Definite Integration
Grade 12
Question:
<p>Let \(a_n=\int_0^{n\pi}\frac{|\sin x|}{x}\,dx\). Which is true? [JEE Advanced 2010]</p>
\(a_n\) is monotone increasing
\(a_n\to 0\)
\(a_n\to\pi/2\)
\(a_n\) diverges to \(+\infty\)
Step-by-Step Solution
Key Concept: aₙ₊_1 - aₙ = \int_{n\pi}^{(n+1)\pi} |sin x|/x dx > 0 since |sin x|/x > 0. Also the series \Sigma\int_{n\pi}^{(n+1)\pi}|sin x|/x dx \geq \Sigma 2/((n+1)\pi) diverges.
<div class='solution'>
<p>$a_{n+1}-a_n=\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{x}dx>0$. So $\{a_n\}$ is strictly increasing. ✓(A)</p>
<p>Lower bound: $\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{x}dx\ge\frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin x|dx=\frac{2}{(n+1)\pi}$.</p>
<p>$a_n\ge\frac{2}{\pi}\sum_{k=0}^{n-1}\frac{1}{k+1}=\frac{2}{\pi}H_n\to\infty$. So $a_n\to+\infty$. ✓(D)</p>
<p>Both A and D are correct; if only one answer: A (monotone) and D (diverges).</p>
Correct Answer: A