Mathematics - Definite Integration Question with Solution | TestHub

$4\alpha \left\{{\int }_{-1}^0{e}^{\alpha x}dx+{\int }_0^2{e}^{-\alpha x}dx\right\}=5$

$\Rightarrow 4\alpha \left\{{\left(\frac{e^{\alpha x}}{\alpha }\right)}_{-1}^0+\frac{e^{-\alpha x}}{-\alpha }{\left.\right|}_0^2\right\}=5$

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

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

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

$\Rightarrow 4t^2+4t-3=0\Rightarrow \left(2t+3\right)\left(2t-1\right)=0$

$\Rightarrow e^{-\alpha }=\frac{1}{2}\Rightarrow \alpha =\ln 2$