What is the integral from 0 to 2pi of e^{-2ix} ?

The integral from 0 to 2pi of e^{-2ix} is 0

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integral from 0 to 2pi of e^{-2ix}

\int_{0}^{2 π} e^{- 2 i x} d x

0

Solution steps

\int_{0}^{2 π} e^{- 2 i x} d x

Simplify

e^{- 2 i x} : cos (2 x) - i sin (2 x)

= \int_{0}^{2 π} cos (2 x) - i sin (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2 π} cos (2 x) d x - \int_{0}^{2 π} i sin (2 x) d x

\int_{0}^{2 π} cos (2 x) d x = 0

\int_{0}^{2 π} i sin (2 x) d x = 0

= 0 - 0

Simplify

= 0

What is the integral from 0 to 2pi of e^{-2ix} ?
The integral from 0 to 2pi of e^{-2ix} is 0