integral from 1 to 5 of M/(e^M)

解答

\int_{1}^{5} \frac{M}{e ^{M}} d M

\frac{1}{e ln ( e )} + \frac{1}{e ln ^{2} ( e )} - \frac{5}{e ^{5} ln ( e )} - \frac{1}{e ^{5} ln ^{2} ( e )}

求解步骤

\int_{1}^{5} \frac{M}{e ^{M}} d M

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{M}{e ^{M}} = (M) e^{- M}

= \int_{1}^{5} M e^{- M} d M

使用换元积分法

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

使用分布积分法

= - [\frac{e ^{u} u}{ln ( e )} - \int \frac{e ^{u}}{ln ( e )} d u]_{- 5}^{- 1}

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

= - [\frac{e ^{u} u}{ln ( e )} - \frac{e ^{u}}{ln ^{2} ( e )}]_{- 5}^{- 1}

带入上下限计算后得:

- \frac{1}{e ln ( e )} - \frac{1}{e ln ^{2} ( e )} + \frac{5}{e ^{5} ln ( e )} + \frac{1}{e ^{5} ln ^{2} ( e )}

= - (- \frac{1}{e ln ( e )} - \frac{1}{e ln ^{2} ( e )} + \frac{5}{e ^{5} ln ( e )} + \frac{1}{e ^{5} ln ^{2} ( e )})

化简

= \frac{1}{e ln ( e )} + \frac{1}{e ln ^{2} ( e )} - \frac{5}{e ^{5} ln ( e )} - \frac{1}{e ^{5} ln ^{2} ( e )}

\int \frac{1}{x ( a + b x )} d x

x \to 1 lim (\frac{x - 8}{3 x - 4})

\int (x^{2} - 5 x) e^{x} d x

5 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{\partial}{\partial x} (2 x^{2} yz)