# How do you integrate t^5e^(-t)dt?

Feb 15, 2015

Hello !

The standard method to integrate $t \setminus \mapsto {t}^{n} {e}^{a t}$ is the integration by parts :

$\setminus {\int}_{a}^{b} u \left(t\right) v ' \left(t\right) \mathrm{dt} = {\left[u \left(t\right) v \left(t\right)\right]}_{a}^{b} - \setminus {\int}_{a}^{b} u ' \left(t\right) v \left(t\right) \mathrm{dt}$

This formula is proved if you remark that $\left(u v\right) ' = u ' v + u v '$ and if you integrate that on $\left[a , b\right]$.

Here, consider ${I}_{n} = \setminus {\int}_{0}^{x} {t}^{n} {e}^{- t} \mathrm{dt}$, $u \left(t\right) = {t}^{n}$ and $v \left(t\right) = - {e}^{- t}$. You have $u ' \left(t\right) = n {t}^{n - 1}$ and $v ' \left(t\right) = {e}^{- t}$, therefore :

$\setminus {\int}_{0}^{x} {t}^{n} {e}^{- t} \mathrm{dt} = \setminus \int u \left(t\right) v ' \left(t\right) \mathrm{dt} = {\left[- {t}^{n} {e}^{- t}\right]}_{0}^{x} - \setminus {\int}_{0}^{x} n {t}^{n - 1} \left(- {e}^{- t}\right) \mathrm{dt}$

or, easier, ${I}_{n} = - {x}^{n} {e}^{- x} + n {I}_{n - 1}$.

So, you can find ${I}_{5}$ step by step :

${I}_{5} = - {x}^{5} {e}^{- x} + 5 {I}_{4}$
${I}_{4} = - {x}^{4} {e}^{- x} + 4 {I}_{3}$
${I}_{3} = - {x}^{3} {e}^{- x} + 3 {I}_{2}$
${I}_{2} = - {x}^{2} {e}^{- x} + 2 {I}_{1}$
${I}_{1} = - x {e}^{- x} + {I}_{0}$
${I}_{0} = \setminus {\int}_{0}^{x} {e}^{- t} \mathrm{dt} = - {e}^{- x} + 1$

therefore,

${I}_{1} = - x {e}^{- x} - {e}^{- x} + 1 = - \left(x + 1\right) {e}^{- x} + 1$
${I}_{2} = - {x}^{2} {e}^{- x} + 2 {I}_{1} = - \left({x}^{2} + 2 x + 2\right) {e}^{- x} + 2$
${I}_{3} = - {x}^{3} {e}^{- x} + 3 {I}_{2} = - \left({x}^{3} + 3 {x}^{2} + 6 x + 6\right) {e}^{- x} + 6$
${I}_{4} = - {x}^{4} {e}^{- x} + 4 {I}_{3} = - \left({x}^{4} + 4 {x}^{3} + 12 {x}^{2} + 24 x + 24\right) {e}^{- x} + 24$

and finally,

${I}_{5} = - \left({x}^{5} + 5 {x}^{4} + 20 {x}^{3} + 60 {x}^{2} + 120 x + 120\right) {e}^{- x} + 120$

Feb 15, 2015

$I = - {e}^{-} t \left({t}^{5} + 5 {t}^{4} + 20 {t}^{3} + 60 {t}^{2} + 120 t + 120\right) + c$

We have to use for five times (!) the integration by parts, that says:

$\int f \left(t\right) g ' \left(t\right) \mathrm{dt} = f \left(t\right) g \left(t\right) - \int g \left(t\right) f ' \left(t\right) \mathrm{dt}$

We can assume that $f \left(t\right)$ is, step by step, the polynomial function and $g ' \left(t\right)$ is the exponential function

First step:

$f \left(t\right) = {t}^{5}$ and $g ' \left(t\right) = {e}^{-} x$

so:

$f ' \left(t\right) = 5 {t}^{4}$ and $g ' \left(t\right) = - {e}^{-} x$.

$I = \int {t}^{5} {e}^{-} t \mathrm{dt} = - {t}^{5} {e}^{-} t - \int 5 {t}^{4} \left(- {e}^{-} t\right) \mathrm{dt} =$

$= - {t}^{5} {e}^{-} t + 5 \int {t}^{4} {e}^{-} t \mathrm{dt}$.

Second step:

$f \left(t\right) = {t}^{4}$ and $g ' \left(t\right) = {e}^{-} x$

so:

$f ' \left(t\right) = 4 {t}^{3}$ and $g ' \left(t\right) = - {e}^{-} x$.

$I = - {t}^{5} {e}^{-} t + 5 \left[- {t}^{4} {e}^{-} t - \int 4 {t}^{3} \left(- {e}^{-} t\right) \mathrm{dt}\right] =$

$= - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t + 20 \int {t}^{3} {e}^{-} t \mathrm{dt}$.

Third step:

$f \left(t\right) = {t}^{3}$ and $g ' \left(t\right) = {e}^{-} x$

so:

$f ' \left(t\right) = 3 {t}^{2}$ and $g ' \left(t\right) = - {e}^{-} x$.

$I = - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t + 20 \left[- {t}^{3} {e}^{-} t - \int 3 {t}^{2} \left(- {e}^{-} t\right) \mathrm{dt}\right] =$

$= - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t + 60 \int {t}^{2} {e}^{-} t \mathrm{dt}$.

Fourth step:

$f \left(t\right) = {t}^{2}$ and $g ' \left(t\right) = {e}^{-} x$

so:

$f ' \left(t\right) = 2 t$ and $g ' \left(t\right) = - {e}^{-} x$.

$I = - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t + 60 \left[- {t}^{2} {e}^{-} t - \int 2 t \left(- {e}^{-} t\right) \mathrm{dt}\right] =$

$= - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t - 60 {t}^{2} {e}^{-} t + 120 \int t {e}^{-} t \mathrm{dt}$.

Fifth step:

$f \left(t\right) = t$ and $g ' \left(t\right) = {e}^{-} x$

so:

$f ' \left(t\right) = 1$ and $g ' \left(t\right) = - {e}^{-} x$.

$I = - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t - 60 {t}^{2} {e}^{-} t + 120 \left[- t {e}^{-} t - \int 1 \left(- {e}^{-} t\right) \mathrm{dt}\right] =$

$= - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t - 60 {t}^{2} {e}^{-} t - 120 t {e}^{t} + 120 \int {e}^{-} t \mathrm{dt} =$

$= - {t}^{5} {e}^{-} t - 5 {t}^{4} {e}^{-} t - 20 {t}^{3} {e}^{-} t - 60 {t}^{2} {e}^{-} t - 120 t {e}^{t} - 120 {e}^{-} t + c =$

$= - {e}^{-} t \left({t}^{5} + 5 {t}^{4} + 20 {t}^{3} + 60 {t}^{2} + 120 t + 120\right) + c =$