📝 Improper integrals calculus (41 MCQs)

📖 Explanation: An improper integral can have an infinite limit of integration or an infinite discontinuity within the interval. Options A, C, and D have finite intervals but contain discontinuities. Option B has an infinite upper limit, making it improper due to the interval's infinite extent.

📖 Explanation: This is a standard $p$-integral with $p=2 > 1$. The antiderivative is $-1/x$, and evaluating from 1 to $b$ and taking the limit as $b \to \infty$ gives 1. This is a foundational result for understanding convergence.

📖 Explanation: The integrand has a vertical asymptote at $x=1$. To evaluate, we split the integral at 1. Evaluating each improper integral, both tend to infinity. A common misconception is that symmetry around the asymptote implies convergence, which is not true for this integrand.

📖 Explanation: The integrand has an infinite discontinuity at $x=0$. We evaluate the limit as $t \to 0^+$ of $\int_{t}^{1} x^{-1/2} \, dx = 2(1 - \sqrt{t})$. The limit is 2, so the improper integral converges.

📖 Explanation: The function $1/x$ has an infinite discontinuity at $x=0$, which lies within the interval of integration. The Fundamental Theorem of Calculus cannot be applied directly. Splitting the integral shows that both parts diverge, so the integral does not converge to 0.

📖 Explanation: A $p$-integral of the form $\int_{1}^{\infty} x^{-p} \, dx$ converges for $p>1$. Options A and B have $p=1$ and $p=1/2$, so they diverge. Option D diverges as $p=1$ on a finite interval with a discontinuity.

📖 Explanation: The integral converges to $1/(p-1)$ for $p>1$. Setting $1/(p-1) = 2$ gives $p-1 = 0.5$, so $p=1.5$. This requires solving a simple equation after applying the convergence formula.

📖 Explanation: The area is $\int_{1}^{\infty} x^{-2} \, dx = 1$, which is finite. The volume of revolution is $\pi \int_{1}^{\infty} x^{-4} \, dx = \pi/3$, which is also finite. This connects two different Easys of improper integrals.

📖 Explanation: The area is $\int_{0}^{1} x^{-1/2} \, dx$, which converges to 2. The graph has a vertical asymptote at $x=0$, but the area under the curve is finite. This demonstrates that a function can be unbounded but still have a finite integral.

📖 Explanation: Using integration by parts with $u = \ln x$ and $dv = x^{-2} dx$, the integral becomes $\left[ -\frac{\ln x}{x} \right]_{1}^{b} + \int_{1}^{b} x^{-2} dx$. The limit as $b \to \infty$ is $0 + 1 = 1$. This combines integration techniques with improper limits.

📖 Explanation: The integral evaluates to $\lim_{b \to \infty} (1 - e^{-b}) = 1$. The misconception is confusing the limit of the integrand with the value of the integral. Even though the function tends to 0, the area under the curve is finite and non-zero.

📖 Explanation: For large $x$, $\sqrt{x^3 + x} \approx x^{3/2}$. Thus, the integrand behaves like $x^{-3/2}$, which is a convergent $p$-integral since $p=1.5 > 1$. This is a direct Easy of the Limit Hard Test.

📖 Explanation: The volume is $\pi \int_{0}^{\infty} e^{-2x} \, dx = \pi \cdot (1/2) = \pi/2$. This tests the ability to set up and evaluate a volume integral with an infinite limit.

📖 Explanation: The limit $\lim_{b \to \infty} \int_{0}^{b} \sin x \, dx = \lim (1 - \cos b)$ does not exist. The function oscillates, and the area does not settle to a finite value. Improper integrals of oscillating functions require careful analysis.

📖 Explanation: The area is $\int_{1}^{\infty} x^{-1} \, dx$, which diverges to infinity. The curve $y = 1/x$ approaches the x-axis slowly, accumulating infinite area. This is a classic example of a function that tends to zero but whose integral diverges.

📖 Explanation: An infinite discontinuity occurs when the function becomes unbounded at a point within the interval. Option B has a vertical asymptote at $x=1$. Option A and C have infinite intervals, while D is continuous on [1,2].

📖 Explanation: The integral evaluates to $\lim_{b \to \infty} \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{b} = \lim_{b \to \infty} \left( \frac{1}{2} - \frac{1}{2} e^{-2b} \right) = \frac{1}{2}$.

📖 Explanation: The antiderivative is $\frac{1}{2} \ln(x^2+1)$. Evaluating from 0 to $b$ and taking the limit gives $\frac{1}{2} \ln(b^2+1)$, which tends to infinity. Thus, the integral diverges.

📖 Explanation: The integral evaluates to $k$. Setting $k = 3$ gives the solution. This is a straightforward parameter-finding problem.

📖 Explanation: The integral is improper at $x=0$. The correct evaluation requires the limit $\lim_{t \to 0^+} \int_{t}^{1} x^{-1/2} dx$. Without this step, the process is not rigorous, though the final answer happens to be 2 in this case.

📖 Explanation: Near $x=0$, $1/\sqrt{x}$ is less than $1/x^{0.9}$ for any exponent less than 1. A direct Hard with the convergent $p$-integral $\int_{0}^{1} x^{-1/2} dx$ is not helpful. Comparing to a constant is not standard; the Hard test usually uses a known convergent/divergent function with similar behavior near the singularity.

📖 Explanation: The arc length formula gives $\int_{0}^{1} \sqrt{1 + \left(\frac{3}{2}x^{1/2}\right)^2} dx = \int_{0}^{1} \sqrt{1 + \frac{9}{4}x} dx$. This is a proper integral. A similar problem can be constructed where the derivative is unbounded, requiring improper integral techniques.

📖 Explanation: $\int_{1}^{\infty} x^{-2} dx$ converges to 1. $\int_{1}^{\infty} x^{-1} dx$ diverges. The boundary between convergence and divergence for $p$-integrals is at $p=1$.

📖 Explanation: The Hard test states that if $f(x) \le g(x)$ and $\int f$ diverges, then $\int g$ diverges. If $\int f$ converges, $\int g$ is not guaranteed to converge, as $g$ could be larger. This is a common point of confusion.

📖 Explanation: The volume of revolution is $\pi \int_{1}^{\infty} x^{-4} dx = \pi/3$, which is finite. The shape is a funnel that extends infinitely but has finite volume.

📖 Explanation: This is a standard arctangent integral. $\int_{0}^{\infty} \frac{1}{x^2+1} dx = \left[ \tan^{-1} x \right]_{0}^{\infty} = \pi/2$.

📖 Explanation: For $\int_{0}^{1} x^{-p} dx$, convergence occurs for $p<1$. For $p=0.5$, the integral is $\int_{0}^{1} x^{-1/2} dx = 2$.

📖 Explanation: Distance is $\int_{0}^{\infty} e^{-t} dt = 1$. This models a physical scenario where the particle's speed decays exponentially, resulting in a finite total distance.

📖 Explanation: The expression $\infty - (-\infty)$ is indeterminate. The correct approach is to evaluate $\lim_{b \to \infty} \ln b - \lim_{a \to 0^+} \ln a$, both of which are infinite, so the integral diverges.

📖 Explanation: $\int_{0}^{1} x^{-1/2} dx$ converges to 2. Options A and D diverge, and B diverges because $p=1/2 < 1$ on an infinite interval.

📖 Explanation: If the limit of $f(x)$ as $x \to \infty$ were a non-zero constant $L$, the integral would diverge (similar to integrating a constant). Therefore, for the integral to converge, $L$ must be 0.

📖 Explanation: The integral is improper, but it converges. The antiderivative is $2\sqrt{x}$, and evaluating the limit as $t \to 0^+$ gives $2$. This is a classic example of a convergent improper integral with a vertical asymptote.

📖 Explanation: Using substitution $u = x^2$, the integral becomes $\frac{1}{2} \int_{0}^{\infty} e^{-u} du = \frac{1}{2}$. This requires both substitution and knowledge of the exponential integral.

📖 Explanation: For $\int_{0}^{1} x^{-p} dx$, convergence occurs when $p<1$. For $p=1$, the integral is $\ln x$, which diverges. For $p=1/2$, the integral is $2\sqrt{x}$, which converges. This subtle difference is a common source of error.

📖 Explanation: The volume is $\pi \int_{1}^{\infty} x^{-3} dx$, which converges because $p=3>1$. The area under the curve also converges, but that is not the direct reason the volume is finite. The specific $p$-value for the volume integral is what ensures finiteness.

📖 Explanation: This integral is split at 0: $\int_{-\infty}^{0} + \int_{0}^{\infty} = \pi/2 + \pi/2 = \pi$. This is an even function, and the total area under the curve from $-\infty$ to $\infty$ is $\pi$.

📖 Explanation: The integral is linear. Multiplying the integrand by a constant multiplies the integral by that same constant: $2 \times 5 = 10$.

📖 Explanation: For finite intervals with singularities, $\int_{0}^{1} x^{-2} dx$ diverges. For infinite intervals, $\int_{1}^{\infty} x^{-2} dx$ converges to 1. The behavior depends on the location of the singularity and the interval type.

📖 Explanation: The area $\int_{1}^{\infty} x^{-1} dx$ diverges. The volume $\pi \int_{1}^{\infty} x^{-2} dx$ converges to $\pi$. This is a famous paradox (Gabriel's Horn) that demonstrates the difference between area and volume convergence.

📖 Explanation: Using substitution $u = \ln x$, the integral becomes $\int_{0}^{\infty} \frac{1}{u} du$, which diverges. This is a classic example of an integral that diverges slowly.

📖 Explanation: The integral $\int_{0}^{\infty} \frac{\sin x}{x} dx$ is conditionally convergent and equals $\pi/2$. This is a classic result in analysis and is often used in signal processing. It tests understanding of convergence of non-positive functions.