Definite Integration
Squeeze Theorem + Integral
Grade None

Question:

<p>For \(f(x)=\dfrac{\sin x}{x}\), which is correct? \(\int_n^{n+1}f(x)\,dx\to 0\) as \(n\to\infty\). [JEE Advanced 2013]</p>
True — |\intₙ^(n+1) f| \leq 1/n \to 0
False — integral oscillates
True — f\to 0
Cannot determine

Step-by-Step Solution

Key Concept: |\intₙ^(n+1) sinx/x dx| \leq \intₙ^(n+1) |sinx|/x dx \leq \intₙ^(n+1) 1/n dx = 1/n \to 0.
<div class='solution'> <p>\(\left|\int_n^{n+1}\frac{\sin x}{x}dx\right|\le\int_n^{n+1}\frac{|\sin x|}{x}dx\le\int_n^{n+1}\frac{1}{n}dx=\frac{1}{n}\to 0\quad\text{as }n\to\infty\)</p> <p>So the integral → 0. ✓</p> </div>
Correct Answer: A

Master Definite Integration with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free