What is the integral from-1 to 0 of e^{x+ix+i} ?

The integral from-1 to 0 of e^{x+ix+i} is (-1+e(cos(1)+sin(1)))/(2e)+i(1+e(-cos(1)+sin(1)))/(2e)

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integral from-1 to 0 of e^{x+ix+i}

\int_{- 1}^{0} e^{x + i x + i} d x

\frac{- 1 + e ( cos ( 1 ) + sin ( 1 ) )}{2 e} + i \frac{1 + e ( - cos ( 1 ) + sin ( 1 ) )}{2 e}

Solution steps

\int_{- 1}^{0} e^{x + i x + i} d x

Compute the indefinite integral:

\int e^{x + i x + i} d x = \frac{1}{2} (e^{x} cos (1 + x) + e^{x} sin (1 + x)) + i \frac{1}{2} (- e^{x} cos (1 + x) + e^{x} sin (1 + x)) + C

Compute the boundaries:

\int_{- 1}^{0} e^{x + i x + i} d x = \frac{cos ( 1 ) + sin ( 1 )}{2} + i \frac{- cos ( 1 ) + sin ( 1 )}{2} - (\frac{1}{2 e} - i \frac{1}{2 e})

= \frac{cos ( 1 ) + sin ( 1 )}{2} + i \frac{- cos ( 1 ) + sin ( 1 )}{2} - (\frac{1}{2 e} - i \frac{1}{2 e})

Simplify

= \frac{- 1 + e ( cos ( 1 ) + sin ( 1 ) )}{2 e} + i \frac{1 + e ( - cos ( 1 ) + sin ( 1 ) )}{2 e}

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What is the integral from-1 to 0 of e^{x+ix+i} ?
The integral from-1 to 0 of e^{x+ix+i} is (-1+e(cos(1)+sin(1)))/(2e)+i(1+e(-cos(1)+sin(1)))/(2e)