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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= \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

\int_{0}^{4} (x + 1)^{3} d x

\int_{- 2}^{3} (5 + x - 6 x^{2}) d x

\int_{\frac{π}{4}}^{\frac{π}{2}} sin^{5} (x) d x

\int_{0}^{t} \frac{1}{x} d x

\int_{0}^{\frac{1}{2}} x arccos (x) d x

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