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

Dec 16, 2017

first, integrate the indefinite integral $\int {x}^{3} {e}^{-} x \mathrm{dx}$

with some integration by parts, you should get $- \frac{{x}^{3} + 3 {x}^{2} + 6 x + 6}{e} ^ x + C$ (althought the C doesnt matter here)

the answer is $- \frac{{x}^{3} + 3 {x}^{2} + 6 x + 6}{e} ^ x$ as $x \rightarrow \infty$ subtracted by $- \frac{{x}^{3} + 3 {x}^{2} + 6 x + 6}{e} ^ x$ at $x = 0$

since ${e}^{x}$ increases much faster than the polynomial in the numerator, $- \frac{{x}^{3} + 3 {x}^{2} + 6 x + 6}{e} ^ x$ as $x \rightarrow \infty$ is $0$.
$- \frac{{x}^{3} + 3 {x}^{2} + 6 x + 6}{e} ^ x$ at $x = 0$ is $- 6$ from direct substitution.

so the answer is $0 - \left(- 6\right) = 6$

Dec 16, 2017

See below.

#### Explanation:

Filling in some of the extra steps...

You can use integration by parts.

$u v - \int v \mathrm{du}$

Let:

$u = {x}^{3} \text{ } \mathrm{du} = 3 {x}^{2} \mathrm{dx}$

$\mathrm{dv} = {e}^{- x} \text{ } v = - {e}^{- x}$

Substituting into the above expression:

$\implies - {x}^{3} {e}^{- x} + 3 \int {x}^{2} {e}^{- x} \mathrm{dx}$

Apply again:

$u = {x}^{2} \text{ } \mathrm{du} = 2 x \mathrm{dx}$

$\mathrm{dv} = {e}^{- x} \text{ } v = - {e}^{- x}$

$\implies - {x}^{3} {e}^{- x} + 3 \left[- {x}^{2} {e}^{- x} + 2 \int x {e}^{- x} \mathrm{dx}\right]$

Finally:

$u = x \text{ } \mathrm{du} = \mathrm{dx}$

$\mathrm{dv} = {e}^{- x} \text{ } v = - {e}^{- x}$

$\implies - {x}^{3} {e}^{- x} + 3 \left[- {x}^{2} {e}^{- x} + 2 \left\{- x {e}^{- x} + \int {e}^{- x} \mathrm{dx}\right\}\right]$

$- {x}^{3} - 3 {x}^{2} - 6 x - 6$

Simplify:

$- {e}^{- x} \left({x}^{3} + 3 {x}^{2} + 6 x + 6\right)$

Evaluate from $0 \to \infty$.

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

$\implies 0 - \left[- \frac{1}{e} ^ 0 \left({\left(0\right)}^{3} + 3 \left({0}^{2}\right) + 6 \left(0\right) + 6\right)\right]$

Note that ${e}^{0} = 1$.

$\implies 1 \left(6\right) = 6$