Question

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

Answer 1

Hello !

The standard method to integrate `t\mapsto t^n e^{at}` is the integration by parts :

`\int_a^b u(t)v'(t)dt = [u(t)v(t)]_a^b - \int_a^b u'(t)v(t) dt`

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

Here, consider `I_n = \int_0^x t^{n} e^{-t}dt`, `u(t) = t^n` and `v(t) = -e^{-t}`. You have `u'(t) = nt^{n-1}` and `v'(t) = e^{-t}`, therefore :

`\int_0^x t^n e^{-t} dt = \int u(t)v'(t)dt = [-t^n e^{-t}]_0^x - \int_0^x nt^{n-1} (-e^{-t}) dt`

or, easier, `I_n = -x^n e^{-x} + nI_{n-1}`.

So, you can find `I_5` step by step :

`I_5 = -x^5e^{-x} + 5I_4`
`I_4 = -x^4e^{-x} + 4I_3`
`I_3 = -x^3e^{-x} + 3I_2`
`I_2 = -x^2e^{-x} + 2I_1`
`I_1 = -xe^{-x} + I_0`
`I_0 = \int_0^x e^{-t}dt = -e^{-x} + 1`

therefore,

`I_1 = -x e^{-x} - e^{-x} + 1 = -(x+1)e^{-x} + 1`
`I_2 = -x^2 e^{-x} + 2I_1= -(x^2+2x+2)e^{-x}+2`
`I_3 = -x^3e^{-x} + 3I_2 = -(x^3 +3x^2+6x+6)e^{-x}+6`
`I_4 = -x^4e^{-x} + 4I_3 = -(x^4+4x^3+12x^2+24x+24)e^{-x} + 24`

and finally,

`I_5 =-(x^5+5x^4+20x^3+60x^2+120x+120)e^{-x}+120`

Answer 2

The answer is:

`I=-e^-t(t^5+5t^4+20t^3+60t^2+120t+120)+c`

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

`intf(t)g'(t)dt=f(t)g(t)-intg(t)f'(t)dt`

We can assume that `f(t)` is, step by step, the polynomial function and `g'(t)` is the exponential function

First step:

`f(t)=t^5` and `g'(t)=e^-x`

so:

`f'(t)=5t^4` and `g'(t)=-e^-x`.

`I=intt^5e^-tdt=-t^5e^-t-int5t^4(-e^-t)dt=`

`=-t^5e^-t+5intt^4e^-tdt`.

Second step:

`f(t)=t^4` and `g'(t)=e^-x`

so:

`f'(t)=4t^3` and `g'(t)=-e^-x`.

`I=-t^5e^-t+5[-t^4e^-t-int4t^3(-e^-t)dt]=`

`=-t^5e^-t-5t^4e^-t+20intt^3e^-tdt`.

Third step:

`f(t)=t^3` and `g'(t)=e^-x`

so:

`f'(t)=3t^2` and `g'(t)=-e^-x`.

`I=-t^5e^-t-5t^4e^-t+20[-t^3e^-t-int3t^2(-e^-t)dt]=`

`=-t^5e^-t-5t^4e^-t-20t^3e^-t+60intt^2e^-tdt`.

Fourth step:

`f(t)=t^2` and `g'(t)=e^-x`

so:

`f'(t)=2t` and `g'(t)=-e^-x`.

`I=-t^5e^-t-5t^4e^-t-20t^3e^-t+60[-t^2e^-t-int2t(-e^-t)dt]=`

`=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t+120intte^-tdt`.

Fifth step:

`f(t)=t` and `g'(t)=e^-x`

so:

`f'(t)=1` and `g'(t)=-e^-x`.

`I=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t+120[-te^-t-int1(-e^-t)dt]=`

`=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t-120te^t+120inte^-tdt=`

`=-t^5e^-t-5t^4e^-t-20t^3e^-t-60t^2e^-t-120te^t-120e^-t+c=`

`=-e^-t(t^5+5t^4+20t^3+60t^2+120t+120)+c=`