What is the integral from 0 to 1 of (x^2-x)e^{-x} ?

The integral from 0 to 1 of (x^2-x)e^{-x} is -3/e+1

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integral from 0 to 1 of (x^2-x)e^{-x}

\int_{0}^{1} (x^{2} - x) e^{- x} d x

- \frac{3}{e} + 1

Decimal Notation

- 0.10363 \dots

Solution steps

\int_{0}^{1} (x^{2} - x) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} - x) - \int - e^{- x} (2 x - 1) d x]_{0}^{1}

= [- e^{- x} (x^{2} - x) - \int - e^{- x} (2 x - 1) d x]_{0}^{1}

\int - e^{- x} (2 x - 1) d x = 2 e^{- x} x + e^{- x}

= [- e^{- x} (x^{2} - x) - (2 e^{- x} x + e^{- x})]_{0}^{1}

Expand

- e^{- x} (x^{2} - x) - (2 e^{- x} x + e^{- x}) : - e^{- x} x^{2} - e^{- x} x - e^{- x}

= [- e^{- x} x^{2} - e^{- x} x - e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{3}{e} + 1

= - \frac{3}{e} + 1

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What is the integral from 0 to 1 of (x^2-x)e^{-x} ?
The integral from 0 to 1 of (x^2-x)e^{-x} is -3/e+1