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

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

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

\int_{0}^{2 π} e^{2 i x} d x

0

Solution steps

\int_{0}^{2 π} e^{2 i x} d x

Apply imaginary number rule:

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

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

\int_{0}^{2 π} cos (2 x) d x = 0

\int_{0}^{2 π} i sin (2 x) d x = 0

= 0 + 0

Simplify

= 0

f (x) = \frac{5}{t ^{2}}

x \to 4 lim (\frac{13 x - 52}{2 x - 8})

\int \frac{x cos ( x ) - sin ( x )}{x ^{2}} d x

x \to 5 lim (3 x^{2} - 6 x + 8)

\frac{1}{x ^{2}} (1.1)

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