The Fundamental Theorem of Calculus is not applicable because the integrand is discontinuous on the interval $[-2, 2]$. The Fundamental Theorem of Calculus (Part 2) requires the integrand to be continuous on the closed interval of integration. The function $f(x) = \frac{1}{x^4}$ has an infinite discontinuity at $x=0$, which is within the interval $[-2, 2]$. Therefore, the theorem cannot be applied directly. This is an improper integral which, in this case, diverges. The student's calculation, while mechanically following the FTC steps, leads to a meaningless result.

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