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options -nπ -(n +1) π -2n π -2(n +1) π

options
  1. -nπ
  2. -(n +1) π
  3. -2n π
  4. -2(n +1) π

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{x}[sint]dt
x\in (2n\pi ,(4n+1)\pi )
I = \int_{0}^{2n\pi }[sint]dt + \int_{2n\pi }^{(4n+1)\pi }[sint]dt
[sint] is a periodic with a period of 2pi.
I = \int_{0}^{2n\pi }[sint]dt + \int_{2n\pi }^{(4n+1)\pi }[sint]dt
I = n\int_{0}^{2\pi }[sint]dt + \int_{2n\pi }^{4n\pi }[sint]dt + \int_{4n\pi }^{(4n+1)\pi }[sint]dt
[sint] = 0, 0\leq x\leq \pi/2 [sint] = 1,x = \pi/2 [sint] = 0, \pi /2\leq x\leq \pi [sint] = -1, \pi \leq x\leq 2\pi
I = n(0-1)_{\pi }^{2\pi} + \int_{2n\pi }^{4n\pi }[sint]dt
I = -n\pi + \int_{2n\pi }^{4n\pi }[sint]dt
I = -n\pi + (2n-n)\int_{0 }^{2\pi }[sint]dt
I = -n\pi + n\int_{\pi }^{2\pi }(-1)dt
I = -2n\pi ?

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