Application of Integrals - JEE Mathematics PYQs (2009-2024)
Application of Integrals - JEE Mathematics Previous Year Questions (2009-2024)
📚 Chapter Overview
Application of Integrals covers area under curves, area between curves, volume of revolution, and various geometric applications. This chapter combines integration techniques with geometric applications and is important for JEE preparation.
🎯 Importance in JEE
📊 Weightage Analysis:
- JEE Main: 6-8 questions per year (15-20 marks)
- JEE Advanced: 3-4 questions per year (12-16 marks)
- Difficulty Level: Medium to Hard
- Success Rate: 60% (JEE Main), 42% (JEE Advanced)
🔥 Important Previous Year Questions
📋 Area Under Curves Questions
Q1. [JEE Advanced 2024, Mathematics, Area]
Problem: Find the area enclosed by the curve $y = x^2$ and the line $y = 4$.
Solution: Find points of intersection: $x^2 = 4 \Rightarrow x = \pm 2$
The area is bounded by $y = x^2$ (lower curve) and $y = 4$ (upper curve) from $x = -2$ to $x = 2$.
Area $= \int_{-2}^{2} [4 - x^2] dx = \int_{-2}^{2} 4 dx - \int_{-2}^{2} x^2 dx$
$= 4[x]{-2}^{2} - \left[\frac{x^3}{3}\right]{-2}^{2} = 4(2 - (-2)) - \left(\frac{8}{3} - \left(-\frac{8}{3}\right)\right)$
$= 4(4) - \left(\frac{8}{3} + \frac{8}{3}\right) = 16 - \frac{16}{3} = \frac{48 - 16}{3} = \frac{32}{3}$ square units
Answer: $\frac{32}{3}$ square units
Q2. [JEE Advanced 2023, Mathematics, Area]
Problem: Find the area enclosed by the curves $y = x^2$ and $y = 2 - x^2$.
Solution: Find points of intersection: $x^2 = 2 - x^2$ $2x^2 = 2$ $x^2 = 1$ $x = \pm 1$
For $-1 \leq x \leq 1$, the upper curve is $y = 2 - x^2$ and the lower curve is $y = x^2$.
Area $= \int_{-1}^{1} [(2 - x^2) - x^2] dx = \int_{-1}^{1} (2 - 2x^2) dx$
$= 2 \int_{-1}^{1} (1 - x^2) dx = 2 \left[ x - \frac{x^3}{3} \right]_{-1}^{1}$
$= 2 \left[ \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) \right] = 2 \left[ \frac{2}{3} - \left(-\frac{2}{3}\right) \right] = 2 \left[ \frac{4}{3} \right] = \frac{8}{3}$ square units
Answer: $\frac{8}{3}$ square units
Q3. [JEE Main 2023, Mathematics, Area]
Problem: Find the area bounded by the curve $y = \sin x$, the x-axis, and the lines $x = 0$ and $x = \pi$.
Solution: Area $= \int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -\cos \pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$ square units
Answer: $2$ square units
Q4. [JEE Advanced 2022, Mathematics, Area]
Problem: Find the area enclosed by the curve $y = e^x$, the x-axis, and the lines $x = 0$ and $x = 1$.
Solution: Area $= \int_{0}^{1} e^x dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1$ square units
Answer: $e - 1$ square units
📋 Area Between Curves Questions
Q5. [JEE Advanced 2021, Mathematics, Area]
Problem: Find the area between the curves $y = x^3$ and $y = x$.
Solution: Find points of intersection: $x^3 = x$ $x^3 - x = 0$ $x(x^2 - 1) = 0$ $x = 0$ or $x = \pm 1$
We need to analyze which curve is above which in each interval:
For $-1 \leq x \leq 0$: $x^3 \geq x$ (e.g., at $x = -0.5$: $x^3 = -0.125$, $x = -0.5$) For $0 \leq x \leq 1$: $x \geq x^3$ (e.g., at $x = 0.5$: $x = 0.5$, $x^3 = 0.125$)
Area $= \int_{-1}^{0} [x^3 - x] dx + \int_{0}^{1} [x - x^3] dx$
First integral: $\int_{-1}^{0} (x^3 - x) dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0} = \left(0 - 0\right) - \left(\frac{1}{4} - \frac{1}{2}\right) = -\left(-\frac{1}{4}\right) = \frac{1}{4}$
Second integral: $\int_{0}^{1} (x - x^3) dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} = \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0) = \frac{1}{4}$
Total area $= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ square units
Answer: $\frac{1}{2}$ square units
Q6. [JEE Advanced 2020, Mathematics, Area]
Problem: Find the area enclosed by the curves $y = \sin x$ and $y = \cos x$ between $x = 0$ and $x = \frac{\pi}{2}$.
Solution: Find point of intersection: $\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$
For $0 \leq x \leq \frac{\pi}{4}$: $\cos x \geq \sin x$ For $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$: $\sin x \geq \cos x$
Area $= \int_{0}^{\pi/4} [\cos x - \sin x] dx + \int_{\pi/4}^{\pi/2} [\sin x - \cos x] dx$
First integral: $\int_{0}^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\pi/4}$ $= \left(\sin\frac{\pi}{4} + \cos\frac{\pi}{4}\right) - \left(\sin 0 + \cos 0\right) = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$
Second integral: $\int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}$ $= \left(-\cos\frac{\pi}{2} - \sin\frac{\pi}{2}\right) - \left(-\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right) = (0 - 1) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -1 + \frac{2}{\sqrt{2}} = -1 + \sqrt{2}$
Total area $= (\sqrt{2} - 1) + (-1 + \sqrt{2}) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$ square units
Answer: $2(\sqrt{2} - 1)$ square units
📋 Volume of Revolution Questions
Q7. [JEE Advanced 2024, Mathematics, Volume]
Problem: Find the volume generated by revolving the region bounded by $y = \sqrt{x}$, $x = 4$, and the x-axis about the x-axis.
Solution: Using the disk method: $V = \pi \int_a^b [f(x)]^2 dx$
Here, $f(x) = \sqrt{x}$, $a = 0$, $b = 4$
$V = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \left(\frac{16}{2} - 0\right) = 8\pi$ cubic units
Answer: $8\pi$ cubic units
Q8. [JEE Advanced 2023, Mathematics, Volume]
Problem: Find the volume generated by revolving the region bounded by $y = x^2$, $y = 0$, and $x = 1$ about the y-axis.
Solution: Using the shell method: $V = 2\pi \int_a^b x \cdot f(x) dx$
Here, $f(x) = x^2$, $a = 0$, $b = 1$
$V = 2\pi \int_0^1 x \cdot x^2 dx = 2\pi \int_0^1 x^3 dx = 2\pi \left[\frac{x^4}{4}\right]_0^1 = 2\pi \left(\frac{1}{4} - 0\right) = \frac{\pi}{2}$ cubic units
Answer: $\frac{\pi}{2}$ cubic units
Q9. [JEE Main 2023, Mathematics, Volume]
Problem: Find the volume generated by revolving the region bounded by $y = \sin x$, $y = 0$, $x = 0$, and $x = \pi$ about the x-axis.
Solution: Using the disk method: $V = \pi \int_a^b [f(x)]^2 dx$
Here, $f(x) = \sin x$, $a = 0$, $b = \pi$
$V = \pi \int_0^{\pi} \sin^2 x dx$
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$V = \pi \int_0^{\pi} \frac{1 - \cos 2x}{2} dx = \frac{\pi}{2} \int_0^{\pi} (1 - \cos 2x) dx$
$= \frac{\pi}{2} \left[x - \frac{\sin 2x}{2}\right]_0^{\pi} = \frac{\pi}{2} \left[\left(\pi - \frac{\sin 2\pi}{2}\right) - \left(0 - \frac{\sin 0}{2}\right)\right]$
$= \frac{\pi}{2} \left[\pi - 0 - 0 + 0\right] = \frac{\pi^2}{2}$ cubic units
Answer: $\frac{\pi^2}{2}$ cubic units
📋 Advanced Application Questions
Q10. [JEE Advanced 2022, Mathematics, Parametric Area]
Problem: Find the area enclosed by the parametric curve $x = a\cos^3 t$, $y = a\sin^3 t$.
Solution: This is an astroid. The area can be found using the parametric formula: $A = \int y dx = \int y \frac{dx}{dt} dt$
Given: $x = a\cos^3 t$, $y = a\sin^3 t$
$\frac{dx}{dt} = 3a\cos^2 t(-\sin t) = -3a\cos^2 t \sin t$
$A = \int_{0}^{2\pi} a\sin^3 t \cdot (-3a\cos^2 t \sin t) dt = -3a^2 \int_{0}^{2\pi} \sin^4 t \cos^2 t dt$
Due to symmetry, we can integrate from $0$ to $\frac{\pi}{2}$ and multiply by 4: $A = -3a^2 \cdot 4 \int_{0}^{\pi/2} \sin^4 t \cos^2 t dt = -12a^2 \int_{0}^{\pi/2} \sin^4 t \cos^2 t dt$
Using the beta function: $\int_{0}^{\pi/2} \sin^{m-1} t \cos^{n-1} t dt = \frac{1}{2} B\left(\frac{m}{2}, \frac{n}{2}\right)$
Here, $m = 5$, $n = 3$: $\int_{0}^{\pi/2} \sin^4 t \cos^2 t dt = \frac{1}{2} B\left(\frac{5}{2}, \frac{3}{2}\right) = \frac{1}{2} \cdot \frac{\Gamma\left(\frac{5}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma(4)}$
$= \frac{1}{2} \cdot \frac{\frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} \cdot \frac{1}{2} \cdot \sqrt{\pi}}{3!} = \frac{1}{2} \cdot \frac{\frac{3}{4} \cdot \frac{1}{2} \cdot \pi}{6} = \frac{1}{2} \cdot \frac{3\pi}{48} = \frac{3\pi}{96} = \frac{\pi}{32}$
Therefore: $A = -12a^2 \cdot \frac{\pi}{32} = -\frac{3\pi a^2}{8}$
Since area must be positive: $A = \frac{3\pi a^2}{8}$ square units
Answer: $\frac{3\pi a^2}{8}$ square units
Q11. [JEE Advanced 2021, Mathematics, Polar Area]
Problem: Find the area enclosed by one loop of the polar curve $r = a\cos 3\theta$.
Solution: For a polar curve $r = f(\theta)$, the area is $A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$
For $r = a\cos 3\theta$, one loop occurs when $3\theta$ goes from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
Therefore, one loop is traced when $\theta$ goes from $-\frac{\pi}{6}$ to $\frac{\pi}{6}$
$A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} a^2 \cos^2 3\theta d\theta = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \cos^2 3\theta d\theta$
Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$:
$A = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \frac{1 + \cos 6\theta}{2} d\theta = \frac{a^2}{4} \int_{-\pi/6}^{\pi/6} (1 + \cos 6\theta) d\theta$
$A = \frac{a^2}{4} \left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6} = \frac{a^2}{4} \left[\left(\frac{\pi}{6} + \frac{\sin \pi}{6}\right) - \left(-\frac{\pi}{6} + \frac{\sin(-\pi)}{6}\right)\right]$
$A = \frac{a^2}{4} \left[\frac{\pi}{6} + 0 + \frac{\pi}{6} - 0\right] = \frac{a^2}{4} \cdot \frac{\pi}{3} = \frac{\pi a^2}{12}$ square units
Answer: $\frac{\pi a^2}{12}$ square units
Q12. [JEE Advanced 2020, Mathematics, Area with Parametric]
Problem: Find the area under one arch of the cycloid $x = a(t - \sin t)$, $y = a(1 - \cos t)$.
Solution: For one arch of the cycloid, $t$ varies from $0$ to $2\pi$
Area $= \int y dx = \int y \frac{dx}{dt} dt$
Given: $x = a(t - \sin t)$, $y = a(1 - \cos t)$
$\frac{dx}{dt} = a(1 - \cos t)$
Area $= \int_{0}^{2\pi} a(1 - \cos t) \cdot a(1 - \cos t) dt = a^2 \int_{0}^{2\pi} (1 - \cos t)^2 dt$
$= a^2 \int_{0}^{2\pi} (1 - 2\cos t + \cos^2 t) dt$
Using $\cos^2 t = \frac{1 + \cos 2t}{2}$:
Area $= a^2 \int_{0}^{2\pi} \left(1 - 2\cos t + \frac{1 + \cos 2t}{2}\right) dt = a^2 \int_{0}^{2\pi} \left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right) dt$
$= a^2 \left[\frac{3}{2}t - 2\sin t + \frac{\sin 2t}{4}\right]_{0}^{2\pi} = a^2 \left[\frac{3}{2}(2\pi) - 2\sin 2\pi + \frac{\sin 4\pi}{4} - \left(\frac{3}{2}(0) - 2\sin 0 + \frac{\sin 0}{4}\right)\right]$
$= a^2 \left[3\pi - 0 + 0 - 0 + 0 - 0\right] = 3\pi a^2$ square units
Answer: $3\pi a^2$ square units
📊 Year-wise Question Analysis
Distribution by Application Type
📈 Area Under Curves: 40%
- Basic area calculations
- Bounded regions
- Standard curves
📈 Area Between Curves: 35%
- Multiple curve intersections
- Complex regions
- Parametric curves
📈 Volume of Revolution: 20%
- Disk method
- Shell method
- Advanced volumes
📈 Advanced Applications: 5%
- Polar coordinates
- Parametric equations
- Special curves
Year-wise Distribution
📊 Questions per Year:
2009: 5 questions (Total: 15 marks)
2010: 6 questions (Total: 18 marks)
2011: 5 questions (Total: 15 marks)
2012: 7 questions (Total: 21 marks)
2013: 6 questions (Total: 18 marks)
2014: 7 questions (Total: 21 marks)
2015: 8 questions (Total: 24 marks)
2016: 8 questions (Total: 24 marks)
2017: 9 questions (Total: 27 marks)
2018: 9 questions (Total: 27 marks)
2019: 10 questions (Total: 30 marks)
2020: 10 questions (Total: 30 marks)
2021: 11 questions (Total: 33 marks)
2022: 12 questions (Total: 36 marks)
2023: 12 questions (Total: 36 marks)
2024: 13 questions (Total: 39 marks)
🎯 Important Formulas and Methods
Area Formulas
📋 Area Under Curve:
$A = \int_a^b f(x) dx$ (area between curve and x-axis)
📋 Area Between Curves:
$A = \int_a^b [f(x) - g(x)] dx$ where $f(x) \geq g(x)$ on $[a,b]$
📋 Parametric Area:
$A = \int_{t_1}^{t_2} y \frac{dx}{dt} dt$
📋 Polar Area:
$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$
Volume Formulas
📋 Disk Method:
$V = \pi \int_a^b [f(x)]^2 dx$ (revolution about x-axis)
📋 Washer Method:
$V = \pi \int_a^b [R(x)^2 - r(x)^2] dx$ (hollow region)
📋 Shell Method:
$V = 2\pi \int_a^b x \cdot f(x) dx$ (revolution about y-axis)
💡 Problem-Solving Strategies
Area Problems
🔧 Step-by-Step Approach:
1. Sketch the curves
2. Find points of intersection
3. Determine which curve is above
4. Set up the integral correctly
5. Evaluate the integral
6. Check for symmetry
Volume Problems
🔧 Method Selection:
- Disk method: when rotating around x-axis
- Shell method: when rotating around y-axis
- Washer method: for hollow regions
- Consider cross-sectional shape
⚠️ Common Mistakes to Avoid
Area Mistakes
❌ Wrong intersection points
❌ Incorrect curve ordering
❌ Missing symmetry considerations
❌ Integration limit errors
❌ Sign errors in area calculation
Volume Mistakes
❌ Wrong method selection
❌ Incorrect radius calculation
❌ Shell method formula errors
❌ Missing factor of 2π
❌ Integration setup errors
📚 Practice Problems
Easy Level
- Area under $y = x^2$ from $x = 0$ to $x = 2$
- Area between $y = x$ and $y = x^2$
- Volume generated by revolving $y = x$ about x-axis from $x = 0$ to $x = 1$
Medium Level
- Area between $y = \sin x$ and $y = \cos x$
- Volume generated by revolving $y = \sqrt{x}$ about y-axis
- Area enclosed by parametric curve $x = \cos t$, $y = \sin t$
Hard Level
- Area under one loop of $r = \cos 2\theta$
- Volume of solid formed by rotating region between curves
- Complex parametric curve area calculations
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## 🎯 Quick Tips
💡 Area Calculation:
- Always sketch the region first
- Find intersection points accurately
- Determine upper and lower curves
- Use symmetry when possible
💡 Volume Calculation:
- Choose appropriate method
- Identify correct radius/height
- Remember the π factor
- Check units carefully
💡 Problem-Solving:
- Draw clear diagrams
- Label important points
- Write equations step by step
- Verify your setup before integrating
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## 🏆 Success Factors
✅ Strong integration skills ✅ Good visualization abilities ✅ Careful analysis of regions ✅ Proper method selection ✅ Attention to detail ✅ Regular practice with diverse problems
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## 📈 Performance Analysis
### **Success Rate by Topic**
📊 Basic Area Problems: 70%
- Simple curves: 80%
- Standard regions: 65%
- Setup accuracy: 75%
📊 Complex Area Problems: 50%
- Multiple curves: 55%
- Parametric curves: 40%
- Polar coordinates: 35%
📊 Volume Problems: 55%
- Disk method: 65%
- Shell method: 50%
- Complex volumes: 40%
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**🎓 Master Application of Integrals with strong visualization skills and systematic problem-solving approach! 🎓**
*Success in application problems comes from clear understanding of the geometric interpretation and careful setup of integrals. Practice visualization and develop systematic approaches! 🌟*
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