What is the integral from 0 to 16 of xe^{-x} ?

The integral from 0 to 16 of xe^{-x} is -(17)/(e^{16)}+1

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integral from 0 to 16 of xe^{-x}

\int_{0}^{16} x e^{- x} d x

- \frac{17}{e ^{16}} + 1

Decimal Notation

0.99999 \dots

Solution steps

\int_{0}^{16} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{16}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{16}

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

= [- e^{- x} x - e^{- x}]_{0}^{16}

Compute the boundaries:

- \frac{17}{e ^{16}} + 1

= - \frac{17}{e ^{16}} + 1

x \to 0 - lim (\frac{11}{tan ( x )})

\int 8 x ln (x) d x

\int \frac{( 2 x - 4 )}{( x - 2 ) ^{2}} d x

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

\frac{d y}{d x} = 3 y - x^{2} y

What is the integral from 0 to 16 of xe^{-x} ?
The integral from 0 to 16 of xe^{-x} is -(17)/(e^{16)}+1