Question

How do you evaluate the definite integral `int_0^(oo)x^3e^(-x)dx`?

Answer 1

first, integrate the indefinite integral `intx^3e^-xdx`

with some integration by parts, you should get `-(x^3+3x^2+6x+6)/e^x+C` (althought the C doesnt matter here)

the answer is `-(x^3+3x^2+6x+6)/e^x` as `xrarroo` subtracted by `-(x^3+3x^2+6x+6)/e^x` at `x=0`

since `e^x` increases much faster than the polynomial in the numerator, `-(x^3+3x^2+6x+6)/e^x` as `xrarroo` is `0`.
`-(x^3+3x^2+6x+6)/e^x` at `x=0` is `-6` from direct substitution.

so the answer is `0-(-6)=6`

Answer 2

See below.

Explanation:

Filling in some of the extra steps...

You can use integration by parts.

`uv-intvdu`

Let:

`u=x^3 " " du=3x^2dx`

`dv=e^(-x)" " v=-e^(-x)`

Substituting into the above expression:

`=>-x^3e^(-x)+3intx^2e^(-x)dx`

Apply again:

`u=x^2 " " du=2xdx`

`dv=e^(-x)" " v=-e^(-x)`

`=>-x^3e^(-x)+3[-x^2e^(-x)+2intxe^(-x)dx]`

Finally:

`u=x " " du=dx`

`dv=e^(-x)" " v=-e^(-x)`

`=>-x^3e^(-x)+3[-x^2e^(-x)+2{-xe^(-x)+inte^(-x)dx}]`

`-x^3-3x^2-6x-6`

Simplify:

`-e^(-x)(x^3+3x^2+6x+6)`

Evaluate from `0->oo`.

Note that `e^(-x)=1/e^x`. When we plug in "`oo`", think of this as one divided by a very big number (to say the least). This is approximately zero.

`=>0-[-1/e^0((0)^3+3(0^2)+6(0)+6)]`

Note that `e^0=1`.

`=>1(6)=6`