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受欢迎的微积分 >

integral of (2w^2-w)e^{7w-1}

解答

\int (2 w^{2} - w) e^{7 w - 1} d w

解答

\frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1}) + C

求解步骤

\int (2 w^{2} - w) e^{7 w - 1} d w

使用换元积分法

= \int \frac{e ^{u} ( 2 u ^{2} - 3 u - 5 )}{343} d u

提出常数:

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

= \frac{1}{343} \cdot \int e^{u} (2 u^{2} - 3 u - 5) d u

乘开

e^{u} (2 u^{2} - 3 u - 5) : 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{u}

= \frac{1}{343} \cdot \int 2 e^{u} u^{2} - 3 e^{u} u - 5 e^{u} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{343} (\int 2 e^{u} u^{2} d u - \int 3 e^{u} u d u - \int 5 e^{u} d u)

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

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

\int 5 e^{u} d u = 5 e^{u}

= \frac{1}{343} (2 (u^{2} e^{u} - 2 (e^{u} u - e^{u})) - 3 (e^{u} u - e^{u}) - 5 e^{u})

u = 7 w - 1

代回

= \frac{1}{343} (2 ((7 w - 1)^{2} e^{7 w - 1} - 2 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1})) - 3 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1}) - 5 e^{7 w - 1})

化简

\frac{1}{343} (2 ((7 w - 1)^{2} e^{7 w - 1} - 2 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1})) - 3 (e^{7 w - 1} (7 w - 1) - e^{7 w - 1}) - 5 e^{7 w - 1}) : \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1})

= \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1})

解答补常数

= \frac{1}{343} (98 e^{7 w - 1} w^{2} - 77 e^{7 w - 1} w + 11 e^{7 w - 1}) + C

作图

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