Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 6 and Unit 8; FL MA.912.C.4.6, C.4.8-C.4.10; CCSS.MP2, MP3, MP5, MP6; Integrated Calculus II Standards CAL2-IMP, CAL2-NUM; TEKS 19 TAC section 111.55

CAL2.45 Improper Integrals with Vertical Asymptotes Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Calculus II Math worksheet provides structured practice on CAL2.45 Improper Integrals with Vertical Asymptotes. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.45 Improper Integrals with Vertical Asymptotes - Choose the correct one-sided limit setup for an integrand unbounded on a finite interval. Standards: AP Calculus BC Unit 6 and Unit 8; FL MA.912.C.4.6, C.4.8-C.4.10; CCSS.MP2, MP3, MP5, MP6; Integrated Calculus II Standards CAL2-IMP, CAL2-NUM; TEKS 19 TAC section 111.55

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Calculus II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related CAL2.45 Improper Integrals with Vertical Asymptotes skill assessments:

Problem 1Difficulty: Mastery

Which setup is the full improper-integral definition for ∫_{0}^2π 1/(sqrt(x)(2π - x)^(1/3)) dx, with unbounded behavior at x=0 and x=2π?

Alim_{t→0^+} ∫_{t}^π 1/(sqrt(x)(2π - x)^(1/3)) dx + lim_{s→2π^-} ∫_{π}^s 1/(sqrt(x)(2π - x)^(1/3)) dx

B∫_{0}^2π 1/(sqrt(x)(2π - x)^(1/3)) dx

Clim_{t→0^-} ∫_{t}^π 1/(sqrt(x)(2π - x)^(1/3)) dx + lim_{s→2π^+} ∫_{π}^s 1/(sqrt(x)(2π - x)^(1/3)) dx

Dlim_{t→0^+} ∫_{t}^π 1/(sqrt(x)(2π - x)^(1/3)) dx

Solution & Explanation:The integrand is unbounded at x=0 and x=2π. The interval must be split so that each vertical asymptote is approached only from within its neighboring subintervals. Every one-sided limit in the setup must converge; omitting a side or using an outside approach is a common error.

Problem 2Difficulty: Mastery

For ∫_{-2}² 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx, choose the setup that approaches each vertical asymptote from inside the interval at x=-2, x=0, x=1, and x=2.

Alim_{t→-2^+} ∫_{t}^-1 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{s→0^-} ∫_{-1}^s 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{u→0^+} ∫_{u}^1/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{v→1^-} ∫_{1/2}^v 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{w→1^+} ∫_{w}^3/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx

Blim_{t→-2^+} ∫_{t}^-1 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{s→0^-} ∫_{-1}^s 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{u→0^+} ∫_{u}^1/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{v→1^-} ∫_{1/2}^v 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{w→1^+} ∫_{w}^3/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{r→2^-} ∫_{3/2}^r 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx

C∫_{-2}² 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx

Dlim_{t→-2^-} ∫_{t}^-1 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{s→0^+} ∫_{-1}^s 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{u→0^-} ∫_{u}^1/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{v→1^+} ∫_{1/2}^v 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{w→1^-} ∫_{w}^3/2 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx + lim_{r→2^+} ∫_{3/2}^r 1/(sqrt(x + 2)(x² - x)sqrt(2 - x)) dx

Solution & Explanation:The integrand is unbounded at x=-2, x=0, x=1, and x=2. The interval must be split so that each vertical asymptote is approached only from within its neighboring subintervals. Every one-sided limit in the setup must converge; omitting a side or using an outside approach is a common error.

Problem 3Difficulty: Mastery

Which choice correctly avoids an unsplit or principal-value shortcut for ∫_{0}¹ 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx, whose integrand is unbounded at x=0, x=1/2, and x=1?

Alim_{t→0^-} ∫_{t}^1/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{s→1/2^+} ∫_{1/4}^s 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{u→1/2^-} ∫_{u}^3/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{v→1^+} ∫_{3/4}^v 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx

Blim_{t→0^+} ∫_{t}^1/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{s→1/2^-} ∫_{1/4}^s 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{u→1/2^+} ∫_{u}^3/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx

Clim_{t→0^+} ∫_{t}^1/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{s→1/2^-} ∫_{1/4}^s 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{u→1/2^+} ∫_{u}^3/4 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx + lim_{v→1^-} ∫_{3/4}^v 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx

D∫_{0}¹ 1/(sqrt(x)(x - 1/2)² sqrt(1 - x)) dx

Solution & Explanation:The integrand is unbounded at x=0, x=1/2, and x=1. The interval must be split so that each vertical asymptote is approached only from within its neighboring subintervals. Every one-sided limit in the setup must converge; omitting a side or using an outside approach is a common error.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Calculus II Math students. It aligns with educational standard AP Calculus BC Unit 6 and Unit 8; FL MA.912.C.4.6, C.4.8-C.4.10; CCSS.MP2, MP3, MP5, MP6; Integrated Calculus II Standards CAL2-IMP, CAL2-NUM; TEKS 19 TAC section 111.55 (Choose the correct one-sided limit setup for an integrand unbounded on a finite interval. Standards: AP Calculus BC Unit 6 and Unit 8; FL MA.912.C.4.6, C.4.8-C.4.10; CCSS.MP2, MP3, MP5, MP6; Integrated Calculus II Standards CAL2-IMP, CAL2-NUM; TEKS 19 TAC section 111.55).