AP Calculus BC Quiz: Evaluating Improper Integrals
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Evaluating Improper IntegralsQuestion 1 of 20

Compute ∫−∞011+x2 dx\int_{-\infty}^{0} \frac{1}{1+x^2}\,dx; does it converge, and what is the value?

Converges to π2\tfrac{\pi}{2}
Converges to π\pi
Diverges
Converges to 00
Converges to 11
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AP Calculus BC Quiz

AP Calculus BC Quiz: Evaluating Improper Integrals

Practice Evaluating Improper Integrals in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Evaluating Improper Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Compute ∫−∞011+x2 dx\int_{-\infty}^{0} \frac{1}{1+x^2}\,dx; does it converge, and what is the value?

  1. Converges to π2\tfrac{\pi}{2} (correct answer)
  2. Converges to π\pi
  3. Diverges
  4. Converges to 00
  5. Converges to 11

Explanation: This problem requires evaluating an improper integral with an infinite lower limit. To evaluate ∫−∞011+x2 dx\int_{-\infty}^{0} \frac{1}{1+x^2}\,dx, we write it as lim⁡t→−∞∫t011+x2 dx\lim_{t \to -\infty} \int_{t}^{0} \frac{1}{1+x^2}\,dx. The antiderivative of 11+x2\frac{1}{1+x^2} is arctan⁡(x)\arctan(x), so we have lim⁡t→−∞[arctan⁡(x)]t0=lim⁡t→−∞(arctan⁡(0)−arctan⁡(t))=lim⁡t→−∞(0−arctan⁡(t))\lim_{t \to -\infty} [\arctan(x)]_{t}^{0} = \lim_{t \to -\infty} (\arctan(0) - \arctan(t)) = \lim_{t \to -\infty} (0 - \arctan(t)). As t→−∞t \to -\infty, we have arctan⁡(t)→−π2\arctan(t) \to -\frac{\pi}{2}, so the integral converges to 0−(−π2)=π20 - (-\frac{\pi}{2}) = \frac{\pi}{2}. A common error is thinking arctan⁡(t)→−π\arctan(t) \to -\pi as t→−∞t \to -\infty, but the range of arctangent is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). When evaluating improper integrals involving arctangent, remember its horizontal asymptotes are at ±π2\pm\frac{\pi}{2}.

Question 2

A model spikes at x=0x=0; analyze ∫011x dx\int_{0}^{1} \frac{1}{x}\,dx for convergence and, if applicable, its value.

  1. Converges to 00
  2. Converges to 11
  3. Converges to ln⁡1\ln 1
  4. Converges to ∞\infty
  5. Diverges (correct answer)

Explanation: This problem requires evaluating the improper integral ∫011x dx\int_{0}^{1} \frac{1}{x}\,dx, which has a vertical asymptote at x=0x = 0. We rewrite this as lim⁡a→0+∫a11x dx\lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x}\,dx. The antiderivative of 1x\frac{1}{x} is ln⁡∣x∣\ln|x|, so we have lim⁡a→0+[ln⁡∣x∣]a1=lim⁡a→0+(ln⁡1−ln⁡a)=lim⁡a→0+(0−ln⁡a)=lim⁡a→0+(−ln⁡a)\lim_{a \to 0^+} [\ln|x|]_{a}^{1} = \lim_{a \to 0^+} (\ln 1 - \ln a) = \lim_{a \to 0^+} (0 - \ln a) = \lim_{a \to 0^+} (-\ln a). As a→0+a \to 0^+, we have ln⁡a→−∞\ln a \to -\infty, so −ln⁡a→+∞-\ln a \to +\infty, meaning the integral diverges. Students might confuse this with the convergent integral ∫011x dx\int_{0}^{1} \frac{1}{\sqrt{x}}\,dx, but the key difference is that 1x\frac{1}{x} approaches infinity too rapidly near x=0x = 0. For improper integrals with vertical asymptotes at x=0x = 0, ∫0c1xp dx\int_{0}^{c} \frac{1}{x^p}\,dx converges if and only if p<1p < 1; here p=1p = 1, so it diverges.

Question 3

A decay model uses h(x)=e−xh(x)=e^{-x} for x≥0x\ge0; does ∫0∞h(x) dx\int_{0}^{\infty} h(x)\,dx converge, and to what value?

  1. Converges to 11 (correct answer)
  2. Converges to 00
  3. Diverges
  4. Converges to ee
  5. Converges to 1e\tfrac{1}{e}

Explanation: Evaluating improper integrals involves determining whether they converge and finding their value if they do, often by taking limits. To assess ∫ from 0 to ∞ of e^{-x} dx, express it as the limit as b approaches infinity of the integral from 0 to b. The antiderivative is -e^{-x}, so evaluate lim_{b→∞} [-e^{-b} + e^{0}] which is 0 + 1 = 1. Thus, the integral converges to 1. A tempting distractor might be to think it converges to 0 because e^{-∞}=0, but forgetting the +1 from the lower bound leads to that error. A general strategy for improper integrals with exponential decay is to recognize their rapid convergence and compute limits directly.

Question 4

A signal has intensity I(x)=11+x2I(x)=\frac{1}{1+x^2} for x≥0x\ge0; does ∫0∞11+x2 dx\int_{0}^{\infty}\frac{1}{1+x^2}\,dx converge, and to what value?

  1. Converges to π2\frac{\pi}{2} (correct answer)
  2. Converges to π\pi
  3. Converges to 11
  4. Diverges
  5. Converges to 12\frac{1}{2}

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 0 to ∞ of 1/(1+x21+x^2) dx, replace the upper limit with b and take the limit as b approaches infinity of the integral from 0 to b. The antiderivative is arctan(x), so evaluating from 0 to b gives arctan(b) - arctan(0) = arctan(b). As b approaches infinity, arctan(b) approaches π/2, so the integral converges to π/2. A tempting distractor is 'Converges to π' by confusing it with the full range from -∞ to ∞, but this fails as the integral is only from 0 to ∞. A transferable strategy for improper integrals at infinity is to find the antiderivative and evaluate the limit, checking if it approaches a finite value.

Question 5

For f(x)=1(1+x)3f(x)=\frac{1}{(1+x)^{3}} on [0,∞)[0,\infty), does ∫0∞1(1+x)3 dx\int_{0}^{\infty}\frac{1}{(1+x)^3}\,dx converge, and to what value?

  1. Converges to 12\frac{1}{2} (correct answer)
  2. Converges to 11
  3. Diverges
  4. Converges to 32\frac{3}{2}
  5. Converges to 13\frac{1}{3}

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫0∞1(1+x)3 dx\int_0^\infty \frac{1}{(1+x)^3} \, dx, replace the upper limit with b and take the limit as b approaches infinity of the integral from 0 to b. The antiderivative is −12(1+x)2-\frac{1}{2(1+x)^2}, so evaluating from 0 to b gives −12(1+b)2−(−12(1+0)2)=−12(1+b)2+12-\frac{1}{2(1+b)^2} - \left( -\frac{1}{2(1+0)^2} \right) = -\frac{1}{2(1+b)^2} + \frac{1}{2}. As b approaches infinity, the first term approaches 0, so the integral converges to 12\frac{1}{2}. A tempting distractor is 'Converges to 1' by miscounting the power in the antiderivative, but this fails as the correct exponent leads to 12\frac{1}{2}. A transferable strategy for rational functions at infinity is to ensure the degree of the denominator exceeds the numerator by more than 1 for convergence.

Question 6

A wave envelope is A(x)=sin⁡xxA(x)=\frac{\sin x}{x} for x≥1x\ge1; does ∫1∞sin⁡xx dx\int_{1}^{\infty} \frac{\sin x}{x}\,dx converge, and to what value?

  1. Diverges
  2. Converges to π2\frac{\pi}{2}
  3. Converges to ∫1∞sin⁡x dx\int_{1}^{\infty} \sin x\,dx
  4. Converges to a finite value (not expressible elementarily) (correct answer)
  5. Converges to 00

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 1 to ∞ of (sin x)/x dx, recognize it as the Dirichlet integral, which is known to converge but lacks an elementary antiderivative. The convergence can be established using the Dirichlet test or comparison with 1/x^2 for large x, showing it approaches a finite value. This value is the sine integral Si(∞) = π/2 - Si(1), but it's not expressible elementarily. A tempting distractor is 'Diverges' due to the oscillating nature, but this fails as the 1/x decay ensures conditional convergence. A transferable strategy for oscillating improper integrals is to use tests like Dirichlet's for convergence without needing the exact value.

Question 7

A stress function is s(x)=1x2/3s(x)=\frac{1}{x^{2/3}} near 00; does ∫01x−2/3 dx\int_{0}^{1} x^{-2/3}\,dx converge, and to what value?

  1. Diverges
  2. Converges to 32\frac{3}{2}
  3. Converges to 33 (correct answer)
  4. Converges to 23\frac{2}{3}
  5. Converges to 13\frac{1}{3}

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 0 to 1 of x^{-2/3} dx, replace the lower limit with a approaching 0 from above and compute the limit of the integral from a to 1. The antiderivative is 3 x^{1/3}, so evaluating from a to 1 gives 3(1)^{1/3} - 3(a)^{1/3} = 3 - 3a^{1/3}. As a approaches 0^+, this becomes 3 - 0 = 3, so the integral converges to 3. A tempting distractor is 'Diverges' assuming the singularity at 0 causes divergence, but this fails since the exponent -2/3 > -1 ensures convergence for p-integrals near 0. A transferable strategy for improper integrals near 0 is to use the p-test: ∫ from 0 to b of x^{-p} dx converges if p < 1.

Question 8

For f(x)=1(x−1)1/3f(x)=\frac{1}{(x-1)^{1/3}} on (1,8](1,8], does ∫18(x−1)−1/3 dx\int_{1}^{8} (x-1)^{-1/3}\,dx converge, and to what value?

  1. Converges to 32 72/3\frac{3}{2}\,7^{2/3} (correct answer)
  2. Diverges
  3. Converges to 23 72/3\frac{2}{3}\,7^{2/3}
  4. Converges to 3 72/33\,7^{2/3}
  5. Converges to ln⁡7\ln 7

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 1 to 8 of (x-1)^{-1/3} dx, replace the lower limit with a approaching 1^+ and compute the limit of the integral from a to 8. The antiderivative is (3/2) (x-1)^{2/3}, so evaluating from a to 8 gives (3/2)(8-1)^{2/3} - (3/2)(a-1)^{2/3} = (3/2)7^{2/3} - (3/2)(a-1)^{2/3}. As a approaches 1^+, (a-1)^{2/3} approaches 0, so the integral converges to (3/2)7^{2/3}. A tempting distractor is 'Diverges' assuming the cube root singularity causes issues, but this fails since the exponent -1/3 > -1 ensures convergence. A transferable strategy for power singularities is to apply the generalized p-test for finite limits.

Question 9

A field has singularity at x=0x=0 with f(x)=ln⁡xf(x)=\ln x; does ∫01ln⁡x dx\int_{0}^{1} \ln x\,dx converge, and to what value?

  1. Diverges
  2. Converges to 11
  3. Converges to −1-1 (correct answer)
  4. Converges to 00
  5. Converges to ln⁡1\ln 1

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫01ln⁡x dx\int_0^1 \ln x \, dx, replace the lower limit with a approaching 0^+ and compute the limit of the integral from a to 1, using integration by parts with u=ln⁡xu = \ln x, dv=dxdv = dx. This gives [xln⁡x−x][x \ln x - x] from a to 1 = (1ln⁡1−1)(1 \ln 1 - 1) - lim⁡a→0+(aln⁡a−a)\lim_{a \to 0^+} (a \ln a - a), where lim⁡a→0+aln⁡a=0\lim_{a \to 0^+} a \ln a = 0 and −a=0-a = 0. Thus, -1 - 0 = -1, so the integral converges to -1. A tempting distractor is 'Diverges' due to ln⁡x→−∞\ln x \to -\infty as x→0+x \to 0^+, but this fails as the integral converges via the limiting behavior. A transferable strategy for logarithmic singularities is to use integration by parts and evaluate boundary limits carefully.

Question 10

For f(x)=1x(ln⁡x)2f(x)=\frac{1}{x(\ln x)^2} on x≥2x\ge2, determine whether ∫2∞1x(ln⁡x)2 dx\int_{2}^{\infty} \frac{1}{x(\ln x)^2}\,dx converges and its value.

  1. Diverges
  2. Converges to 1ln⁡2\frac{1}{\ln 2} (correct answer)
  3. Converges to ln⁡(ln⁡x)∣2∞\ln(\ln x)\big|_{2}^{\infty}
  4. Converges to ln⁡2\ln 2
  5. Converges to 12\frac{1}{2}

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 2 to ∞ of 1/(x (ln x)^2) dx, replace the upper limit with b and take the limit as b approaches infinity, using substitution u = ln x, du = dx/x. This transforms to ∫ from ln 2 to ln b of u^{-2} du, with antiderivative -1/u, evaluating to -1/ln b + 1/ln 2. As b approaches infinity, ln b → ∞, so -1/ln b → 0, and the integral converges to 1/ln 2. A tempting distractor is 'Diverges' confusing it with 1/(x ln x), but this fails as the extra (ln x) in the denominator ensures convergence. A transferable strategy for logarithmic integrals is to use substitution with u = ln x to simplify and apply p-test analogs.

Question 11

A potential is V(x)=11−xV(x)=\frac{1}{1-x} on [0,1)[0,1); determine whether ∫0111−x dx\int_{0}^{1} \frac{1}{1-x}\,dx converges and its value.

  1. Converges to ln⁡1\ln 1
  2. Diverges (correct answer)
  3. Converges to ln⁡2\ln 2
  4. Converges to 11
  5. Converges to 00

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 0 to 1 of 1/(1-x) dx, replace the upper limit with b approaching 1^- and compute the limit of the integral from 0 to b. The antiderivative is -ln|1-x|, so evaluating from 0 to b gives -ln|1-b| - (-ln|1-0|) = -ln(1-b) + ln 1 = -ln(1-b). As b approaches 1^-, ln(1-b) approaches -∞, so -(-∞) = +∞, and the integral diverges. A tempting distractor is 'Converges to ln 2' by using a different bound or sign error, but this fails due to the logarithmic divergence. A transferable strategy for rational singularities is to recognize logarithmic divergence patterns near poles.

Question 12

Near x=0x=0, a density is ρ(x)=1x\rho(x)=\frac{1}{\sqrt{x}}; determine whether ∫041x dx\int_{0}^{4} \frac{1}{\sqrt{x}}\,dx converges and its value.

  1. Diverges
  2. Converges to 44 (correct answer)
  3. Converges to 22
  4. Converges to 88
  5. Converges to 11

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 0 to 4 of 1/√x dx, replace the lower limit with a approaching 0^+ and compute the limit of the integral from a to 4. The antiderivative is 2√x, so evaluating from a to 4 gives 2√4 - 2√a = 4 - 2√a. As a approaches 0^+, 2√a approaches 0, so the integral converges to 4. A tempting distractor is 'Diverges' due to the singularity at x=0, but this fails since the exponent -1/2 > -1 allows convergence near 0. A transferable strategy for improper integrals with singularities at endpoints is to evaluate the limit of the antiderivative carefully.

Question 13

A distribution is f(x)=1(x+2)2f(x)=\frac{1}{(x+2)^2} for x≥0x\ge0; does ∫0∞1(x+2)2 dx\int_{0}^{\infty} \frac{1}{(x+2)^2}\,dx converge and to what value?

  1. Converges to 12\frac{1}{2} (correct answer)
  2. Diverges
  3. Converges to 22
  4. Converges to 14\frac{1}{4}
  5. Converges to 11

Explanation: This problem tests the skill of evaluating improper integrals. To evaluate the improper integral ∫0∞1(x+2)2 dx\int_0^\infty \frac{1}{(x+2)^2} \, dx, express it as the limit as b approaches infinity of the integral from 0 to b of (x+2)−2(x+2)^{-2} dx. The antiderivative is −1x+2-\frac{1}{x+2}. Evaluating from 0 to b gives −1b+2+12-\frac{1}{b+2} + \frac{1}{2}, and as b approaches infinity, this limit is 12\frac{1}{2}, so the integral converges to 12\frac{1}{2}. A tempting distractor is choice B, which claims it diverges, but this fails because the integrand decays like 1/x21/x^2 at infinity, which converges. A transferable strategy for improper integrals at infinity is to compute the antiderivative and evaluate the limit carefully, checking convergence criteria like the p-test for power functions.

Question 14

A signal has f(x)=sec⁡2xf(x)=\sec^2 x on [0,π2)[0,\frac{\pi}{2}); does ∫0π/2sec⁡2x dx\int_{0}^{\pi/2} \sec^2 x\,dx converge and to what value?

  1. Converges to 11
  2. Converges to 00
  3. Converges to tan⁡(π2)\tan\left(\frac{\pi}{2}\right)
  4. Diverges (correct answer)
  5. Converges to π2\frac{\pi}{2}

Explanation: This question tests the skill of evaluating improper integrals. To evaluate the improper integral ∫ from 0 to π/2 of sec^2 x dx, replace the upper limit with b approaching (π/2)^- and compute the limit of the integral from 0 to b. The antiderivative is tan x, so evaluating from 0 to b gives tan b - tan 0 = tan b. As b approaches π/2^-, tan b approaches +∞, so the integral diverges. A tempting distractor is 'Converges to π/2' by confusing with the integral of sec x or another function, but this fails as sec^2 x integrates to tan x, which diverges at π/2. A transferable strategy for trigonometric improper integrals is to check the behavior near asymptotes using known antiderivatives.

Question 15

A particle's speed is v(t)=1t3/2v(t)=\frac{1}{t^{3/2}} for t≥1t\ge1; does ∫1∞v(t) dt\int_{1}^{\infty} v(t)\,dt converge, and to what value?​

  1. Diverges
  2. Converges to 22 (correct answer)
  3. Converges to 23\frac{2}{3}
  4. Converges to 11
  5. Converges to 00

Explanation: This problem requires evaluating an improper integral with an infinite upper limit. To evaluate ∫1∞1t3/2 dt\int_{1}^{\infty} \frac{1}{t^{3/2}}\,dt, we replace infinity with a variable limit bb and take the limit as b→∞b \to \infty: lim⁡b→∞∫1bt−3/2 dt\lim_{b \to \infty} \int_{1}^{b} t^{-3/2}\,dt. The antiderivative of t−3/2t^{-3/2} is t−1/2−1/2=−2t−1/2\frac{t^{-1/2}}{-1/2} = -2t^{-1/2}, so we get lim⁡b→∞[−2t−1/2]1b=lim⁡b→∞(−2b−1/2+2)=0+2=2\lim_{b \to \infty} [-2t^{-1/2}]_{1}^{b} = \lim_{b \to \infty} (-2b^{-1/2} + 2) = 0 + 2 = 2. A common error is forgetting the negative sign when integrating t−3/2t^{-3/2}, which would incorrectly give −2-2 instead of 22. For improper integrals with infinite limits, always convert to a limit of a proper integral and carefully track signs when finding antiderivatives.

Question 16

For p(x)=1xp(x)=\frac{1}{x} on [1,∞)[1,\infty), does ∫1∞p(x) dx\int_{1}^{\infty} p(x)\,dx converge, and to what value?​

  1. Converges to 11
  2. Converges to 00
  3. Converges to ln⁡(∞)−ln⁡(1)\ln(\infty)-\ln(1)
  4. Diverges (correct answer)
  5. Converges to ln⁡2\ln 2

Explanation: This problem asks whether the improper integral of 1x\frac{1}{x} converges on [1,∞)[1,\infty), a fundamental example in calculus. To evaluate ∫1∞1x dx\int_{1}^{\infty} \frac{1}{x}\,dx, we write it as lim⁡b→∞∫1b1x dx=lim⁡b→∞[ln⁡∣x∣]1b=lim⁡b→∞(ln⁡b−ln⁡1)=lim⁡b→∞ln⁡b\lim_{b \to \infty} \int_{1}^{b} \frac{1}{x}\,dx = \lim_{b \to \infty} [\ln|x|]_{1}^{b} = \lim_{b \to \infty} (\ln b - \ln 1) = \lim_{b \to \infty} \ln b. Since ln⁡b→∞\ln b \to \infty as b→∞b \to \infty, the integral diverges. A tempting error is to think that since 1x→0\frac{1}{x} \to 0 as x→∞x \to \infty, the integral must converge, but the decay is too slow. The key insight is that ∫1∞1xp dx\int_{1}^{\infty} \frac{1}{x^p}\,dx converges if and only if p>1p > 1; when p=1p = 1, we get logarithmic divergence.

Question 17

A response function is q(x)=1x4/3q(x)=\frac{1}{x^{4/3}} for x≥1x\ge1; does ∫1∞q(x) dx\int_{1}^{\infty} q(x)\,dx converge, and to what value?​

  1. Converges to 33 (correct answer)
  2. Diverges
  3. Converges to 13\frac{1}{3}
  4. Converges to 34\frac{3}{4}
  5. Converges to 31\frac{3}{1}

Explanation: This problem requires evaluating an improper integral with a power function over an infinite interval. To evaluate ∫1∞1x4/3 dx=∫1∞x−4/3 dx\int_{1}^{\infty} \frac{1}{x^{4/3}}\,dx = \int_{1}^{\infty} x^{-4/3}\,dx, we write it as lim⁡b→∞∫1bx−4/3 dx\lim_{b \to \infty} \int_{1}^{b} x^{-4/3}\,dx. The antiderivative of x−4/3x^{-4/3} is x−1/3−1/3=−3x−1/3\frac{x^{-1/3}}{-1/3} = -3x^{-1/3}, so we get lim⁡b→∞[−3x−1/3]1b=lim⁡b→∞(−3b−1/3−(−3))=0+3=3\lim_{b \to \infty} [-3x^{-1/3}]_{1}^{b} = \lim_{b \to \infty} (-3b^{-1/3} - (-3)) = 0 + 3 = 3. A common mistake is incorrectly applying the power rule or forgetting to evaluate at both limits. For improper integrals of the form ∫1∞1xp dx\int_{1}^{\infty} \frac{1}{x^p}\,dx, remember that convergence occurs when p>1p > 1, and the value is 1p−1\frac{1}{p-1} when starting at x=1x = 1.

Question 18

Water drains with rate r(t)=e−tr(t)=e^{-t} liters/min for t≥0t\ge0; does ∫0∞r(t) dt\int_{0}^{\infty} r(t)\,dt converge, and to what value?​

  1. Converges to 11 (correct answer)
  2. Converges to 00
  3. Diverges
  4. Converges to ee
  5. Converges to 1e\frac{1}{e}

Explanation: This problem asks us to evaluate an improper integral representing total water drained over infinite time. To evaluate ∫0∞e−t dt\int_{0}^{\infty} e^{-t}\,dt, we write it as lim⁡b→∞∫0be−t dt\lim_{b \to \infty} \int_{0}^{b} e^{-t}\,dt. The antiderivative of e−te^{-t} is −e−t-e^{-t}, so we get lim⁡b→∞[−e−t]0b=lim⁡b→∞(−e−b−(−e0))=lim⁡b→∞(−e−b+1)=0+1=1\lim_{b \to \infty} [-e^{-t}]_{0}^{b} = \lim_{b \to \infty} (-e^{-b} - (-e^{0})) = \lim_{b \to \infty} (-e^{-b} + 1) = 0 + 1 = 1. A tempting error is to forget the negative sign in the antiderivative of e−te^{-t}, which would incorrectly suggest divergence. For exponential decay integrals, the key insight is that e−t→0e^{-t} \to 0 as t→∞t \to \infty, ensuring convergence when the exponent is negative.

Question 19

For the density f(x)=1(x−2)2f(x)=\frac{1}{(x-2)^2} on [0,5][0,5] with a break at x=2x=2, does ∫05f(x) dx\int_{0}^{5} f(x)\,dx converge?​

  1. Converges to 56\frac{5}{6}
  2. Converges to 13\frac{1}{3}
  3. Diverges (correct answer)
  4. Converges to 32\frac{3}{2}
  5. Converges to 12\frac{1}{2}

Explanation: This problem involves evaluating an improper integral with a vertical asymptote at x=2x = 2 within the interval [0,5][0,5]. Since f(x)=1(x−2)2f(x) = \frac{1}{(x-2)^2} has an infinite discontinuity at x=2x = 2, we must split the integral: ∫051(x−2)2 dx=∫021(x−2)2 dx+∫251(x−2)2 dx\int_{0}^{5} \frac{1}{(x-2)^2}\,dx = \int_{0}^{2} \frac{1}{(x-2)^2}\,dx + \int_{2}^{5} \frac{1}{(x-2)^2}\,dx. For the first part, we evaluate lim⁡c→2−∫0c(x−2)−2 dx=lim⁡c→2−[−(x−2)−1]0c=lim⁡c→2−(−(c−2)−1+12)\lim_{c \to 2^-} \int_{0}^{c} (x-2)^{-2}\,dx = \lim_{c \to 2^-} [-(x-2)^{-1}]_{0}^{c} = \lim_{c \to 2^-} (-(c-2)^{-1} + \frac{1}{2}), which diverges to +∞+\infty as c→2−c \to 2^-. Since one part diverges, the entire integral diverges—a common mistake is evaluating only from one side of the discontinuity. When an integrand has a vertical asymptote inside the interval, always split the integral at that point and check convergence of each piece separately.

Question 20

For x>1x>1, r(x)=1x(ln⁡x)2r(x)=\frac{1}{x(\ln x)^2}; does ∫2∞r(x) dx\int_2^{\infty} r(x)\,dx converge, and to what value?

  1. Converges to 1ln⁡2\frac{1}{\ln 2} (correct answer)
  2. Diverges
  3. Converges to ln⁡2\ln 2
  4. Converges to 12\frac{1}{2}
  5. Converges to 11

Explanation: This problem involves evaluating the improper integral ∫2∞1x(ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx. We use the substitution u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x}dx, and when x=2x = 2, u=ln⁡2u = \ln 2; when x→∞x \to \infty, u→∞u \to \infty. The integral becomes ∫ln⁡2∞1u2 du=lim⁡b→∞∫ln⁡2bu−2 du\int_{\ln 2}^{\infty} \frac{1}{u^2}\,du = \lim_{b \to \infty} \int_{\ln 2}^b u^{-2}\,du. Evaluating gives us lim⁡b→∞[−u−1]ln⁡2b=lim⁡b→∞(−1b−(−1ln⁡2))=0+1ln⁡2=1ln⁡2\lim_{b \to \infty} \left[-u^{-1}\right]_{\ln 2}^b = \lim_{b \to \infty} \left(-\frac{1}{b} - (-\frac{1}{\ln 2})\right) = 0 + \frac{1}{\ln 2} = \frac{1}{\ln 2}. A student might incorrectly get ln⁡2\ln 2 by confusing the substitution. When dealing with logarithmic integrands, substitution often simplifies the problem to a standard power function integral.