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

The integral from 0 to pi of e^{-2x}sin(x) is 1/(5e^{2pi)}+1/5

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integral from 0 to pi of e^{-2x}sin(x)

\int_{0}^{π} e^{- 2 x} sin (x) d x

\frac{1}{5 e ^{2 π}} + \frac{1}{5}

Decimal Notation

0.20037 \dots

Solution steps

\int_{0}^{π} e^{- 2 x} sin (x) d x

Compute the indefinite integral:

\int e^{- 2 x} sin (x) d x = - \frac{e ^{- 2 x} cos ( x )}{5} - \frac{2 e ^{- 2 x} sin ( x )}{5} + C

Compute the boundaries:

\int_{0}^{π} e^{- 2 x} sin (x) d x = \frac{1}{5 e ^{2 π}} - (- \frac{1}{5})

= \frac{1}{5 e ^{2 π}} - (- \frac{1}{5})

Simplify

= \frac{1}{5 e ^{2 π}} + \frac{1}{5}

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What is the integral from 0 to pi of e^{-2x}sin(x) ?
The integral from 0 to pi of e^{-2x}sin(x) is 1/(5e^{2pi)}+1/5