Inequalities and Estimation Techniques for Definite Integrals

Introduction

When exact evaluation of an integral is difficult, inequalities provide bounds and comparisons that can answer multiple‑choice questions efficiently. These techniques are frequently tested in JEE Main and Advanced.

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1. Basic Bounding Inequality

Theorem: If $m \le f(x) \le M$ for all $x \in [a, b]$, then
$m(b-a) \le \int_a^b f(x)\,dx \le M(b-a).$

How to find $m$ and $M$:
For a continuous $f$ on $[a,b]$, $m$ is the minimum and $M$ the maximum of $f$ on that interval.
Check:

1. Critical points where $f'(x)=0$.
2. Endpoints $f(a)$ and $f(b)$.
3. Monotonicity if $f'$ does not change sign.

Example: Estimate $\displaystyle I = \int_0^1 e^{-x^2} dx$.
On $[0,1]$, $e^{-x^2}$ decreases, so $M = e^0 = 1$, $m = e^{-1} = 1/e$.
Thus:
$\frac{1}{e} \le I \le 1.$
A sharper lower bound: $e^{-x^2} \ge 1 - x^2$ gives
$I \ge \int_0^1 (1-x^2)dx = \frac{2}{3}.$

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2. Comparison Theorem

If $f(x) \le g(x)$ on $[a,b]$, then
$\int_a^b f(x)dx \le \int_a^b g(x)dx.$

Example: Show $\displaystyle \int_0^1 \frac{x^{99}}{1+x} dx < \frac{1}{100}$.
Since $\frac{x^{99}}{1+x} < x^{99}$ for $x>0$,
$\int_0^1 \frac{x^{99}}{1+x} dx < \int_0^1 x^{99} dx = \frac{1}{100}.$

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3. Cauchy–Schwarz Inequality for Integrals

Statement: For integrable functions $f,g$ on $[a,b]$,
$\boxed{\left(\int_a^b f(x)g(x)\,dx\right)^2 \le \left(\int_a^b f(x)^2 dx\right)\left(\int_a^b g(x)^2 dx\right)}.$
Equality holds iff $f = \lambda g$ for some constant $\lambda$.

Proof: Consider $\phi(\lambda) = \int_a^b (f + \lambda g)^2 dx \ge 0$ for all $\lambda$. Expanding gives a quadratic in $\lambda$ that must have non‑positive discriminant.

Corollary (taking square roots):
$\left|\int_a^b f(x)g(x)dx\right| \le \sqrt{\int_a^b f(x)^2 dx} \; \sqrt{\int_a^b g(x)^2 dx}.$

Example: Prove $\displaystyle \left(\int_0^\pi x\sin x\,dx\right)^2 \le \frac{\pi^4}{6}$.
By Cauchy‑Schwarz with $f(x)=x,\; g(x)=\sin x$:
$\left(\int_0^\pi x\sin x\,dx\right)^2 \le \int_0^\pi x^2 dx \;\int_0^\pi \sin^2 x\,dx = \frac{\pi^3}{3} \cdot \frac{\pi}{2} = \frac{\pi^4}{6}.$
(Actually $\int_0^\pi x\sin x\,dx = \pi$, so LHS $= \pi^2$, and $\pi^2 < \frac{\pi^4}{6}$ for $\pi > \sqrt{6}$, which is true.)

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4. Absolute‑Value Inequality

$\left|\int_a^b f(x)dx\right| \le \int_a^b |f(x)| dx.$

Example: For $\displaystyle I = \int_0^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x} dx$,
using $|\sin x - \cos x| \le \sqrt{2}$ and $1+\sin x\cos x \ge 1$, we get $|I| \le \sqrt{2} \cdot \frac{\pi}{2}$.

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5. Mean Value Theorem for Integrals

If $f$ is continuous on $[a,b]$, there exists $c \in (a,b)$ such that
$\int_a^b f(x)dx = f(c)(b-a).$

Corollary – Average value:
$f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)dx.$

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6. Worked JEE‑Style Examples

Example 1 (Basic Bounds)

Show $\displaystyle \frac{1}{2} < \int_0^1 \frac{dx}{1+x^2+x^4} < 1$.

Solution: Let $f(x) = \frac{1}{1+x^2+x^4}$.
On $[0,1]$, denominator increases, so $f$ decreases.
$f(0)=1,\; f(1)=\frac13$. Hence:
$\frac13 \cdot 1 < I < 1 \cdot 1.$
For a better lower bound, note $1+x^2+x^4 \le 1+x^2+x^2 = 1+2x^2$, so
$f(x) \ge \frac{1}{1+2x^2}.$
Then $I \ge \int_0^1 \frac{dx}{1+2x^2} = \frac{1}{\sqrt2}\tan^{-1}\sqrt2 \approx 0.615 > \frac12$.
Thus $\frac12 < I < 1$.

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Example 2 (Cauchy‑Schwarz Application)

Prove $\displaystyle \left(\int_0^1 e^x\sin x\,dx\right)^2 < \frac{e^2-1}{2} \cdot \frac{1-\cos 2}{4}$.

Solution: Take $f(x)=e^x,\; g(x)=\sin x$. Then:
$\left(\int_0^1 e^x\sin x\,dx\right)^2 \le \int_0^1 e^{2x}dx \;\int_0^1 \sin^2 x\,dx.$
Now $\int_0^1 e^{2x}dx = \frac{e^2-1}{2}$,
$\int_0^1 \sin^2 x\,dx = \frac12 \int_0^1 (1-\cos 2x)dx = \frac12\left(1 - \frac{\sin 2}{2}\right)$.
Since $e^x$ and $\sin x$ are not proportional, the inequality is strict.

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Example 3 (JEE Advanced 2012 – Multiple Correct)

Let $S = \int_0^1 e^{-x^2} dx$. Which are true?
(A) $S \ge \frac1e$
(B) $S \ge 1 - \frac1e$
(C) $S \le \frac14\left(1 + \frac1{\sqrt e}\right)$
(D) $S \le \frac1{\sqrt2} + \frac1{\sqrt e}\left(1 - \frac1{\sqrt2}\right)$.

Analysis:

• For (A): On $[0,1]$, $e^{-x^2} \ge e^{-1}$, so $S \ge \frac1e$. ✓
• For (B): Since $x^2 \le x$ on $[0,1]$, $e^{-x^2} \ge e^{-x}$, so
  $S \ge \int_0^1 e^{-x}dx = 1 - \frac1e$. ✓
• For (C) & (D): Need careful upper‑bound estimates (trapezoidal rule on subintervals). Detailed checking shows (C) false, (D) true.

Answer: (A), (B), (D).

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Example 4 (JEE Main Pattern)

The value of $\displaystyle I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$ lies in:
(A) $[0, \frac{\pi}{16}]$
(B) $[\frac{\pi}{16}, \frac{\pi}{8}]$
(C) $[\frac{\pi}{8}, \frac{\pi}{4}]$
(D) $[\frac{\pi}{4}, \frac{\pi}{2}]$.

Solution: Let $f(x) = \frac{\sin x + \cos x}{3 + \sin 2x}$.
On $[0,\pi/4]$: $\sin 2x$ increases from 0 to 1, so denominator ∈ $[3,4]$.
Numerator: $\sin x + \cos x = \sqrt2\sin(x+\pi/4)$ ∈ $[1, \sqrt2]$.
Thus $\frac14 \le f(x) \le \frac{\sqrt2}{3}$. Multiply by interval length $\pi/4$:
$\frac{\pi}{16} \le I \le \frac{\sqrt2\,\pi}{12} \approx 0.37.$
Since $\pi/8 \approx 0.39$, we have $I \in [\pi/16, \pi/8]$.

Answer: (B).

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7. Quick‑Reference Summary

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8. Tips for JEE Problems

1. Identify monotonicity – if $f'$ has constant sign, the extremes occur at endpoints.
2. Use known inequalities like $\sin x < x < \tan x$ for $0<x<\pi/2$, $e^x \ge 1+x$, etc.
3. For Cauchy‑Schwarz, look for products $\int f(x)g(x)dx$; choose $f,g$ cleverly.
4. Break the interval – sometimes bounds on subintervals give sharper estimates.
5. Check special points – endpoints, midpoints, and points where derivative vanishes.

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9. Practice Problems with Hints

1. Prove $\displaystyle 1 < \int_0^1 \sqrt{1+x^2}\,dx < \sqrt2$.
   Hint: $\sqrt{1+x^2}$ is increasing.

2. Show $\displaystyle \frac{\pi}{8} < \int_0^{\pi/4} \tan x\,dx < \frac{\pi}{4}$.
   Hint: Use $\tan x > x$ on $(0,\pi/4)$ and $\tan x < 1$.

3. Using Cauchy‑Schwarz, prove $\displaystyle \int_0^1 f(x)^2 dx \ge 1$ if $\int_0^1 f(x)dx = 1$.
   Hint: Take $g(x)=1$.

4. Estimate $\displaystyle \int_0^{\pi/2} \frac{\sin x}{x} dx$.
   Hint: On $(0,\pi/2]$, $\frac{2}{\pi} \le \frac{\sin x}{x} \le 1$.

5. Prove $\displaystyle \int_0^1 \frac{x^n}{1+x} dx < \frac{1}{n+1}$ for $n>0$.
   Hint: $\frac{1}{1+x} < 1$.

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10. Answers to Practice Problems

1. $f(0)=1,\; f(1)=\sqrt2$; $f$ increasing ⇒ $1 \cdot 1 < I < \sqrt2 \cdot 1$.
2. $\tan x > x$ gives $I > \int_0^{\pi/4} x\,dx = \pi^2/32 > \pi/8$; $\tan x < 1$ gives $I < \pi/4$.
3. By C‑S: $1 = \left(\int_0^1 1\cdot f\,dx\right)^2 \le \int_0^1 1\,dx \cdot \int_0^1 f^2 dx = 1\cdot \int_0^1 f^2 dx$.
4. $\frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \le I \le 1 \cdot \frac{\pi}{2} = \pi/2$; actual value ≈ 1.370.
5. $\frac{x^n}{1+x} < x^n$; integrate: $\int_0^1 x^n dx = 1/(n+1)$.

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Master these inequalities through practice. They allow you to bound integrals quickly, verify answers, and solve “comparison” questions common in JEE.