Thread: having 2pi as the upper limit of an interval
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Old October 7th, 2009, 07:31 PM
walleye walleye is offline
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Default having 2pi as the upper limit of an interval
i cant make this work
e.g.

\int_0^2\pi \theta d\theta

the pi is mean to be the upper limit of the integral, not inside the integrand
any helps?


also i cant find out how to do the inequality sign
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