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

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

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

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

- \frac{1}{2 e}

Decimal

- 0.18393 \dots

Solution steps

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

Expand

x (x^{2} - 1) e^{- x^{2}} : e^{- x^{2}} x^{3} - e^{- x^{2}} x

= \int_{0}^{1} e^{- x^{2}} x^{3} - e^{- x^{2}} x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} e^{- x^{2}} x^{3} d x - \int_{0}^{1} e^{- x^{2}} x d x

\int_{0}^{1} e^{- x^{2}} x^{3} d x = - \frac{- e + 2}{2 e}

\int_{0}^{1} e^{- x^{2}} x d x = \frac{e - 1}{2 e}

= - \frac{- e + 2}{2 e} - \frac{e - 1}{2 e}

Simplify

- \frac{- e + 2}{2 e} - \frac{e - 1}{2 e} : - \frac{1}{2 e}

= - \frac{1}{2 e}

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x \to 0 + lim (ln (x^{x}))

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