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

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

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integral from 0 to pi of e^{it}

\int_{0}^{π} e^{i t} d t

2 i

Solution steps

\int_{0}^{π} e^{i t} d t

Apply u-substitution

: \int_{0}^{iπ} - i e^{u} d u

= \int_{0}^{iπ} - 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}^{iπ} e^{u} d u

Use the common integral:

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

= - i [e^{u}]_{0}^{iπ}

Compute the boundaries:

- 2

= - i (- 2)

Simplify

= 2 i

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