# Calculus 2. Is this the right answer?

## The question is: Find the area of the region bounded by $y = x {e}^{- {x}^{2} / 2}$ and its asymptote. For my final answer, I got an area equal to 1 and I was hoping if someone could tell me if this is correct. Thanks in advance!

Mar 15, 2017

${\int}_{0}^{\infty} \left(x {e}^{- {x}^{2} / 2}\right) \mathrm{dx} = 1$

The area is 1.

#### Explanation:

$y = x {e}^{- {x}^{2} / 2}$
graph{xe^(-x^2/2) [-6.244, 6.243, -3.12, 3.123]}

Since the problem is referring to the curve bounded within $\text{quadrant I}$, we are solving for:
${\int}_{0}^{\infty} \left(x {e}^{- {x}^{2} / 2}\right) \mathrm{dx}$

(Fix notation in advance by writing as a limit of a definite integral bounded by $0$ and $Q$)
$= {\lim}_{Q \to \infty} \left[{\int}_{0}^{Q} \left(x {e}^{- {x}^{2} / 2}\right) \mathrm{dx}\right]$

$= {\lim}_{Q \to \infty} {\left[- {e}^{- {x}^{2} / 2}\right]}_{0}^{Q}$

$= {\lim}_{Q \to \infty} \left[\left(- {e}^{- {Q}^{2} / 2}\right) - \left(- {e}^{- {0}^{2} / 2}\right)\right]$

$= {\lim}_{Q \to \infty} \left[- {e}^{- {Q}^{2} / 2} + 1\right]$

$= {\lim}_{Q \to \infty} \left[\frac{- 1}{{e}^{\left({Q}^{2} / 2\right)}} + 1\right]$

$= 1$