What is the integral from 0 to 1 of e^{-x}sin(pix) ?

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

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integral from 0 to 1 of e^{-x}sin(pix)

\int_{0}^{1} e^{- x} sin (π x) d x

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

Decimal Notation

0.39535 \dots

Solution steps

\int_{0}^{1} e^{- x} sin (π x) d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

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