Application Of Integrals Class 12 Exam Prep Revision — Grandmaster Guide
Ayush (Founder)
Exam Strategist
- ∫(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
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