Periodic Function Integrals

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Exploiting the periodicity of the absolute sine function, evaluate the definite integral of the absolute value of sine x taken over the interval from zero to two \pi.

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Because the absolute sine function |\sin x| is periodic with period \pi rather than the usual 2\pi, the integral over [0, 2\pi] equals twice the integral over a single period [0, \pi]. The governing property is \int_0^{nT} f = n\int_0^T f for a function of period T, which lets us replace a wide integral by a multiple of one period. On [0, \pi] the ordinary sine is non-negative, so |\sin x| = \sin x and \int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 2. Therefore the full integral equals 2	imes 2 = 4. Option 0 wrongly applies signed cancellation, which the absolute value explicitly prevents by reflecting the negative arch upward. Option 2 covers only a single arch and forgets the second period. Option 2\pi confuses the interval length with the accumulated area. This periodicity reduction is a key time-saving property in JEE Advanced integration. As a final plausibility check, the integrand stays non-negative everywhere thanks to the absolute value, so the result must exceed the single-arch value 2, and 4 correctly reflects the two equal positive humps spanning the full interval.

Periodic Function Integrals

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