integral from 0 to 1 of e^{-st}(t)

解答

\int_{0}^{1} e^{- s t} (t) d t

\frac{- s e ^{- s} - e ^{- s} + 1}{s ^{2}}

求解步骤

\int_{0}^{1} e^{- s t} t d t

使用换元积分法

= \int_{0}^{- s} \frac{e ^{u} u}{s ^{2}} d u

提出常数:

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

= \frac{1}{s ^{2}} \cdot \int_{0}^{- s} e^{u} u d u

使用分布积分法

= \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- s}

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

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s}

化简

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

带入上下限计算后得:

- s e^{- s} - e^{- s} + 1

= \frac{- s e ^{- s} - e ^{- s} + 1}{s ^{2}}

49 y^{^{''}} - 70 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 3

k = 1 \sum \infty (sin (100))^{k}

\int ln (x) \cdot x d x

x \to 0 lim (\frac{e ^{x} - 1}{19 x})

x \to - 2 lim (3 x)