What is the integral from 1 to 5 of M/(e^M) ?

The integral from 1 to 5 of M/(e^M) is 1/(eln(e))+1/(eln^2(e))-5/(e^5ln(e))-1/(e^5ln^2(e))

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

\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 )}

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

= \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

Apply Integration By Parts

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

= - [\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}

Compute the boundaries:

- \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 )})

Simplify

= \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

\frac{d y}{d t} + e^{t} y = - 20 e^{t}

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

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

2 x, \frac{x ^{2}}{4}

What is the integral from 1 to 5 of M/(e^M) ?
The integral from 1 to 5 of M/(e^M) is 1/(eln(e))+1/(eln^2(e))-5/(e^5ln(e))-1/(e^5ln^2(e))