If the area of the bounded region R=(x,y): max 0, log_ex≤ y≤ 2^x, 1 2≤ x≤ 2 is, α ( log_e2)^-1+β ( log_e2)+γ then the value of (α +β -2γ )^2 is equal…

$$R={(x,y):\mathrm{max}(0,{\mathrm{log}}_{e}x)\leq y\leq {2}^{x},\frac{1}{2}\leq x\leq 2}$$R=(x,y):max(0,loge​x)≤y≤2x,21​≤x≤2

So, the area of the region is

$${\int }_{\frac{1}{2}}^{2}{2}^{x}dx-{\int }_{1}^{2}\ell nxdx$$∫21​2​2xdx−∫12​ℓnxdx

$$={[\frac{{2}^{x}}{\mathrm{ln}2}]}_{1/2}^{2}-{[x\mathrm{ln}x-x]}_{1}^{2}$$=[ln22x​]1/22​−[xlnx−x]12​

$$=\frac{({2}^{2})-{2}^{1/2}}{{\mathrm{log}}_{e}2}-(2\mathrm{ln}2-1)$$=loge​2(22)−21/2​−(2ln2−1)

$$=\frac{({2}^{2}-\sqrt{2})}{{\mathrm{log}}_{e}2}-2\mathrm{ln}2+1$$=loge​2(22−2​)​−2ln2+1

$$\therefore \alpha ={2}^{2}-\sqrt{2},\beta =-2,\gamma =1$$∴α=22−2​,β=−2,γ=1

So, ${(\alpha +\beta -2\gamma )}^{2}$(α+β−2γ)2

$$={({2}^{2}-\sqrt{2}-2-2)}^{2}$$=(22−2​−2−2)2

$$={(-\sqrt{2})}^{2}=2$$=(−2​)2=2