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人気のある微分積分 >

integral from 0 to 5 of 1/96 y^3e^{-y/2}

解

\int_{0}^{5} \frac{1}{96} y^{3} e^{- \frac{y}{2}} d y

解

- \frac{- 48 e ^{\frac{5}{2}} + 443}{48 e ^{\frac{5}{2}}}

+1

十進法表記

0.24242 \dots

解答ステップ

\int_{0}^{5} \frac{1}{96} y^{3} e^{- \frac{y}{2}} d y

定数を除く:

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

= \frac{1}{96} \cdot \int_{0}^{5} y^{3} e^{- \frac{y}{2}} d y

u置換積分法を適用する

= \frac{1}{96} \cdot \int_{0}^{- \frac{5}{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{5}{2}}^{0} 16 e^{u} u^{3} d u)

定数を除く:

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

= \frac{1}{96} (- 16 \cdot \int_{- \frac{5}{2}}^{0} e^{u} u^{3} d u)

部分積分を適用する

= \frac{1}{96} (- 16 [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- \frac{5}{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{5}{2}}^{0})

簡素化

\frac{1}{96} (- 16 [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- \frac{5}{2}}^{0}) : - \frac{1}{6} [e^{u} u^{3} - 3 (e^{u} u^{2} - 2 (e^{u} u - e^{u}))]_{- \frac{5}{2}}^{0}

= - \frac{1}{6} [e^{u} u^{3} - 3 (e^{u} u^{2} - 2 (e^{u} u - e^{u}))]_{- \frac{5}{2}}^{0}

限度を計算する:

- 6 + \frac{443}{8 e ^{\frac{5}{2}}}

= - \frac{1}{6} (- 6 + \frac{443}{8 e ^{\frac{5}{2}}})

簡素化

= - \frac{- 48 e ^{\frac{5}{2}} + 443}{48 e ^{\frac{5}{2}}}

グラフ

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