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

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

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integral from 0 to 2pi of e^{-2iθ}

\int_{0}^{2 π} e^{- 2 i θ} d θ

0

Solution steps

\int_{0}^{2 π} e^{- 2 i θ} d θ

Simplify

e^{- 2 i θ} : cos (2 θ) - i sin (2 θ)

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

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 θ) d θ - \int_{0}^{2 π} i sin (2 θ) d θ

\int_{0}^{2 π} cos (2 θ) d θ = 0

\int_{0}^{2 π} i sin (2 θ) d θ = 0

= 0 - 0

Simplify

= 0

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