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受欢迎的微积分 >

integral of-(x^3)/(3e^{x/2)}

解答

\int - \frac{x ^{3}}{3 e ^{\frac{x}{2}}} d x

解答

- \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

求解步骤

\int - \frac{x ^{3}}{3 e ^{\frac{x}{2}}} d x

提出常数:

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

= - \frac{1}{3} \cdot \int \frac{x ^{3}}{e ^{\frac{x}{2}}} d x

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{3}}{e ^{\frac{x}{2}}} = (x^{3}) e^{- \frac{x}{2}}

= - \frac{1}{3} \cdot \int x^{3} e^{- \frac{x}{2}} d x

使用换元积分法

= - \frac{1}{3} \cdot \int 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}{3} \cdot 16 \cdot \int e^{u} u^{3} d u

使用分布积分法

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

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

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

u = - \frac{x}{2}

代回

= - \frac{1}{3} \cdot 16 ((- \frac{x}{2})^{3} e^{- \frac{x}{2}} - 3 ((- \frac{x}{2})^{2} e^{- \frac{x}{2}} - 2 (e^{- \frac{x}{2}} (- \frac{x}{2}) - e^{- \frac{x}{2}})))

化简

- \frac{1}{3} \cdot 16 ((- \frac{x}{2})^{3} e^{- \frac{x}{2}} - 3 ((- \frac{x}{2})^{2} e^{- \frac{x}{2}} - 2 (e^{- \frac{x}{2}} (- \frac{x}{2}) - e^{- \frac{x}{2}}))) : - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

= - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

解答补常数

= - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

作图

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