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

解

\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 )}

解答ステップ

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

指数の規則を適用する:

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

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

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

u置換積分法を適用する

= \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

部分積分を適用する

= - [\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}

限度を計算する:

- \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 )})

簡素化

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