integral from 1 to 2 of 12(y-1)x^3e^{2x(y-1)}

解答

\int_{1}^{2} 12 (y - 1) x^{3} e^{2 x (y - 1)} d y

3 x (2 x e^{2 x} - e^{2 x} + 1)

求解步骤

\int_{1}^{2} 12 (y - 1) x^{3} e^{2 x (y - 1)} d y

提出常数:

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

= 12 x^{3} \cdot \int_{1}^{2} (- 1 + y) e^{(- 1 + y) \cdot 2 x} d y

使用换元积分法

= 12 x^{3} \cdot \int_{0}^{1} e^{2 xu} u d u

使用分布积分法

= 12 x^{3} [\frac{1}{2} e^{2 xu} u^{2} - \int x e^{2 xu} u^{2} d u]_{0}^{1}

\int x e^{2 xu} u^{2} d u = \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}}

= 12 x^{3} [\frac{1}{2} e^{2 xu} u^{2} - \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}}]_{0}^{1}

化简

\frac{1}{2} e^{2 xu} u^{2} - \frac{e ^{2 xu} ( 2 x ^{2} u ^{2} - 2 xu + 1 )}{4 x ^{2}} : \frac{2 x e ^{2 xu} u - e ^{2 xu}}{4 x ^{2}}

= 12 x^{3} [\frac{2 x e ^{2 xu} u - e ^{2 xu}}{4 x ^{2}}]_{0}^{1}

带入上下限计算后得:

\frac{2 x e ^{2 x} - e ^{2 x}}{4 x ^{2}} + \frac{1}{4 x ^{2}}

= 12 x^{3} (\frac{2 x e ^{2 x} - e ^{2 x}}{4 x ^{2}} + \frac{1}{4 x ^{2}})

化简

= 3 x (2 x e^{2 x} - e^{2 x} + 1)

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