The value of

$4 \alpha\left\{\int_{-1}^{0} e^{\alpha x} d x+\int_{0}^{2} e^{-\alpha x} d x\right\}=5$

$\Rightarrow 4 \alpha\left\{\left.\frac{e^{\alpha x}}{\alpha}\right|_{-1} ^{0}+\left.\frac{e^{-\alpha x}}{-\alpha}\right|_{0} ^{2}\right\}=5$

$\Rightarrow 4 \alpha\left\{\left(\frac{1-e^{-\alpha}}{\alpha}\right)-\left(\frac{e^{-2 \alpha}-1}{\alpha}\right)\right\}=5$

$\Rightarrow 4\left(2-e^{-\alpha}-e^{-2 \alpha}\right)=5$

Put $e^{-\alpha}=t$

$\Rightarrow 4 \mathrm{t}^{2}+4 \mathrm{t}-3=0 \quad \Rightarrow \quad(2 \mathrm{t}+3)(2 \mathrm{t}-1)=0$

$\Rightarrow \quad e^{-\alpha}=\frac{1}{2} \Rightarrow \alpha=\log _{e} 2$