What is the integral from 0 to 1 of 2x^3e^{-x^2} ?

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

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integral from 0 to 1 of 2x^3e^{-x^2}

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

1 - \frac{2}{e}

Decimal

0.26424 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int_{0}^{- 1} \frac{e ^{u} u}{2} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 1}^{0} \frac{e ^{u} u}{2} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 1}^{0}

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

\int e^{u} d u = e^{u}

= 2 (- \frac{1}{2} [e^{u} u - e^{u}]_{- 1}^{0})

Simplify

2 (- \frac{1}{2} [e^{u} u - e^{u}]_{- 1}^{0}) : - [e^{u} u - e^{u}]_{- 1}^{0}

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

Compute the boundaries:

- 1 + \frac{2}{e}

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

Simplify

= 1 - \frac{2}{e}

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