The value of $\alpha$ for which
$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$, is:

$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$ <br><br>$\Rightarrow$ $$4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $$ = 5 <br><br>$\Rightarrow$ $$4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$$ = 5 <br><br>$\Rightarrow$ $4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$ = 0 <br><br>Let e^-$\alpha$ = t <br><br>$\therefore$ 4t² + 4t - 3 = 0 <br><br>$\Rightarrow$ t = ${1 \over 2}$ <br><br>$\therefore$ e^-$\alpha$ = ${1 \over 2}$ <br><br>$\Rightarrow$ $\alpha$ = ${\log _e}2$