📝 Functions defined by integrals FTC (26 MCQs)

📖 Explanation: The Fundamental Theorem of Calculus Part 2 states that S'(x) = \sin\left(\frac{\pi x^2}{2}\right). Substituting $x=2$ gives $\sin\left(\frac{\pi \cdot 4}{2}\right) = \sin(2\pi)$. This tests whether students can correctly apply the theorem rather than attempting to evaluate the integral itself, which is impossible using elementary functions.

📖 Explanation: The integrand $e^{-t^2}$ is positive for all $t$, so the integral from 0 to $x$ is positive for $x>0$. By the FTC, F'(x) = e^{-x^2} > 0, so $F$ is increasing, not decreasing. $F(0)=0$, so $x=0$ is an absolute minimum, not a maximum. The function is not elementary as it cannot be expressed using elementary functions.

📖 Explanation: The integral $\int_{2}^{x} \frac{1}{t}\,dt = [\ln t]_{2}^{x} = \ln x - \ln 2$. The student's claim would only be true if the lower limit was 1. This error highlights the importance of the lower limit of integration when defining functions via integrals. The constant difference is $\ln 2$, not an arbitrary integration constant.

📖 Explanation: The derivative of an integral with a constant lower limit and a variable upper limit is found by substituting the upper limit into the integrand. Since the constant factor $\frac{2}{\sqrt{\pi}}$ multiplies the integral, the derivative is $\frac{2}{\sqrt{\pi}} e^{-x^2}$. Option B correctly applies the FTC, while other options confuse the integrand or forget the chain rule.

📖 Explanation: This requires applying the chain rule to the FTC. The upper limit is $g(x) = x^2$, so H'(x) = \cos(g(x)) \cdot g'(x) = \cos(x^2) \cdot 2x. Students often forget the chain rule and incorrectly choose option A. This tests the understanding of differentiating functions defined by integrals with variable limits.

📖 Explanation: The function $F(x) = \int_{1}^{x} \frac{\sin t}{t}\,dt$ is the sine integral, a non-elementary function. It is differentiable by the FTC on its domain $(0, \infty)$, and its derivative is $\frac{\sin x}{x}$. The second derivative can be computed by differentiating $\frac{\sin x}{x}$. Only the claim that it is elementary is false.

📖 Explanation: Acceleration a(t) = v'(t). By the FTC, v'(t) = \cos(t^2). At $t = \sqrt{\pi}$, $a = \cos(\pi) = -1$. This problem connects the concept of derivatives of integral-defined functions to physics, requiring students to apply the FTC in a modeling context. Option B is a common distractor if one assumes acceleration is zero.

📖 Explanation: An antiderivative of $\ln x$ is $x\ln x - x + C$. Option A is a definite integral, not an antiderivative (it is a specific number for a given x). Option B is a valid antiderivative, and Option C represents its general family. This question tests the distinction between definite integrals, indefinite integrals, and the fundamental theorem's Easy to find antiderivatives.

📖 Explanation: If $f(t)$ changes from positive to negative at $x=2$, then F'(x) = f(x) changes sign from positive to negative, so $F$ has a local maximum at $x=2$. $F$ decreases where $f$ is negative, so (2,4) is decreasing. Concavity depends on f'(x), not directly from the sign of f. This requires interpreting the behavior of an integral-defined function based on its derivative.

📖 Explanation: Since F'(x) = \frac{1}{1+x^2} and G'(x) = F'(x) + 0, both are antiderivatives. By definition, $G(x) = F(x) + 3$. This consolidates the understanding that adding a constant to an integral-defined function produces another function with the same derivative, illustrating the general antiderivative concept.

📖 Explanation: $F(0) = \int_{0}^{0} f(t)\,dt = 0$. It is often thought of as the 'constant' that disappears when evaluating a definite integral, but here it's a specific value from the antiderivative. This clarifies the relationship between definite and indefinite integrals and the role of the lower limit. Mistaking it for an arbitrary constant is a common misconception.

📖 Explanation: The integral $\int_{1}^{x} \frac{1}{t}\,dt$ is the definition of the natural logarithm. By the FTC, it is differentiable (and therefore continuous) on its domain. Options A, B, and C are all correct characterizations. This question tests the foundational definition of logarithms via integrals and the resulting properties.

📖 Explanation: By the definition of the derivative and the FTC, \lim_{x \to 0} \frac{F(x) - F(0)}{x} = F'(0) = \cos(0^2) = 1. Since $F(0)=0$, this limit is exactly F'(0). This problem combines the definition of the derivative with the FTC, testing a deeper understanding of both concepts. Options A and D are common misconceptions.

📖 Explanation: The growth rate is P'(t) = \frac{500}{1+t^2}. At $t=1$, P'(1) = 250. This is an Easy problem that requires differentiating the integral-defined function and interpreting the result in a real-world context (people per year). Students must also handle the units correctly, recognizing that the derivative gives a rate.

📖 Explanation: The function $F(x) = \int_{0}^{x} e^{t^2}\,dt$ has derivative $e^{x^2}$ by the FTC, so it is an antiderivative. This is a key Easy of Part 2 of the FTC, showing that even when an antiderivative cannot be expressed in elementary terms, it can be defined as an integral. Option A is the correct conceptual representation of an antiderivative.

📖 Explanation: The integral limits need to be reversed or the property of definite integrals applied. The derivative is $-\sin(x^3)$. This is a common error because students often forget to apply the negative sign when the variable is in the lower limit. Correct Easy is: $\frac{d}{dx}\int_{x}^{1} f(t)\,dt = -f(x)$.

📖 Explanation: If $f(t)$ is negative, then F&#039;(x) = f(x) < 0, so $F(x)$ is decreasing. The value of $F(x)$ will be negative (since integrating negative values yields a negative cumulative area), but the question specifically asks about its behavior, which is decreasing. This interprets the geometric meaning of the derivative of an area function.

📖 Explanation: F&#039;(x) = \sqrt{1+x^2}. The second derivative is the derivative of this, which is $\frac{x}{\sqrt{1+x^2}}$. This question requires applying the FTC to find the first derivative and then differentiating again. It tests understanding of higher-order derivatives of integral-defined functions. Option A is a common mistake if one forgets to differentiate the derivative.

📖 Explanation: The definite integral $\int_{0}^{\pi} \sin t\,dt = 2$, which is positive. The student's conclusion of $-1$ is incorrect. The integral gives the net signed area. While the total positive area from 0 to $\pi$ is 2. The error lies in not evaluating the integral correctly or misunderstanding signed vs. total area.

📖 Explanation: By the FTC, $F(x)$ is indeed the antiderivative of $f(x)$ with $F(a) = \int_{a}^{a} f(t)\,dt = 0$. Any other antiderivative $G(x) = F(x) + C$ will have $G(a) = C$. Therefore, $F(x) = G(x) - G(a)$, so the constant is $C = -F(a)$. It is also the cumulative net area. This question ties together the multiple interpretations of a function defined by an integral.

📖 Explanation: The integrand is positive, continuous, and $\lim_{t\to\infty} \frac{1}{1+t^4}=0$. This implies the improper integral converges, so $F(x)$ is bounded above. This is a conceptual question that uses the properties of the integral-defined function to reason about its behavior at infinity. It requires synthesizing multiple facts about the integrand to make a global statement.

📖 Explanation: F&#039;(x) = \cos x, so F&#039;(\pi) = -1. $F(\pi) = \int_{0}^{\pi} \cos t\,dt = 0$. The tangent line is $y - 0 = -1(x - \pi)$, or $y = -x + \pi$. This problem combines the FTC with finding tangent lines, requiring students to find both the function value and the derivative at a point.

📖 Explanation: The derivative is $\sin(x^2) \cdot 2x$. The missing factor is $2x$, which comes from the chain rule. This is a classic error in applying the FTC with variable limits. The question targets this specific misconception by making the incorrect answer a tempting distractor and asking to identify the missing piece.

📖 Explanation: $F(x) = \ln x$. Therefore, $F(e^3) = \ln(e^3) = 3$. This question tests whether students recognize the integral as the definition of the natural logarithm. It avoids direct computation, focusing on the fundamental relationship between the integral and the logarithmic function. Option C is the correct conceptual answer.

📖 Explanation: Relative extrema of $C(x)$ occur when C&#039;(x) = \cos\left(\frac{\pi x^2}{2}\right) = 0. This occurs when $\frac{\pi x^2}{2} = \frac{\pi}{2} + n\pi$, so $x^2 = 1 + 2n$, and $x = \sqrt{2n+1}$. This question demands solving the derivative equation, interpreting the trigonometric condition, and correctly identifying all critical points. It's a Easy problem.

📖 Explanation: If $f(t) = 2t$, then $F(x) = \int_{0}^{x} 2t\,dt = [t^2]_{0}^{x} = x^2$. This requires finding the function from its derivative's geometric description, applying the FTC, and evaluating the integral. It connects geometry, algebra, and calculus concepts to determine the specific function.