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The area bounded by the curve $y = \sin x$, the $x$-axis, between $x = 0$ and $x = 2\pi$ is
A$0$
B$4$
C$2$
D$1$
Answer & Solution
Correct answer: B. $4$
1. On $[0, 2\pi]$, $\sin x \geq 0$ for $[0, \pi]$ and $\sin x \leq 0$ for $[\pi, 2\pi]$.
2. Area $= \int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right|$.
3. $\int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = 2$; the second piece also has magnitude $2$.
4. Total area $= 2 + 2 = 4$.
5. Without absolute value the two halves cancel to $0$, the common trap.
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.6_
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