What is the integral from 0 to 2 of 1/(e^{pix)} ?

The integral from 0 to 2 of 1/(e^{pix)} is (-1+e^{2pi})/(pie^{2pi)}

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integral from 0 to 2 of 1/(e^{pix)}

\int_{0}^{2} \frac{1}{e ^{π x}} d x

\frac{- 1 + e ^{2 π}}{π e ^{2 π}}

Decimal Notation

0.31771 \dots

Solution steps

\int_{0}^{2} \frac{1}{e ^{π x}} d x

Apply u-substitution

: \int_{1}^{e^{π 2}} \frac{1}{π u ^{2}} d u

= \int_{1}^{e^{π 2}} \frac{1}{π u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π} \cdot \int_{1}^{e^{π 2}} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{e^{π 2}}

= \frac{1}{π} [- \frac{1}{u}]_{1}^{e^{π 2}}

Compute the boundaries:

- \frac{1}{e ^{2 π}} + 1

= \frac{1}{π} (- \frac{1}{e ^{2 π}} + 1)

Simplify

= \frac{- 1 + e ^{2 π}}{π e ^{2 π}}

\frac{d y}{d x} + 4 y = e^{- 4 x}

\frac{\partial}{\partial x} (3 x^{2} - 3)

\frac{d}{d x} (a^{2} arcsin (\frac{x}{a}))

\int \frac{x - 1}{x ^{2} ( x + 1 ) ^{2}} d x

y^{2} d y + x^{3} d x = 0

What is the integral from 0 to 2 of 1/(e^{pix)} ?
The integral from 0 to 2 of 1/(e^{pix)} is (-1+e^{2pi})/(pie^{2pi)}