What is the integral from 0 to 4 of xe^{-x} ?

The integral from 0 to 4 of xe^{-x} is -5/(e^4)+1

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integral from 0 to 4 of xe^{-x}

\int_{0}^{4} x e^{- x} d x

- \frac{5}{e ^{4}} + 1

Decimal Notation

0.90842 \dots

Solution steps

\int_{0}^{4} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{4}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{4}

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

= [- e^{- x} x - e^{- x}]_{0}^{4}

Compute the boundaries:

- \frac{5}{e ^{4}} + 1

= - \frac{5}{e ^{4}} + 1

y = x^{3}, y = 8, x = - 1

x \to 0 lim (2^{x})

2 x^{3} ln (10 x)

y^{^{'''}} + y^{^{''}} + 24 y^{^{'}} - 26 y = 0

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

What is the integral from 0 to 4 of xe^{-x} ?
The integral from 0 to 4 of xe^{-x} is -5/(e^4)+1