# Question 084f4

Nov 7, 2016

Definitivelly ${x}^{8} < {e}^{\sqrt{x}}$ by far

#### Explanation:

${\int}_{0}^{\infty} {x}^{8} {e}^{- \sqrt{x}} \mathrm{dx} = 2 {\int}_{0}^{\infty} {y}^{17} {e}^{- y} \mathrm{dy}$

$\exists {y}_{0} : \setminus {y}^{17} < {e}^{\frac{y}{2}} , \forall y > {y}_{0}$

${\int}_{0}^{\infty} {x}^{8} {e}^{- \sqrt{x}} \mathrm{dx} \le 2 {\int}_{0}^{{y}_{0}} {y}^{17} {e}^{- y} \mathrm{dy} + 2 {\int}_{{y}_{0}}^{\infty} {e}^{- \frac{y}{2}} \mathrm{dx} =$
$= 2 {\int}_{0}^{{y}_{0}} {y}^{17} {e}^{- y} \mathrm{dy} - 4 {\left[{e}^{- \frac{y}{2}}\right]}_{{y}_{0}}^{\infty} = 2 {\int}_{0}^{{y}_{0}} {y}^{17} {e}^{- y} \mathrm{dy} + 4 {e}^{- {y}_{0} / 2}$

$\int {y}^{n} {e}^{- y} \mathrm{dy}$ can be calculated by parts

inty^n e^(-y)dy=-e^(-y)*(y^n+ny^(n-1)+...+n!)+c#

Nov 8, 2016

Given integral $= {\int}_{0}^{\infty} {y}^{17} {e}^{-} y \mathrm{dy}$
$f \left(y\right) = {y}^{17} {e}^{-} y = {e}^{17 \ln y - y} \setminus \setminus$
$g \left(y\right) = \left(1 - \frac{17}{y}\right) \cdot f \left(y\right)$
$c = 1$

#### Explanation:

${\lim}_{y \to + \infty} f \frac{x}{g} \left(x\right) = 1$

${\int}_{0}^{\infty} g \left(y\right) \mathrm{dy} = - {\int}_{0}^{\infty} \left(\frac{17}{y} - 1\right) {e}^{17 \ln y - y} =$
$= - {\left[{e}^{17 \ln y - y}\right]}_{0}^{\infty} = 0$