📝 Logarithmic and Other Functions Defined by Integrals (24 MCQs)

📖 Explanation: Since $1/2 < 1$, the integral $\int_1^{1/2} 1/t \, dt$ is negative. The Fundamental Theorem of Calculus gives $\ln(1/2) = - \int_{1/2}^1 1/t \, dt$, which is the negative of the area under the curve from $1/2$ to $1$. Option A confuses the sign, Option C confuses the interval, and Option D is incorrect because the integral starts at 1, not 0.

📖 Explanation: The sign of a definite integral depends on both the integrand and the limits. While $1/t$ is positive for $t>0$, the integral from 1 to x is negative when $0 < x < 1$ because the upper limit is less than the lower limit. Option C correctly identifies the core error: ignoring the effect of the limits on the integral's sign. Option B is false because the integrand is indeed positive, and Option A is partially true but misses the key point about integration limits.

📖 Explanation: Since velocity is the derivative of position, s&#039;(t) = v(t) = 1/t. Integrating gives $s(t) = \ln t + C$ for $t>0$. Using the initial condition $s(1)=3$, we get $3 = \ln 1 + C$, so $C = 3$. Thus, $s(t) = \ln t + 3$. Option C incorrectly subtracts the constant, and Option A forgets the initial condition. Option D uses the absolute value, which is unnecessary for $t>0$.

📖 Explanation: By definition, a definite integral from 0 to 2 represents the net signed area. Since $\sin(\pi t^2/2)$ can be negative, the integral is not simply the 'area' (which is always positive). Option A is incorrect because it ignores the sign of the function. Option C and D confuse the function with its integral and derivative, respectively, as per the Fundamental Theorem of Calculus.

📖 Explanation: The Fundamental Theorem of Calculus Part 2 states: \frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x)) g&#039;(x). The student correctly substituted the upper limit into the integrand but forgot to multiply by the derivative of the upper limit, which is $2x$. The correct answer is $\frac{2x}{1+x^4}$. Option B is false because the antiderivative is not needed here. Option D is irrelevant as the lower limit is constant.

📖 Explanation: By the Fundamental Theorem of Calculus, F&#039;(x) = f(x). Since $f(x)$ is positive for $x>1$, F&#039;(x) > 0, so $F(x)$ is increasing. Since $f(x)$ is increasing, F&#039;&#039;(x) = f&#039;(x) > 0, so $F(x)$ is concave up. Option A reverses the monotonicity. Option C incorrectly states concave down. Option D is incorrect for both properties.

📖 Explanation: Using the property $\int_x^1 f(t)dt = - \int_1^x f(t)dt$, we have $F(x) = - \int_1^x 1/t \, dt$. Differentiating gives F&#039;(x) = -1/x. The common mistake is forgetting the negative sign that arises from swapping the limits of integration. Option A is the derivative of $\int_1^x 1/t \, dt$, not $\int_x^1 1/t \, dt$. Options C and D are incorrect and likely arise from an improper Easy of the power rule.

📖 Explanation: The standard proof involves splitting the integral $\int_1^{ab} 1/t \, dt = \int_1^a 1/t \, dt + \int_a^{ab} 1/t \, dt$. For the second integral, substituting $t=au$, $dt = a \, du$, and changing limits from $t=a$ to $u=1$ and $t=ab$ to $u=b$, gives $\int_1^b 1/u \, du$. Thus, $\ln(ab) = \ln a + \ln b$. Option B is a correct approach but less direct. Option C ignores the substitution needed for the second integral. Option D is a different, albeit correct, proof.

📖 Explanation: The error function is used in probability and statistics. As $x \to \infty$, the integral $\int_0^\infty e^{-t^2} dt = \sqrt{\pi}/2$, which is a known result (Gaussian integral). Thus, $\lim_{x \to \infty} \text{erf}(x) = \frac{2}{\sqrt{\pi}} \cdot \frac{\sqrt{\pi}}{2} = 1$. Option A and C are incorrect values; Option D is a common misconception for functions defined by integrals.

📖 Explanation: The rate of change of the infected population is P&#039;(t) = r(t). Thus, $P(t) = \int \frac{1}{t+1} dt = \ln(t+1) + C$. Since $P(0)=0$, we get $0 = \ln(1) + C$, so $C=0$. Thus, $P(t) = \ln(t+1)$. Option B is an indefinite integral without applying the initial condition. Option C has an incorrect constant. Option D is the derivative of $-1/(t+1)$, which is not the integral.

📖 Explanation: First, F&#039;(x) = 2x \sin(x^4) by the Fundamental Theorem of Calculus and the chain rule. Then, F&#039;&#039;(x) = 2 \sin(x^4) + 2x \cdot \cos(x^4) \cdot 4x^3 = 2 \sin(x^4) + 8x^4 \cos(x^4). Evaluating at $x=0$ gives F&#039;&#039;(0) = 2 \sin(0) + 0 = 0. Options B and C are incorrect and might arise from misapplying the chain rule or differentiating the product incorrectly. Option D is a common trap when functions are defined by integrals, but here the derivative exists and is well-defined.

📖 Explanation: A function defined by an integral, like $\text{erf}(x)$, is precisely defined for every $x$ by the value of the definite integral. This is a rigorous mathematical definition. While it might lack a simple closed-form, it is not less 'complete.' Option A is subjective. Option B is false, as continuous integrands guarantee differentiability. Option D is false, as functions like $\text{erf}(x)$ are computationally intensive to evaluate, requiring numerical methods.

📖 Explanation: Option B uses the formal definition of a limit to infinity. It establishes that for any large number $M$, we can choose n such that $n \ln 2 > M$, and then for any $x > 2^n$, $\ln x > M$. This is a rigorous proof. Option A is an intuitive but not formal argument. Option C is a known theorem but not a proof from the definition. Option D is a logically flawed argument; a derivative approaching zero does not imply the function approaches infinity.

📖 Explanation: The Fundamental Theorem of Calculus states that F&#039;(x) = \sin x / x. However, this function is undefined at $x=0$ because of division by zero. The integral is defined and differentiable for all x, but the derivative formula is only valid where the integrand is continuous, which excludes $x=0$. Option A incorrectly states it's valid for all x. Option C is false; $\sin t / t$ is continuous for all $t \neq 0$, so the integral is defined on all intervals excluding zero. Option D is a common misconception; the derivative exists, even if the antiderivative isn't elementary.

📖 Explanation: Option C is the most elegant and theoretically sound proof. The derivative of $\ln x$ is $1/x$, which is positive for $x>0$, proving $\ln x$ is strictly increasing. Therefore, for $8>4$, $\ln 8 > \ln 4$. Option A is circular, as it requires knowing $\ln 2 > 0$. Option B relies on the monotonicity of $1/t$, which is true but less direct. Option D is a geometric argument, not a rigorous algebraic one.

📖 Explanation: We have C&#039;(x) = \cos(\pi x^2/2). Setting C&#039;(x)=0 gives $\pi x^2/2 = \pi/2 + n\pi$, so $x^2 = 1 + 2n$. For $n=0$, $x=1$. For $x=1$, C&#039;&#039;(x) = -\pi x \sin(\pi x^2/2). At $x=1$, C&#039;&#039;(1) = -\pi \sin(\pi/2) = -\pi < 0, which indicates a relative maximum. Option A is incorrect because $x=\sqrt{2}$ would be a minimum. Option C and D are also incorrect as they are not critical points.

📖 Explanation: The integral $\int_1^{-2} \frac{1}{t} dt$ is improper because the function is discontinuous at $t=0$, which lies between the limits. It does not converge to a real number. Option A is incorrect because $1/t$ is integrable on any interval that does not contain 0. Option B is partially true but misses the key point that the integral diverges. Option D is false; limits can be swapped, but the integral would be $\int_1^{-2}$, which is improper.

📖 Explanation: Formula (11) assumes the integrand $f(t)$ is continuous on an open interval containing the path from $x_0$ to $x$. The function $f(x)=|x|$ is continuous everywhere, so the formula applies. Actually, all these functions are continuous on appropriate intervals. The question tests understanding that the formula applies to all continuous f, including those without elementary antiderivatives. Options A, B, and C all have continuous integrands on the relevant intervals. Option D is the trick question; it is also continuous and solvable.

📖 Explanation: The second derivative of $\ln x$ is $-1/x^2$, which is negative for all $x > 0$. A negative second derivative means the function is concave down. Option A is incorrect; it would require a positive second derivative. Option C is false because F&#039;&#039;(x) never changes sign. Option D is false because the function is clearly concave down.

📖 Explanation: Acceleration is the derivative of velocity: v&#039;(t) = a(t) = 1/t. Integrating gives $v(t) = \ln t + C$. Using the initial condition $v(1)=2$, we get $2 = \ln 1 + C$, so $C=2$. Thus, $v(t) = \ln t + 2$. Option C unnecessarily uses the absolute value, which is valid but not simpler. Option B has an incorrect constant. Option D is the general solution without applying the initial condition.

📖 Explanation: The student correctly applied the chain rule in the first part but forgot it in the second. The correct derivative is G&#039;(x) = \frac{\sin x^2}{x^2} \cdot 2x = \frac{2 \sin x^2}{x}. Option B is incorrect; they did substitute $x^2$, but didn't multiply by its derivative. Option C is false; while the integrand has a removable discontinuity at 0, the integral is differentiable. Option D is false.

📖 Explanation: Using the properties of integrals and logarithms: $f(x) = \int_{x}^{x^2} \frac{1}{t} dt = \ln t |_{x}^{x^2} = \ln(x^2) - \ln(x) = 2\ln x - \ln x = \ln x$. Wait, let's re-evaluate. The correct Easy: $\int_x^{x^2} 1/t \, dt = \ln t \big|_x^{x^2} = \ln(x^2) - \ln(x) = 2\ln x - \ln x = \ln x$. Option A is correct. Option C is the negative of the correct answer. Option D uses an improper integral starting at 0. Option B is incorrect in its evaluation.

📖 Explanation: The average value of $f(t)$ on [3,5] is $\frac{1}{5-3} \int_3^5 f(t) dt$. Using the Fundamental Theorem of Calculus, $\int_3^5 f(t) dt = F(5) - F(3) = 9 - 5 = 4$. Thus, the average value is $\frac{1}{2} \cdot 4 = 2$. Option B is the integral itself, not the average. Option C and D are arbitrary values derived from incorrect sums.

📖 Explanation: Both methods are numerical approximations of the true value of $\ln 2$. A calculator typically uses more sophisticated algorithms and can compute with high precision, while the midpoint rule with $n=10$ is a relatively coarse approximation. Option A is false because both are approximations. Option B is false; it's an approximation, not exact. Option D is generally false; calculator approximations are usually very accurate.