Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 8; FL MA.912.C.5.5-C.5.7; CCSS.HSM; CCSS.HSN.Q.1-3; CCSS.MP2, MP4, MP6, MP7; Integrated Calculus II Standards CAL2-APP; TEKS 19 TAC section 111.55

CAL2.58 Area with Multiple Intersections Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Calculus II Math worksheet provides structured practice on CAL2.58 Area with Multiple Intersections. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.58 Area with Multiple Intersections - Select the correct split integrals when the top and bottom functions change over the interval. Standards: AP Calculus BC Unit 8; FL MA.912.C.5.5-C.5.7; CCSS.HSM; CCSS.HSN.Q.1-3; CCSS.MP2, MP4, MP6, MP7; Integrated Calculus II Standards CAL2-APP; TEKS 19 TAC section 111.55.

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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 CAL2.58 Area with Multiple Intersections skill assessments:

Problem 1Difficulty: Mastery

For the continuous curves u(x)=x²+1+(x+6)(x+4)(x+1)x(x-3)(x-5) and v(x)=x²+1 on [-7,7], the only intersection x-values are -6, -4, -1, 0, 3, and 5. Which expression correctly sets up the total area between the curves?

A∫_{-7}^-6 [u(x)-v(x)] dx + ∫_{-6}^-4 [u(x)-v(x)] dx + ∫_{-4}^-1 [u(x)-v(x)] dx + ∫_{-1}⁰ [u(x)-v(x)] dx + ∫_{0}³ [u(x)-v(x)] dx + ∫_{3}⁵ [u(x)-v(x)] dx + ∫_{5}⁷ [u(x)-v(x)] dx

B∫_{-7}^-6 [v(x)-u(x)] dx + ∫_{-6}^-4 [v(x)-u(x)] dx + ∫_{-4}^-1 [v(x)-u(x)] dx + ∫_{-1}⁰ [v(x)-u(x)] dx + ∫_{0}³ [v(x)-u(x)] dx + ∫_{3}⁵ [v(x)-u(x)] dx + ∫_{5}⁷ [v(x)-u(x)] dx

C∫_{-7}^-6 [v(x)-u(x)] dx + ∫_{-6}^-4 [u(x)-v(x)] dx + ∫_{-4}^-1 [v(x)-u(x)] dx + ∫_{-1}⁰ [u(x)-v(x)] dx + ∫_{0}³ [v(x)-u(x)] dx + ∫_{3}⁵ [u(x)-v(x)] dx + ∫_{5}⁷ [v(x)-u(x)] dx

D∫_{-7}^-6 [u(x)-v(x)] dx + ∫_{-6}^-4 [v(x)-u(x)] dx + ∫_{-4}^-1 [u(x)-v(x)] dx + ∫_{-1}⁰ [v(x)-u(x)] dx + ∫_{0}³ [u(x)-v(x)] dx + ∫_{3}⁵ [v(x)-u(x)] dx + ∫_{5}⁷ [u(x)-v(x)] dx

Solution & Explanation:The area between curves is the sum of upper-minus-lower integrals on subintervals separated by intersection x-values. Here the split points are x=-6, -4, -1, 0, 3, and 5; u(x) is above v(x) on (-7,-6), (-4,-1), (0,3), (5,7); v(x) is above u(x) on (-6,-4), (-1,0), (3,5). Therefore the correct setup is ∫_{-7}^-6 [u(x)-v(x)] dx + ∫_{-6}^-4 [v(x)-u(x)] dx + ∫_{-4}^-1 [u(x)-v(x)] dx + ∫_{-1}⁰ [v(x)-u(x)] dx + ∫_{0}³ [u(x)-v(x)] dx + ∫_{3}⁵ [v(x)-u(x)] dx + ∫_{5}⁷ [u(x)-v(x)] dx. A common mistake is to integrate u(x)-v(x) across the whole interval, which gives signed net area instead of total area.

Problem 2Difficulty: Mastery

The table for continuous functions R_1(x) and R_2(x) lists the sign of R_1(x)-R_2(x) across [-2,14]: R_2(x)>R_1(x) on (-2,-1); R_1(x)>R_2(x) on (-1,2); R_2(x)>R_1(x) on (2,6); R_1(x)>R_2(x) on (6,8); R_2(x)>R_1(x) on (8,10); R_1(x)>R_2(x) on (10,13); R_2(x)>R_1(x) on (13,14). The zeros are x=-1, 2, 6, 8, 10, and 13. Which expression gives the total area between the curves?

A∫_{-2}¹⁴ [R_2(x)-R_1(x)] dx

B∫_{-2}^-1 [R_1(x)-R_2(x)] dx + ∫_{-1}² [R_1(x)-R_2(x)] dx + ∫_{2}⁶ [R_1(x)-R_2(x)] dx + ∫_{6}⁸ [R_1(x)-R_2(x)] dx + ∫_{8}¹⁰ [R_1(x)-R_2(x)] dx + ∫_{10}¹³ [R_1(x)-R_2(x)] dx + ∫_{13}¹⁴ [R_1(x)-R_2(x)] dx

C∫_{-2}^-1 [R_2(x)-R_1(x)] dx + ∫_{-1}² [R_1(x)-R_2(x)] dx + ∫_{2}⁶ [R_2(x)-R_1(x)] dx + ∫_{6}⁸ [R_1(x)-R_2(x)] dx + ∫_{8}¹⁰ [R_2(x)-R_1(x)] dx + ∫_{10}¹³ [R_1(x)-R_2(x)] dx + ∫_{13}¹⁴ [R_2(x)-R_1(x)] dx

D∫_{-2}^-1 [R_2(x)-R_1(x)] dx + ∫_{-1}² [R_2(x)-R_1(x)] dx + ∫_{2}⁶ [R_2(x)-R_1(x)] dx + ∫_{6}⁸ [R_2(x)-R_1(x)] dx + ∫_{8}¹⁰ [R_2(x)-R_1(x)] dx + ∫_{10}¹³ [R_2(x)-R_1(x)] dx + ∫_{13}¹⁴ [R_2(x)-R_1(x)] dx

Solution & Explanation:The area between curves is the sum of upper-minus-lower integrals on subintervals separated by intersection x-values. Here the split points are x=-1, 2, 6, 8, 10, and 13; R_1(x) is above R_2(x) on (-1,2), (6,8), (10,13); R_2(x) is above R_1(x) on (-2,-1), (2,6), (8,10), (13,14). Therefore the correct setup is ∫_{-2}^-1 [R_2(x)-R_1(x)] dx + ∫_{-1}² [R_1(x)-R_2(x)] dx + ∫_{2}⁶ [R_2(x)-R_1(x)] dx + ∫_{6}⁸ [R_1(x)-R_2(x)] dx + ∫_{8}¹⁰ [R_2(x)-R_1(x)] dx + ∫_{10}¹³ [R_1(x)-R_2(x)] dx + ∫_{13}¹⁴ [R_2(x)-R_1(x)] dx. A common mistake is to integrate R_1(x)-R_2(x) across the whole interval, which gives signed net area instead of total area.

Problem 3Difficulty: Mastery

A calculator reports that continuous curves M(x) and N(x) intersect only at x=-3, 0, 1, 5, 9, and 12 on [-4,13]. Testing one point in each subinterval gives: M(x)>N(x) on (-4,-3); N(x)>M(x) on (-3,0); M(x)>N(x) on (0,1); N(x)>M(x) on (1,5); M(x)>N(x) on (5,9); N(x)>M(x) on (9,12); M(x)>N(x) on (12,13). Which expression is the correct exact setup for the area between the curves?

A∫_{-4}¹³ [M(x)-N(x)] dx

B∫_{-4}^-3 [M(x)-N(x)] dx + ∫_{-3}⁰ [N(x)-M(x)] dx + ∫_{0}¹ [M(x)-N(x)] dx + ∫_{1}⁵ [N(x)-M(x)] dx + ∫_{5}⁹ [M(x)-N(x)] dx + ∫_{9}¹² [N(x)-M(x)] dx + ∫_{12}¹³ [M(x)-N(x)] dx

C∫_{-4}¹³ [N(x)-M(x)] dx

D∫_{-4}^-3 [M(x)-N(x)] dx + ∫_{-3}⁰ [M(x)-N(x)] dx + ∫_{0}¹ [M(x)-N(x)] dx + ∫_{1}⁵ [M(x)-N(x)] dx + ∫_{5}⁹ [M(x)-N(x)] dx + ∫_{9}¹² [M(x)-N(x)] dx + ∫_{12}¹³ [M(x)-N(x)] dx

Solution & Explanation:The area between curves is the sum of upper-minus-lower integrals on subintervals separated by intersection x-values. Here the split points are x=-3, 0, 1, 5, 9, and 12; M(x) is above N(x) on (-4,-3), (0,1), (5,9), (12,13); N(x) is above M(x) on (-3,0), (1,5), (9,12). Therefore the correct setup is ∫_{-4}^-3 [M(x)-N(x)] dx + ∫_{-3}⁰ [N(x)-M(x)] dx + ∫_{0}¹ [M(x)-N(x)] dx + ∫_{1}⁵ [N(x)-M(x)] dx + ∫_{5}⁹ [M(x)-N(x)] dx + ∫_{9}¹² [N(x)-M(x)] dx + ∫_{12}¹³ [M(x)-N(x)] dx. A common mistake is to integrate M(x)-N(x) across the whole interval, which gives signed net area instead of total area.

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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 8; FL MA.912.C.5.5-C.5.7; CCSS.HSM; CCSS.HSN.Q.1-3; CCSS.MP2, MP4, MP6, MP7; Integrated Calculus II Standards CAL2-APP; TEKS 19 TAC section 111.55 (Select the correct split integrals when the top and bottom functions change over the interval. Standards: AP Calculus BC Unit 8; FL MA.912.C.5.5-C.5.7; CCSS.HSM; CCSS.HSN.Q.1-3; CCSS.MP2, MP4, MP6, MP7; Integrated Calculus II Standards CAL2-APP; TEKS 19 TAC section 111.55.).