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

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

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

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

- \frac{1}{e}

Decimal

- 0.36787 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

Expand

- e^{- x} (x - 1) - e^{- x} : - e^{- x} x

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

Compute the boundaries:

- \frac{1}{e}

= - \frac{1}{e}

\int_{0}^{2} 2 - x^{2} d x

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

\int_{1}^{2} \frac{e ^{\frac{1}{x}}}{x} d x

\int_{2}^{4} e^{x^{2}} d x

\int_{0}^{1} \frac{1}{( x + 1 ) ^{5}} d x

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