Greatest Integral

Calculus Level 4

∫ 0 2 π [ 2 sin ⁡ ( x ) ] dx = \large \displaystyle\int_0^{2\pi} [ 2\sin(x) ] \text{dx} =

Details :

  • [ . ] represents greatest integer function .
π \pi − 5 π 3 \dfrac{-5\pi}{3} − π -\pi − 2 π -2\pi 5 π 3 \dfrac{5\pi}{3}

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Jon Haussmann
Sep 21, 2015

∫ 0 2 π ⌊ 2 sin ⁡ x ⌋ d x = ∫ 0 π / 6 ⌊ 2 sin ⁡ x ⌋ d x + ∫ π / 6 5 π / 6 ⌊ 2 sin ⁡ x ⌋ d x + ∫ 5 π / 6 π ⌊ 2 sin ⁡ x ⌋ d x + ∫ π 7 π / 6 ⌊ 2 sin ⁡ x ⌋ d x + ∫ 7 π / 6 11 π / 6 ⌊ 2 sin ⁡ x ⌋ d x + ∫ 11 π / 6 2 π ⌊ 2 sin ⁡ x ⌋ d x = ∫ 0 π / 6 0 d x + ∫ π / 6 5 π / 6 1 d x + ∫ 5 π / 6 π 0 d x + ∫ π 7 π / 6 − 1 d x + ∫ 7 π / 6 11 π / 6 − 2 d x + ∫ 11 π / 6 2 π − 1 d x = 2 π 3 − π 6 − 4 π 3 − π 6 = − π . \begin{aligned} \int_0^{2 \pi} \lfloor 2 \sin x \rfloor dx &= \int_0^{\pi/6} \lfloor 2 \sin x \rfloor dx + \int_{\pi/6}^{5 \pi/6} \lfloor 2 \sin x \rfloor dx + \int_{5 \pi/6}^{\pi} \lfloor 2 \sin x \rfloor dx \\ &\quad + \int_{\pi}^{7 \pi/6} \lfloor 2 \sin x \rfloor dx + \int_{7 \pi/6}^{11 \pi/6} \lfloor 2 \sin x \rfloor dx + \int_{11 \pi/6}^{2 \pi} \lfloor 2 \sin x \rfloor dx \\ &= \int_0^{\pi/6} 0 dx + \int_{\pi/6}^{5 \pi/6} 1 dx + \int_{5 \pi/6}^{\pi} 0 dx \\ &\quad + \int_{\pi}^{7 \pi/6} -1 dx + \int_{7 \pi/6}^{11 \pi/6} -2 dx + \int_{11 \pi/6}^{2 \pi} -1 dx \\ &= \frac{2 \pi}{3} - \frac{\pi}{6} - \frac{4 \pi}{3} - \frac{\pi}{6} \\ &= -\pi. \end{aligned}

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@Jon Haussmann I've converted your report over to this solution.

Calvin Lin Staff - 5 years, 8 months ago

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