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

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

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

\int_{0}^{2 π} e^{i x} d x

0

Solution steps

\int_{0}^{2 π} e^{i x} d x

Apply u-substitution

: \int_{0}^{i 2 π} - i e^{u} d u

= \int_{0}^{i 2 π} - i e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - i \cdot \int_{0}^{i 2 π} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - i [e^{u}]_{0}^{i 2 π}

Compute the boundaries:

0

= - i 0

Simplify

= 0

\frac{2 x}{x + 1}, at x = 1

\frac{1}{s ^{2} ( s ^{2} + 2 s + 3 )}

\int x^{6} e^{- 4 x} d x

\int_{0}^{1} \frac{6}{x ^{2} + 1} d x

\int \frac{1}{8 x ^{5}} d x

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