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

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

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

\int_{0}^{2 π} e^{i t} d t

0

Solution steps

\int_{0}^{2 π} e^{i t} d t

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

x \to 0 lim (\frac{x _{- 1}}{x})

6 sin^{2} (t)

f (x) = sin (x) tan (x)

\int sin^{4} (\frac{1}{2}) x d x

\int (3 x + 15) x^{2} + 10 x + 4 d x

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