What is the integral from 0 to 3 of 1/96 y^3e^{-y/2} ?

The integral from 0 to 3 of 1/96 y^3e^{-y/2} is -(-16e^{3/2}+67)/(16e^{3/2)}

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integral from 0 to 3 of 1/96 y^3e^{-y/2}

\int_{0}^{3} \frac{1}{96} y^{3} e^{- \frac{y}{2}} d y

- \frac{- 16 e ^{\frac{3}{2}} + 67}{16 e ^{\frac{3}{2}}}

Decimal

0.06564 \dots

Solution steps

\int_{0}^{3} \frac{1}{96} y^{3} e^{- \frac{y}{2}} d y

Take the constant out:

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

= \frac{1}{96} \cdot \int_{0}^{3} y^{3} e^{- \frac{y}{2}} d y

Apply u-substitution

: \int_{0}^{- \frac{3}{2}} 16 e^{u} u^{3} d u

= \frac{1}{96} \cdot \int_{0}^{- \frac{3}{2}} 16 e^{u} u^{3} d u

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

= \frac{1}{96} (- \int_{- \frac{3}{2}}^{0} 16 e^{u} u^{3} d u)

Take the constant out:

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

= \frac{1}{96} (- 16 \cdot \int_{- \frac{3}{2}}^{0} e^{u} u^{3} d u)

Apply Integration By Parts

: [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- \frac{3}{2}}^{0}

= \frac{1}{96} (- 16 [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- \frac{3}{2}}^{0})

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= \frac{1}{96} (- 16 [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- \frac{3}{2}}^{0})

Simplify

\frac{1}{96} (- 16 [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- \frac{3}{2}}^{0}) : - \frac{1}{6} [e^{u} u^{3} - 3 (e^{u} u^{2} - 2 (e^{u} u - e^{u}))]_{- \frac{3}{2}}^{0}

= - \frac{1}{6} [e^{u} u^{3} - 3 (e^{u} u^{2} - 2 (e^{u} u - e^{u}))]_{- \frac{3}{2}}^{0}

Compute the boundaries:

- 6 + \frac{201}{8 e ^{\frac{3}{2}}}

= - \frac{1}{6} (- 6 + \frac{201}{8 e ^{\frac{3}{2}}})

Simplify

= - \frac{- 16 e ^{\frac{3}{2}} + 67}{16 e ^{\frac{3}{2}}}

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What is the integral from 0 to 3 of 1/96 y^3e^{-y/2} ?
The integral from 0 to 3 of 1/96 y^3e^{-y/2} is -(-16e^{3/2}+67)/(16e^{3/2)}