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

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

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integral from 0 to 2pi of e^{iθ}

\int_{0}^{2 π} e^{i θ} d θ

0

Solution steps

\int_{0}^{2 π} e^{i θ} d θ

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

\int_{0}^{\infty} 4 x e^{- 4 x} d x

x y \cdot d x = (x^{2} + 4) d y

cos (sin (7 x))

h \to 0 lim (\frac{∣ h ∣}{h})

4 x

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