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

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

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integral from 0 to 1 of (xe^{-x})

\int_{0}^{1} (x e^{- x}) d x

- \frac{2}{e} + 1

Decimal

0.26424 \dots

Solution steps

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

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{1}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{1}

\int - e^{- x} d x = e^{- x}

= [- e^{- x} x - e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{2}{e} + 1

= - \frac{2}{e} + 1

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

\int_{0}^{2} 5^{2 x - 1} - 2^{3 x} d x

\int_{- π}^{π} t^{2} d t

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

\int_{0}^{1} e^{y^{2}} y d y

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