Question

8ntergrate by parts |x^3 e^-x dx​

Answer 1

I assume the integral is

`\displaystyle \int x^3 e^(-x) \, dx`

Integrate by parts with

`u = x^3 \implies du = 3x^2 \, dx`

`dv = e^(-x) \, dx \implies v = -e^(-x)`

so that

`\displaystyle \int u \, dv = uv - \int v \, du`

i.e.

`\displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) + 3 \int x^2 e^(-x) \, dx`

Integrate by parts again, this time with

`u = x^2 \implies du = 2x \, dx`

`dv = e^(-x) \, dx \implies v = -e^(-x)`

and so

`\displaystyle \int x^2 e^(-x) \, dx = -x^2 e^(-x) + 2 \int x e^(-x) \, dx`

`\implies \displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) + 3 \left(-x^2 e^(-x) + 2 \int x e^(-x) \, dx\right)`

`\implies \displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) - 3x^2 e^(-x) + 6 \int x e^(-x) \, dx`

Once more, with

`u = x \implies du = dx`

`dv = e^(-x) \, dx \implies v = -e^(-x)`

`\displaystyle \int x e^(-x) \, dx = -xe^(-x) + \int e^(-x) \, dx`

`\implies \displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) - 3x^2 e^(-x) + 6 \left(-xe^(-x) + \int e^(-x) \, dx\right)`

`\implies \displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) - 3x^2 e^(-x) -6xe^(-x) + 6 \int e^(-x) \, dx`

Finally,

`\displaystyle \int e^(-x) \, dx = -e^(-x) + C`

and so

`\displaystyle \int x^3 e^(-x) \, dx = -x^3 e^(-x) - 3x^2 e^(-x) -6xe^(-x) - 6 e^(-x) + C`

or equivalently,

`\displaystyle \int x^3 e^(-x) \, dx = \boxed{-(x^3+3x^2+6x+6) e^(-x) + C}`

We can also approach the integral more generally by considering

`I_n = \displaystyle \int x^n e^(-x) \, dx`

Integrate by parts with

`u = x^n \implies du = nx^(n-1) \, dx`

`dv = e^(-x) \, dx \implies v = -e^(-x)`

Then

`\displaystyle I_n = -x^n e^(-x) + n \int x^(n-1) e^(-x) \, dx = -x^n e^(-x) + n I_(n-1)`

We can solve the recurrence using the initial "value"
`I_0 = -e^(-x)+C` to find a general formula for
`I_n`. Then we get the integral we want
`I_3` "for free".