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Application Of Integrals Class 12 Exam Prep Revision — Grandmaster Guide

A

Ayush (Founder)

Exam Strategist

Last Updated: 2026-04-20
  • ∫(1/x) dx = ln|x| + C
  • ∫(1/√(1-x²)) dx = s∈⁻¹x + C
  • ∫(1/√(x²-1)) dx = ln|x + √(x²-1)| + C or cosh⁻¹x + C
  • ∫(1/√(1+x²)) dx = ln|x + √(1+x²)| + C
  • ∫(1/(a²+x²)) dx = (1/a)tan⁻¹(x/a) + C
  • Area under curve y = f(x) from x = a to x = b is ∫[a,b] f(x) dx
  • Volume of solid formed by revolving area under curve y = f(x) from x = a to x = b about x-axis is ∫[a,b] π(f(x))² dx
  • Volume of solid formed by revolving area under curve y = f(x) from x = a to x = b about y-axis is ∫[a,b] 2πxf(x) dx
  • Surface area of solid formed by revolving curve y = f(x) from x = a to x = b about x-axis is ∫[a,b] 2πf(x)√(1 + (dy/dx)²) dx
  • Length of curve y = f(x) from x = a to x = b is ∫[a,b] √(1 + (dy/dx)²) dx

🪤 The 5 Mistakes That Cost Marks

  • Not checking limits of integration
  • Forgetting constant of integration
  • Incorrectly applying substitution method
  • Not using properties of definite integrals
  • Incorrectly calculating area under curve or volume of solid

✏️ 3 Solved PYQs

  • Question 1: Find area under curve y = x² from x = 0 to x = 2 Step 1: Define limits of integration Step 2: Evaluate ∫[0,2] x² dx = [(1/3)x³] from 0 to 2 Step 3: Calculate area = (1/3)(2³) - (1/3)(0³) = 8/3
  • Question 2: Find volume of solid formed by revolving area under curve y = x² from x = 0 to x = 2 about x-axis Step 1: Define limits of integration Step 2: Evaluate ∫[0,2] π(x²)² dx = π∫[0,2] x⁴ dx = π[(1/5)x⁵] from 0 to 2 Step 3: Calculate volume = π[(1/5)(2⁵) - (1/5)(0⁵)] = 32π/5
  • Question 3: Find length of curve y = x³ from x = 0 to x = 1 Step 1: Find dy/dx = 3x² Step 2: Evaluate ∫[0,1] √(1 + (3x²)²) dx Step 3: Simplify and calculate length

🧠 The One Thing Most Students Get Wrong

  • Most students struggle with applying correct limits of integration and substitution methods
  • Key concept: ∫[a,b] f(x) dx = F(b) - F(a) where F(x) is antiderivative of f(x)
  • Practice problems: focus on applying limits and substitution correctly

👁️ Ayush's Note

  • For JEE Advanced and NEET, focus on shortcuts like using properties of definite integrals and substitution methods
  • Practice solving problems with varying limits of integration and functions
  • Use online resources and practice tests to improve speed and accuracy

🔁 Last 5 Minutes Box

  • Check all limits of integration
  • Ensure correct application of substitution method
  • Verify calculations for area under curve or volume of solid
  • Double-check constant of integration
  • Review properties of definite integrals

📝 Practice MCQs

1. Question: Find area under curve y = 2x from x = 0 to x = 1

A) 1

B) 2

C) 3

D) 4

Answer: A) 1. Explanation: ∫[0,1] 2x dx = [x²] from 0 to 1 = 1² - 0² = 1

2. Question: Find volume of solid formed by revolving area under curve y = x² from x = 0 to x = 1 about x-axis

A) π/5

B) π/3

C) π/2

D) π

Answer: A) π/5. Explanation: ∫[0,1] π(x²)² dx = π∫[0,1] x⁴ dx = π[(1/5)x⁵] from 0 to 1 = π[(1/5)(1⁵) - (1/5)(0⁵)] = π/5

3. Question: Find length of curve y = x³ from x = 0 to x = 1

A) √(1 + 9)

B) √(1 + 3)

C) ∫[0,1] √(1 + (3x²)²) dx

D) ∫[0,1] √(1 + (2x)²) dx

Answer: C) ∫[0,1] √(1 + (3x²)²) dx. Explanation: dy/dx = 3x², so length = ∫[0,1] √(1 + (3x²)²) dx

4. Question: Find area under curve y = 1/x from x = 1 to x = 2

A) ln|2|

B) ln|1/2|

C) ln|2/1|

D) ln|1|

Answer: A) ln|2|. Explanation: ∫[1,2] (1/x) dx = [ln|x|] from 1 to 2 = ln|2| - ln|1| = ln|2|

5. Question: Find volume of solid formed by revolving area under curve y = x from x = 0 to x = 2 about y-axis

A) 8π

B) 4π

C) 2π

D) π

Answer: B) 4π. Explanation: ∫[0,2] 2πx(x) dx = 2π∫[0,2] x² dx = 2π[(1/3)x³] from 0 to 2 = 2π[(1/3)(2³) - (1/3)(0³)] = 8π/3, but using disk method with x = y, ∫[0,2] 2πy dy = 2π[(1/2)y²] from 0 to 2 = π(2²) = 4π


🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

📚 Academic References

Content verified against peer-reviewed research:

  1. �Let the People Rap�: Cultural Rhetorics Pedagogy and Practices U... — Journal of Basic Writing (2019) 🔓 — DOI ↗
  2. Frustration and Hope: Examining Students� Emotional Responses to ... — Journal of Basic Writing (2019) — DOI ↗
  3. Editors' Column — Journal of Basic Writing (2019) — DOI ↗

🔓 = Open Access article

🎬 Watch video explanations on YouTube →


This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.


📚 Related Topics

Continue your revision with these related guides:

  • 📖 Integrals Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Compiler Design Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Computer Networks Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Linear Programming Class 12 Exam Prep Revision — Grandmaster Guide
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Prepared for Scholar
Date: 2026-04-20
CATEGORY: Exam Notes
  • ∫(1/x) dx = ln|x| + C
  • ∫(1/√(1-x²)) dx = s∈⁻¹x + C
  • ∫(1/√(x²-1)) dx = ln|x + √(x²-1)| + C or cosh⁻¹x + C
  • ∫(1/√(1+x²)) dx = ln|x + √(1+x²)| + C
  • ∫(1/(a²+x²)) dx = (1/a)tan⁻¹(x/a) + C
  • Area under curve y = f(x) from x = a to x = b is ∫[a,b] f(x) dx
  • Volume of solid formed by revolving area under curve y = f(x) from x = a to x = b about x-axis is ∫[a,b] π(f(x))² dx
  • Volume of solid formed by revolving area under curve y = f(x) from x = a to x = b about y-axis is ∫[a,b] 2πxf(x) dx
  • Surface area of solid formed by revolving curve y = f(x) from x = a to x = b about x-axis is ∫[a,b] 2πf(x)√(1 + (dy/dx)²) dx
  • Length of curve y = f(x) from x = a to x = b is ∫[a,b] √(1 + (dy/dx)²) dx

🪤 The 5 Mistakes That Cost Marks

  • Not checking limits of integration
  • Forgetting constant of integration
  • Incorrectly applying substitution method
  • Not using properties of definite integrals
  • Incorrectly calculating area under curve or volume of solid

✏️ 3 Solved PYQs

  • Question 1: Find area under curve y = x² from x = 0 to x = 2 Step 1: Define limits of integration Step 2: Evaluate ∫[0,2] x² dx = [(1/3)x³] from 0 to 2 Step 3: Calculate area = (1/3)(2³) - (1/3)(0³) = 8/3
  • Question 2: Find volume of solid formed by revolving area under curve y = x² from x = 0 to x = 2 about x-axis Step 1: Define limits of integration Step 2: Evaluate ∫[0,2] π(x²)² dx = π∫[0,2] x⁴ dx = π[(1/5)x⁵] from 0 to 2 Step 3: Calculate volume = π[(1/5)(2⁵) - (1/5)(0⁵)] = 32π/5
  • Question 3: Find length of curve y = x³ from x = 0 to x = 1 Step 1: Find dy/dx = 3x² Step 2: Evaluate ∫[0,1] √(1 + (3x²)²) dx Step 3: Simplify and calculate length

🧠 The One Thing Most Students Get Wrong

  • Most students struggle with applying correct limits of integration and substitution methods
  • Key concept: ∫[a,b] f(x) dx = F(b) - F(a) where F(x) is antiderivative of f(x)
  • Practice problems: focus on applying limits and substitution correctly

👁️ Ayush's Note

  • For JEE Advanced and NEET, focus on shortcuts like using properties of definite integrals and substitution methods
  • Practice solving problems with varying limits of integration and functions
  • Use online resources and practice tests to improve speed and accuracy

🔁 Last 5 Minutes Box

  • Check all limits of integration
  • Ensure correct application of substitution method
  • Verify calculations for area under curve or volume of solid
  • Double-check constant of integration
  • Review properties of definite integrals

📝 Practice MCQs

1. Question: Find area under curve y = 2x from x = 0 to x = 1

A) 1

B) 2

C) 3

D) 4

Answer: A) 1. Explanation: ∫[0,1] 2x dx = [x²] from 0 to 1 = 1² - 0² = 1

2. Question: Find volume of solid formed by revolving area under curve y = x² from x = 0 to x = 1 about x-axis

A) π/5

B) π/3

C) π/2

D) π

Answer: A) π/5. Explanation: ∫[0,1] π(x²)² dx = π∫[0,1] x⁴ dx = π[(1/5)x⁵] from 0 to 1 = π[(1/5)(1⁵) - (1/5)(0⁵)] = π/5

3. Question: Find length of curve y = x³ from x = 0 to x = 1

A) √(1 + 9)

B) √(1 + 3)

C) ∫[0,1] √(1 + (3x²)²) dx

D) ∫[0,1] √(1 + (2x)²) dx

Answer: C) ∫[0,1] √(1 + (3x²)²) dx. Explanation: dy/dx = 3x², so length = ∫[0,1] √(1 + (3x²)²) dx

4. Question: Find area under curve y = 1/x from x = 1 to x = 2

A) ln|2|

B) ln|1/2|

C) ln|2/1|

D) ln|1|

Answer: A) ln|2|. Explanation: ∫[1,2] (1/x) dx = [ln|x|] from 1 to 2 = ln|2| - ln|1| = ln|2|

5. Question: Find volume of solid formed by revolving area under curve y = x from x = 0 to x = 2 about y-axis

A) 8π

B) 4π

C) 2π

D) π

Answer: B) 4π. Explanation: ∫[0,2] 2πx(x) dx = 2π∫[0,2] x² dx = 2π[(1/3)x³] from 0 to 2 = 2π[(1/3)(2³) - (1/3)(0³)] = 8π/3, but using disk method with x = y, ∫[0,2] 2πy dy = 2π[(1/2)y²] from 0 to 2 = π(2²) = 4π


🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

📚 Academic References

Content verified against peer-reviewed research:

  1. �Let the People Rap�: Cultural Rhetorics Pedagogy and Practices U... — Journal of Basic Writing (2019) 🔓 — DOI ↗
  2. Frustration and Hope: Examining Students� Emotional Responses to ... — Journal of Basic Writing (2019) — DOI ↗
  3. Editors' Column — Journal of Basic Writing (2019) — DOI ↗

🔓 = Open Access article

🎬 Watch video explanations on YouTube →


This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.


📚 Related Topics

Continue your revision with these related guides:

  • 📖 Integrals Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Compiler Design Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Computer Networks Class 12 Exam Prep Revision — Grandmaster Guide
  • 📖 Linear Programming Class 12 Exam Prep Revision — Grandmaster Guide