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

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

Popular >

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

\int_{0}^{2} sin (x) e^{- x} d x

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

Decimal

0.46662 \dots

Solution steps

\int_{0}^{2} sin (x) e^{- x} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int_{- 1}^{0} (4 x^{2} - x - \frac{1}{3}) d x

\int_{100}^{120} - 0.00006 x^{2} + 6 d x

\int_{1}^{8} ln (x) d x

\int_{0}^{π} (cos (x) + 4)^{2} d x

\int_{- 2}^{0} e^{- x} d x

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