Inequalities in Definite Integrals: How to Compare Without Calculating
Master JEE-advanced techniques to compare integrals using properties, inequalities, and smart observations without full computation.
Why Learn Integral Comparison Techniques?
JEE Advanced frequently tests your ability to compare definite integrals without explicit calculation. These problems assess deep conceptual understanding rather than computational skill.
🎯 JEE Examination Pattern
- Single Correct: "Which integral is larger?" type questions
- Multiple Correct: Comparing multiple integrals using properties
- Integer Type: Finding relationships between integrals
- Comprehension: Applying comparison theorems in multi-step problems
🧭 Navigation Guide
Fundamental Comparison Theorem
If $f(x) \geq g(x)$ for all $x \in [a,b]$, then
Equality holds only if $f(x) = g(x)$ for all $x \in [a,b]$
💡 Key Insight
This is the most powerful tool for comparing integrals. The strategy is to:
- Find the difference between the two functions
- Analyze the sign of this difference on the interval
- Use function properties (monotonicity, concavity, symmetry)
Technique 1: Using Monotonicity
Problem Statement
Compare $I_1 = \int_0^1 x^2 dx$ and $I_2 = \int_0^1 x^3 dx$
Solution Using Monotonicity
Step 1: For $x \in [0,1]$, we have $0 \leq x \leq 1$
Step 2: Multiply both sides by $x^2$ (non-negative): $0 \leq x^3 \leq x^2$
Step 3: Therefore, $x^2 \geq x^3$ for all $x \in [0,1]$
Step 4: By comparison theorem: $\int_0^1 x^2 dx \geq \int_0^1 x^3 dx$
Conclusion: $I_1 \geq I_2$
🎯 General Strategy
- If $f(x)$ is increasing and $g(x)$ is decreasing, compare at endpoints
- Use known inequalities: $x^n \geq x^m$ when $n < m$ and $0 \leq x \leq 1$
- For trigonometric functions, use range properties: $\sin x \leq x \leq \tan x$ for $x \in [0,\pi/2)$
Technique 2: Using Symmetry Properties
Problem Statement
Compare $I_1 = \int_0^{\pi/2} \sin x dx$ and $I_2 = \int_0^{\pi/2} \cos x dx$
Solution Using Symmetry
Step 1: Use substitution $x \to \frac{\pi}{2} - x$ in $I_2$:
$$I_2 = \int_0^{\pi/2} \cos x dx = \int_{\pi/2}^0 \cos\left(\frac{\pi}{2} - t\right)(-dt) = \int_0^{\pi/2} \sin t dt = I_1$$
Step 2: Therefore, $I_1 = I_2$ exactly!
🎯 Symmetry Properties to Remember
- $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ (Standard property)
- For even functions: $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$
- For odd functions: $\int_{-a}^a f(x)dx = 0$
- $\int_0^{\pi/2} \sin^n x dx = \int_0^{\pi/2} \cos^n x dx$
Technique 3: Using Function Bounds
Problem Statement
Compare $I_1 = \int_0^1 e^x dx$ and $I_2 = \int_0^1 (1 + x + \frac{x^2}{2}) dx$
Solution Using Series Bounds
Step 1: Recall Taylor series: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
Step 2: For $x \in [0,1]$, all terms are non-negative, so:
$$e^x \geq 1 + x + \frac{x^2}{2}$$
Step 3: By comparison theorem:
$$\int_0^1 e^x dx \geq \int_0^1 \left(1 + x + \frac{x^2}{2}\right) dx$$
Conclusion: $I_1 \geq I_2$
🎯 Common Bounds to Remember
- $1 - x \leq e^{-x} \leq 1$ for $x \in [0,1]$
- $x - \frac{x^3}{6} \leq \sin x \leq x$ for $x \geq 0$
- $1 - \frac{x^2}{2} \leq \cos x \leq 1$ for $x \in [0,\pi/2]$
- $\frac{x}{1+x} \leq \ln(1+x) \leq x$ for $x > -1$
📚 Quick Reference: Integral Inequalities
Basic Inequalities
- If $m \leq f(x) \leq M$ on $[a,b]$, then: $$m(b-a) \leq \int_a^b f(x)dx \leq M(b-a)$$
- $\left|\int_a^b f(x)dx\right| \leq \int_a^b |f(x)|dx$
- Cauchy-Schwarz: $\left(\int_a^b f(x)g(x)dx\right)^2 \leq \int_a^b f^2(x)dx \cdot \int_a^b g^2(x)dx$
Advanced Properties
- First Mean Value Theorem: $$\int_a^b f(x)dx = f(c)(b-a) \text{ for some } c \in [a,b]$$
- If $f$ is convex: $f\left(\frac{a+b}{2}\right) \leq \frac{1}{b-a}\int_a^b f(x)dx \leq \frac{f(a)+f(b)}{2}$
- Chebyshev's inequality for monotonic functions
Technique 4: Smart Substitutions
Problem Statement
Compare $I_1 = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$ and $I_2 = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx$
Solution Using Substitution
Step 1: Use substitution $x \to \frac{\pi}{2} - x$ in $I_1$:
$$I_1 = \int_0^{\pi/2} \frac{\sin\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx = I_2$$
Step 2: Therefore, $I_1 = I_2$
Step 3: Also, $I_1 + I_2 = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$
Step 4: So $I_1 = I_2 = \frac{\pi}{4}$
🎯 Useful Substitutions
- $x \to a - x$ for integrals over $[0,a]$
- $x \to \frac{\pi}{2} - x$ for trigonometric integrals
- $x \to \frac{1}{t}$ for integrals involving reciprocals
- $x \to \tan t$ for integrals with $1+x^2$ in denominator
Technique 5: Using Integral Properties
Problem Statement
Compare $I_1 = \int_0^1 x(1-x)^n dx$ and $I_2 = \int_0^1 x^n(1-x) dx$
Solution Using Beta Function Properties
Step 1: Recognize Beta function form:
$$I_1 = \int_0^1 x^{2-1}(1-x)^{n+1-1} dx = B(2, n+1)$$
$$I_2 = \int_0^1 x^{n+1-1}(1-x)^{2-1} dx = B(n+1, 2)$$
Step 2: Use symmetry: $B(m,n) = B(n,m)$
Step 3: Therefore, $I_1 = I_2$ for all $n$
🎯 Important Integral Properties
- Beta function: $B(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$
- $B(m,n) = B(n,m)$ (Symmetry)
- $\int_0^{\pi/2} \sin^{2m-1}x\cos^{2n-1}x dx = \frac{1}{2}B(m,n)$
- $\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} dx = B(m,n)$
📝 Practice Problems
Test your understanding with these JEE-level problems:
1. Compare without calculation: $\int_0^1 e^{-x^2} dx$ and $\int_0^1 e^{-x} dx$
2. Show that $\int_0^{\pi/2} \sin^n x dx = \int_0^{\pi/2} \cos^n x dx$ using substitution
3. Prove: $\int_0^1 \frac{dx}{\sqrt{1-x^2}} \geq \int_0^1 \frac{dx}{\sqrt{1-x^4}}$
⚠️ Common Mistakes to Avoid
Conceptual Errors
- Assuming monotonicity without verification - Always check if the inequality holds throughout the interval
- Ignoring endpoint behavior - Functions might behave differently at boundaries
- Misapplying symmetry properties - Not all integrals over symmetric intervals are equal
Technical Errors
- Wrong substitution limits - Always check how limits transform
- Forgetting absolute values - When using $|f(x)| \geq f(x)$
- Misremembering standard inequalities - Practice common bounds regularly
🎯 Mastery Checklist
Techniques to Master
- ✓ Function comparison on intervals
- ✓ Symmetry and substitution methods
- ✓ Using known bounds and inequalities
- ✓ Beta and Gamma function properties
- ✓ Mean value theorems for integrals
Problem-Solving Strategy
- ✓ Always check the interval first
- ✓ Look for symmetry patterns
- ✓ Try simple substitutions
- ✓ Use known inequalities
- ✓ Verify with special cases
Ready to Master Integral Inequalities?
Practice with 50+ JEE-level problems with detailed solutions