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

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

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

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

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

Decimal

0.13212 \dots

Solution steps

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

Apply u-substitution

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

= \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

= - \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

= - \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}

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

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

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

Compute the boundaries:

- 1 + \frac{2}{e}

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

Simplify

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

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