integral from 0 to N of e^{-st}e^{-5t}

解答

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

- \frac{e ^{- N (s + 5)} - 1}{s + 5}

求解步骤

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

使用换元积分法

= \int_{1}^{e^{- 5 N}} - \frac{u ^{\frac{s}{5}}}{5} d u

提出常数:

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

= - \frac{1}{5} \cdot \int_{1}^{e^{- 5 N}} u^{\frac{s}{5}} d u

使用幂法则

= - \frac{1}{5} [\frac{5 u ^{\frac{s + 5}{5}}}{s + 5}]_{1}^{e^{- 5 N}}

带入上下限计算后得:

\frac{5 e ^{- N (s + 5)}}{s + 5} - \frac{5}{s + 5}

= - \frac{1}{5} (\frac{5 e ^{- N (s + 5)}}{s + 5} - \frac{5}{s + 5})

化简

= - \frac{e ^{- N (s + 5)} - 1}{s + 5}

\frac{y d y + x d x}{2 1 + x y} = 0

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{6})

\frac{\partial}{\partial x} (4 y^{2} x^{3})

\int_{0}^{1} (x + 1) (x^{2} + 2 x)^{5} d x

\int \frac{- 1800 x}{25 - x ^{2}} d x