<p>The area bounded by the curve <math>y = \sin x</math> between <math>x = 0</math> and <math>x = 2\pi</math> is</p>
Step-by-Step Solution
Key Concept: When finding the area bounded by a curve, we must account for regions where the function is negative by taking absolute values of the integrand.
<p><strong>Solution:</strong></p><p>The graph of <math>y = \sin x</math> can be drawn as shown.</p><p>Required area = Area of OABO + Area BCDB</p><p><math>\displaystyle = \int_0^{\pi} |\sin x| dx + \int_{\pi}^{2\pi} |\sin x| dx</math></p><p>Since <math>\sin x \geq 0</math> for <math>x \in [0, \pi]</math> and <math>\sin x \leq 0</math> for <math>x \in [\pi, 2\pi]</math></p><p><math>\displaystyle = \int_0^{\pi} \sin x dx + \int_{\pi}^{2\pi} (-\sin x) dx</math></p><p><math>= [-\cos x]_0^{\pi} + [\cos x]_{\pi}^{2\pi}</math></p><p><math>= -\cos \pi + \cos 0 + \cos 2\pi - \cos \pi</math></p><p><math>= -(-1) + 1 + 1 - (-1)</math></p><p><math>= 4</math> sq units</p>
Correct Answer: c