integral from 0 to b of e^{ux}a(b-x)

解

\int_{0}^{b} e^{ux} a (b - x) d x

a (\frac{b ( e ^{b u} - 1 )}{u} - \frac{b e ^{b u}}{u} - \frac{- e ^{b u} + 1}{u ^{2}})

解答ステップ

\int_{0}^{b} e^{ux} a (b - x) d x

定数を除く:

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

= a \cdot \int_{0}^{b} (b - x) e^{ux} d x

拡張

(b - x) e^{ux} : b e^{ux} - e^{ux} x

= a \cdot \int_{0}^{b} b e^{ux} - e^{ux} x d x

総和規則を適用する:

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

= a (\int_{0}^{b} b e^{ux} d x - \int_{0}^{b} e^{ux} x d x)

\int_{0}^{b} b e^{ux} d x = \frac{b ( e ^{b u} - 1 )}{u}

\int_{0}^{b} e^{ux} x d x = \frac{b e ^{b u}}{u} + \frac{- e ^{b u} + 1}{u ^{2}}

= a (\frac{b ( e ^{b u} - 1 )}{u} - (\frac{b e ^{b u}}{u} + \frac{- e ^{b u} + 1}{u ^{2}}))

簡素化

= a (\frac{b ( e ^{b u} - 1 )}{u} - \frac{b e ^{b u}}{u} - \frac{- e ^{b u} + 1}{u ^{2}})