What is the integral from 0 to 12 of xe^{-x} ?

The integral from 0 to 12 of xe^{-x} is -(13)/(e^{12)}+1

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integral from 0 to 12 of xe^{-x}

\int_{0}^{12} x e^{- x} d x

- \frac{13}{e ^{12}} + 1

Decimal Notation

0.99992 \dots

Solution steps

\int_{0}^{12} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{12}

= [- e^{- x} x - \int - e^{- x} d x]_{0}^{12}

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

= [- e^{- x} x - e^{- x}]_{0}^{12}

Compute the boundaries:

- \frac{13}{e ^{12}} + 1

= - \frac{13}{e ^{12}} + 1

y = e^{3 x} cos (2 x)

\frac{7}{x ^{2} + 3}

\int_{- 1}^{1} e^{y} - y^{2} - 6 d y

\int cos^{4} (u) d u

\int sin^{5} (t) d t

What is the integral from 0 to 12 of xe^{-x} ?
The integral from 0 to 12 of xe^{-x} is -(13)/(e^{12)}+1