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Solve ∫|sin2pix| dx Limit is from 0 to 1.

Solve
∫|sin2pix| dx
Limit is from 0 to 1.

Grade:12

1 Answers

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

I = \int_{0}^{1}|sin2\pi x|dx
Period of this function is ½.
I = 2\int_{0}^{1/2}|sin2\pi x|dx
And it is +ve b/w 0 and ½.
I = 2\int_{0}^{1/2}(sin2\pi x)dx

I = 2(\frac{-cos2\pi x}{2\pi })_{0}^{1/2}
I = (\frac{-cos2\pi x}{\pi })_{0}^{1/2}
I = \frac{2}{\pi }

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