<p>Evaluate \(\displaystyle\int_0^{n\pi}|\sin x|\,dx\) where \(n\in\mathbb{N}\).</p>
Step-by-Step Solution
Key Concept: |sin x| has period \pi and \int_0^\pi |sin x|dx = 2, so \int_0^(n\pi) = n \cdot 2 = 2n.
<div class='solution'>
<p><strong>Step 1:</strong> |sin x| is periodic with period $\pi$.</p>
<p><strong>Step 2:</strong> $\displaystyle\int_0^\pi|\sin x|\,dx = \int_0^\pi\sin x\,dx = [-\cos x]_0^\pi = 1+1 = 2$</p>
<p><strong>Step 3:</strong> $\displaystyle\int_0^{n\pi}|\sin x|\,dx = n\cdot\int_0^\pi|\sin x|\,dx = n\cdot 2 = \boxed{2n}$</p>
Correct Answer: B