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受欢迎的微积分 >

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

解答

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

解答

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

+1

十进制

0.06564 \dots

求解步骤

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

提出常数:

\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

使用换元积分法

= \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)

提出常数:

\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)

使用分布积分法

= \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})

化简

\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}

带入上下限计算后得:

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

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

化简

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

作图

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