MT2176 Further Calculus: Improper Integrals Quiz

Correct Answer: B

Explanation: This is an improper integral of the first kind because the interval of integration, \( [1, \infty) \), is unbounded. The integrand is continuous on this interval. (Source: Wrede, Chapter 12; Ostaszewski, Section 4.1)

Correct Answer: C

Explanation: This is a p-integral, \( \int_{1}^{\infty} \frac{1}{x^p} \, dx \), with p = 1. According to the p-integral test, the integral diverges if p \le 1. (Source: Wrede, Chapter 12, Special Improper Integrals of the First Kind; Binmore, Section 10.11, Example 17)

Correct Answer: C

Explanation: This is an improper integral of the second kind because the integrand \( f(x) = \frac{1}{\sqrt{x}} \) is unbounded at the lower limit of integration, x = 0. (Source: Wrede, Chapter 12; Ostaszewski, Section 4.1)

Correct Answer: C

Explanation: This is a p-integral of the second kind, \( \int_{0}^{1} \frac{1}{x^p} \, dx \), with p = 2. This type of integral diverges if p \ge 1. (Source: Wrede, Chapter 12, Special Improper Integrals of the Second Kind)

Correct Answer: A

Explanation: This is an exponential integral. We can evaluate it directly: \( \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = \lim_{b \to \infty} (-e^{-b} + 1) = 1 \). Since the limit is finite, the integral converges. (Source: Ostaszewski, Section 4.1.1)

Correct Answer: B

Explanation: We know that \( 0 \le \sin^2 x \le 1 \) for all x. Therefore, \( 0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2} \). Since \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) is a convergent p-integral (p=2 > 1), the Direct Comparison Test implies that \( \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \, dx \) also converges. (Source: Ostaszewski, Section 4.2.1; Wrede, Chapter 12)

Correct Answer: A

Explanation: This is an improper integral of the first kind over \( (-\infty, \infty) \). We must check the convergence of \( \int_{-\infty}^{0} x e^{-x^2} \, dx \) and \( \int_{0}^{\infty} x e^{-x^2} \, dx \) separately. The integrand \( f(x) = x e^{-x^2} \) is an odd function, so \( \int_{-a}^{a} f(x) \, dx = 0 \) for any a. The integral converges to 0. (Source: Wrede, Chapter 12, Example page 267)

Correct Answer: C

Explanation: This integral is of the third kind. It has an infinite interval of integration \( [0, \infty) \) (first kind) and the integrand has a singularity at x=1, which is within the interval (second kind). (Source: Wrede, Chapter 12, Definitions)

Correct Answer: C

Explanation: Let \( f(x) = \frac{x^2+1}{x^4+x} \). For large x, the dominant term in the numerator is \( x^2 \) and in the denominator is \( x^4 \). So we choose \( g(x) = \frac{x^2}{x^4} = \frac{1}{x^2} \). Now we compute the limit of the ratio: \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{(x^2+1)/(x^4+x)}{1/x^2} = \lim_{x \to \infty} \frac{x^4+x^2}{x^4+x} = 1 \). Since the limit is a finite positive number and \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges (p-integral with p=2), the given integral also converges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: C

Explanation: The function \( \tan(x) \) is unbounded as \( x \to (\pi/2)^{-} \). This makes the integral an improper integral of the second kind. We can evaluate it as \( \lim_{b \to (\pi/2)^{-}} \int_{0}^{b} \tan(x) \, dx = \lim_{b \to (\pi/2)^{-}} [-\ln|\cos(x)|]_{0}^{b} = \lim_{b \to (\pi/2)^{-}} (-\ln|\cos(b)| + \ln|\cos(0)|) \). As \( b \to (\pi/2)^{-} \), \( \cos(b) \to 0^{+} \), so \( \ln|\cos(b)| \to -\infty \). Therefore, the integral diverges. (Source: Ostaszewski, Section 4.1)

Correct Answer: B

Explanation: The integrand \(f(x) = 1/x\) has a singularity at x=0, which is within the interval of integration [-1, 1]. This makes it an improper integral of the second kind. (Source: Wrede, Chapter 12)

Correct Answer: B

Explanation: This is a p-integral of the first kind with p = 1/2. Since p \le 1, the integral diverges. (Source: Binmore, Section 10.11, Example 17)

Correct Answer: B

Explanation: The integrand \( \ln(x) \) is unbounded as \( x \to 0^{+} \). We can evaluate it using integration by parts: \( \int \ln(x) \, dx = x\ln(x) - x \). Then \( \lim_{a \to 0^{+}} [x\ln(x) - x]_{a}^{1} = (-1) - \lim_{a \to 0^{+}} (a\ln(a) - a) = -1 \). Since the limit is finite, it converges. (Source: Ostaszewski, Section 4.1)

Correct Answer: B

Explanation: We can use the integral test. Let \( u = \ln x \), so \( du = (1/x) dx \). The integral becomes \( \int_{\ln 2}^{\infty} \frac{1}{u} \, du \), which is a divergent p-integral (p=1). (Source: Wrede, Chapter 12, Problem 12.42)

Correct Answer: A

Explanation: The interval of integration is \( (-\infty, \infty) \), which is unbounded in both directions. The integrand is continuous everywhere. This is a classic example of an improper integral of the first kind. (Source: Binmore, Section 10.11, Example 19)

Correct Answer: A

Explanation: This is a p-integral of the form \( \int_{1}^{\infty} \frac{1}{x^p} \, dx \) where p = -k. The integral converges if p > 1, which means -k > 1, or k < -1. (Source: Binmore, Section 10.11)

Correct Answer: B

Explanation: This is a p-integral of the second kind with p = 3. Since p \ge 1, the integral diverges. (Source: Wrede, Chapter 12)

Correct Answer: B

Explanation: \( \int \frac{1}{1+x^2} \, dx = \arctan(x) \). So, \( \lim_{b \to \infty} [\arctan(x)]_{0}^{b} = \lim_{b \to \infty} (\arctan(b) - \arctan(0)) = \pi/2 - 0 = \pi/2 \). (Source: Binmore, Section 10.11, Example 19)

Correct Answer: A

Explanation: Use the comparison test. For x > 1, \( x^2+x > x^2 \), so \( \frac{1}{x^2+x} < \frac{1}{x^2} \). Since \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges (p=2), the given integral also converges. (Source: Ostaszewski, Section 4.2.1)

Correct Answer: A

Explanation: This is a p-integral of the second kind with p = 1/3. Since p < 1, it converges. \( \int x^{-1/3} \, dx = \frac{3}{2}x^{2/3} \). So, \( \lim_{a \to 0^{+}} [\frac{3}{2}x^{2/3}]_{a}^{1} = \frac{3}{2} - 0 = \frac{3}{2} \). (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: For x > 0, \( e^x+1 > e^x \), so \( \frac{1}{e^x+1} < \frac{1}{e^x} = e^{-x} \). Since \( \int_{0}^{\infty} e^{-x} \, dx \) converges, by the comparison test, \( \int_{0}^{\infty} \frac{dx}{e^x+1} \) also converges. (Source: Ostaszewski, Section 4.2.1)

Correct Answer: B

Explanation: The integrand behaves like \( x/x^3 = 1/x^2 \) for large x. The Limit Comparison Test with \( g(x)=1/x^2 \) will yield a finite positive limit, showing convergence. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: B

Explanation: This is an improper integral of the second kind with a singularity at x=1. We must split it into \( \int_{0}^{1} \frac{1}{(x-1)^2} \, dx + \int_{1}^{2} \frac{1}{(x-1)^2} \, dx \). Both are p-integrals with p=2, which diverge. Since at least one part diverges, the whole integral diverges. (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: Use the substitution \( u = \ln x \), \( du = 1/x \, dx \). The integral becomes \( \int_{1}^{\infty} \frac{1}{u^2} \, du \), which is a convergent p-integral (p=2). (Source: Wrede, Chapter 12, Problem 12.42)

Correct Answer: A

Explanation: This is a standard convergent integral related to the Gamma function. For large x, the exponential decay of \(e^{-x}\) dominates the polynomial growth of \(x^2\), ensuring convergence. Its value is 2! = 2. (Source: Ostaszewski, Section 4.3)

Correct Answer: C

Explanation: The interval is infinite (first kind) and the integrand has a singularity at x=1, which is at the boundary of the interval (second kind). This makes it a mixed or third kind integral. (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: We know that \( 0 < \arctan(x) < \pi/2 \) for x > 0. So, \( \frac{\arctan(x)}{x^2} < \frac{\pi/2}{x^2} \). Since \( \int_{1}^{\infty} \frac{\pi/2}{x^2} \, dx \) converges (p=2), the given integral converges by the comparison test. (Source: Ostaszewski, Section 4.2.1)

Correct Answer: A

Explanation: The integrand is unbounded at x=1. The antiderivative is \(\arcsin(x)\). So, \( \lim_{b \to 1^{-}} [\arcsin(x)]_{0}^{b} = \lim_{b \to 1^{-}} (\arcsin(b) - \arcsin(0)) = \pi/2 - 0 = \pi/2 \). (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: This is \( \int_{0}^{\infty} x e^{-x} \, dx \), which is the Gamma function \(\Gamma(2)\). It is a known convergent integral. Its value is 1! = 1. (Source: Ostaszewski, Section 4.3)

Correct Answer: B

Explanation: For large x, \( \sqrt{x^2-1} \approx \sqrt{x^2} = x \). So we compare with \(g(x) = 1/x\). \( \lim_{x \to \infty} \frac{1/\sqrt{x^2-1}}{1/x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2-1}} = 1 \). Since \( \int_{2}^{\infty} \frac{1}{x} \, dx \) diverges, the given integral also diverges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: B

Explanation: Use the Limit Comparison Test. For large x, 1/x is small, and \(\sin(u) \approx u\) for small u. So we compare with \(g(x) = 1/x\). \( \lim_{x \to \infty} \frac{\sin(1/x)}{1/x} = 1 \). Since \( \int_{1}^{\infty} \frac{1}{x} \, dx \) diverges, the given integral also diverges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: A

Explanation: The Cauchy Principal Value is defined as \( \lim_{\epsilon \to 0^{+}} (\int_{-1}^{-\epsilon} \frac{1}{x} \, dx + \int_{\epsilon}^{1} \frac{1}{x} \, dx) \). This becomes \( \lim_{\epsilon \to 0^{+}} ([\ln|x|]_{-1}^{-\epsilon} + [\ln|x|]_{\epsilon}^{1}) = \lim_{\epsilon \to 0^{+}} ((\ln\epsilon - \ln 1) + (\ln 1 - \ln\epsilon)) = 0 \). Note that the integral itself is divergent. (Source: Wrede, Chapter 12)

Correct Answer: B

Explanation: The integral converges by a test similar to Dirichlet's test (related to integration by parts). However, the integral of the absolute value, \( \int_{0}^{\infty} \frac{|\cos x|}{\sqrt{x}} \, dx \), diverges by comparison with \( \int \frac{\cos^2 x}{\sqrt{x}} \, dx \). Since the integral converges but not absolutely, it is conditionally convergent. (Source: Advanced topic, related to Ostaszewski, Section 4.2.3)

Correct Answer: C

Explanation: This is the standard p-integral test for improper integrals of the second kind. The integral converges if and only if p < 1. (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: This is a third kind integral. We split it at x=1. For \( \int_{0}^{1} \), the integrand is like \(1/\sqrt{x}\) (p=1/2 < 1, converges). For \( \int_{1}^{\infty} \), the integrand is like \(1/x^{3/2}\) (p=3/2 > 1, converges). Since both parts converge, the whole integral converges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: B

Explanation: For x > e, \(\ln x > 1\), so \( \frac{\ln x}{x} > \frac{1}{x} \). Since \( \int_{1}^{\infty} \frac{1}{x} \, dx \) diverges, by the comparison test, the given integral also diverges. (Source: Ostaszewski, Section 4.2.1)

Correct Answer: A

Explanation: Use partial fractions: \( \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \). The integral is \( \lim_{b \to \infty} [\ln|x| - \ln|x+1|]_{1}^{b} = \lim_{b \to \infty} [\ln(\frac{x}{x+1})]_{1}^{b} = \lim_{b \to \infty} (\ln(\frac{b}{b+1}) - \ln(\frac{1}{2})) = \ln(1) - (-\ln 2) = \ln 2 \). (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: Although the integrand looks like it has a singularity at x=0, it does not. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). The integrand is continuous on \([0, \pi]\) if we define its value to be 1 at x=0. Therefore, it is a proper integral. (Source: Ostaszewski, Section 4.1)

Correct Answer: A

Explanation: Split at x=1. Near 0, the term \(x^q\) dominates (since q<1), so the integrand is like \(1/x^q\). The integral converges since q<1. Near \(\infty\), the term \(x^p\) dominates (since p>1), so the integrand is like \(1/x^p\). The integral converges since p>1. Both parts converge, so the whole integral converges. (Source: Combination of p-integral rules)

Correct Answer: A

Explanation: Improper due to \(\ln x\) at x=0. Use integration by parts: \( \int x \ln x \, dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \frac{1}{x} \, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} \). Evaluating \( [\frac{x^2}{2}\ln x - \frac{x^2}{4}]_{a}^{1} \) as \(a \to 0^{+}\) gives \((0 - 1/4) - (0 - 0) = -1/4\). (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: Simplify the integrand: \( \frac{x!}{(x+2)!} = \frac{x!}{(x+2)(x+1)x!} = \frac{1}{(x+2)(x+1)} \). This behaves like \(1/x^2\) for large x. Since \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges, the given integral converges by Limit Comparison Test. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: A

Explanation: This is a famous result. The integral is conditionally convergent. The oscillations of \(\sin(x^2)\) become more rapid as x increases, causing areas of positive and negative contribution to increasingly cancel out. Proving this requires advanced techniques beyond standard comparison tests. (Source: Advanced Calculus topic, mentioned in Wrede)

Correct Answer: B

Explanation: On the interval [0, 1], \(e^x \ge e^0 = 1\). Therefore, \( \frac{e^x}{x} \ge \frac{1}{x} \). Since \( \int_{0}^{1} \frac{1}{x} \, dx \) diverges (p=1), the given integral diverges by the comparison test. (Source: Ostaszewski, Section 4.2.1)

Correct Answer: A

Explanation: This is the same integral as in question 37, just written differently. The integral of \( \frac{1}{x(x+1)} \) converges to \(\ln(2)\). (Source: Wrede, Chapter 12)

Correct Answer: B

Explanation: This is a third kind integral. It has an infinite limit AND a singularity at x=0 within the interval. It must be split into two improper integrals at the point of singularity. (Source: Wrede, Chapter 12)

Correct Answer: B

Explanation: Use Limit Comparison Test. As \(x \to 0\), \(e^x \approx 1+x\), so \(e^x-1 \approx x\). We compare with \(g(x)=1/x\). \( \lim_{x \to 0} \frac{1/(e^x-1)}{1/x} = \lim_{x \to 0} \frac{x}{e^x-1} = 1 \). Since \( \int_{0}^{1} \frac{1}{x} \, dx \) diverges, the given integral diverges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: A

Explanation: \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \). So, \( \lim_{b \to \infty} [-\frac{1}{x}]_{1}^{b} = \lim_{b \to \infty} (-\frac{1}{b} - (-\frac{1}{1})) = 0 + 1 = 1 \). (Source: Binmore, Section 10.11)

Correct Answer: B

Explanation: Use Limit Comparison Test. As \(x \to 0\), \(\tan x \approx x\). So the integrand behaves like \(x/x^2 = 1/x\). Since \( \int_{0}^{1} \frac{1}{x} \, dx \) diverges, the given integral diverges. (Source: Ostaszewski, Section 4.2.2)

Correct Answer: A

Explanation: This is a famous convergent integral. We can split it into \( \int_{0}^{1} e^{-x^2} dx \) (a proper integral) and \( \int_{1}^{\infty} e^{-x^2} dx \). For \(x \ge 1\), \(x^2 \ge x\), so \(e^{-x^2} \le e^{-x}\). Since \( \int_{1}^{\infty} e^{-x} dx \) converges, the second part converges by comparison. The value of the original integral is \(\sqrt{\pi}/2\). (Source: Wrede, Chapter 12)

Correct Answer: A

Explanation: This is an integral with a variable sign integrand. We can test for absolute convergence. We have \( |e^{-x} \sin(x)| \le e^{-x} \). Since \( \int_{0}^{\infty} e^{-x} \, dx \) converges (to 1), by the Direct Comparison Test, the integral \( \int_{0}^{\infty} |e^{-x} \sin(x)| \, dx \) also converges. An absolutely convergent integral is always convergent. (Source: Ostaszewski, Section 4.2.3)