Question

Find $\alpha >0$ for which the inequality${\int }_{-\alpha }^{\alpha }{e}^x$dx > $\frac{3}{2}$ holds true.

Answer 1

${\int }_{-\alpha }^{\alpha }{e}^{x}dx=e^x{|}_{-\alpha }^{\alpha }$

$=e^{\alpha }-\left(e^{-\alpha }\right)$

$=\frac{e^{2\alpha }-1}{e^{\alpha }}>\frac{3}{2}$

$\Rightarrow 2e^{2\alpha }-2>3e^{\alpha }$

$\Rightarrow 2e^{2\alpha }-3e^{\alpha }-2>0$

$\Rightarrow 2e^{2\alpha }-4e^{\alpha }+e^{\alpha }-2>0$

$\Rightarrow 2e^{\alpha }[e^{\alpha }-2]+1[e^{\alpha }-2]>0$

$\Rightarrow \underset{\text{This will always be positive}}{\underset{}{(2e^{\alpha }+1)}}(e^{\alpha }-2)>0$

$\therefore e^{\alpha }-2>0\Rightarrow e^{\alpha }>2$

$\Rightarrow \alpha >\ln 2$.