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".