What is the integral from 0 to 5 of (xe^{-x}) ?

The integral from 0 to 5 of (xe^{-x}) is -6/(e^5)+1

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integral from 0 to 5 of (xe^{-x})

\int_{0}^{5} (x e^{- x}) d x

- \frac{6}{e ^{5}} + 1

Decimal Notation

0.95957 \dots

Solution steps

\int_{0}^{5} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{5}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{5}

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

= [- e^{- x} x - e^{- x}]_{0}^{5}

Compute the boundaries:

- \frac{6}{e ^{5}} + 1

= - \frac{6}{e ^{5}} + 1

\frac{d}{d x} (2 + x x)

y = (x^{5} + 1)^{7}

\frac{d y}{d x} - y = 2 e^{x} y^{2}

\int t \cdot sin (t) d t

\frac{1}{( s + 1 ) ^{4}}

What is the integral from 0 to 5 of (xe^{-x}) ?
The integral from 0 to 5 of (xe^{-x}) is -6/(e^5)+1