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Popular >

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

Solution

\int_{0}^{2 π} e^{- i x} d x

Solution

0

Solution steps

\int_{0}^{2 π} e^{- i x} d x

Simplify

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

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

\int_{0}^{2 π} cos (x) d x = 0

\int_{0}^{2 π} i sin (x) d x = 0

= 0 - 0

Simplify

= 0

Graph

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\int_{- 2}^{2} e^{x + x^{2}} d x

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\int_{2}^{- 2} (- 2 x^{2} + 8) d x

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\int_{0}^{1} (2 t^{3} - \frac{4}{3} t)^{2} d t

Frequently Asked Questions (FAQ)

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