If the area of the bounded region [R = (x,y): 0, _ex y 2^x,1 2 x 2 ] is , [ ( _e2)^ - 1 + …

Question

If the area of the bounded region
$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\}  \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$ is ,
$$\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $$, then the value of $${(\alpha  + \beta  - 2\lambda )^2}$$ is equal to :

Accepted Answer

$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\}  \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$

$$\int\limits_{{1 \over 2}}^2 {{2^x}dx}  - \int\limits_1^2 {\ln xdx} $$

$$ \Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2$$

$$ \Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)$$

$$ \Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1$$

$$\therefore$$ $$\alpha  = {2^2} - \sqrt 2 $$, $$\beta  =  - 2$$, $$\gamma  = 1$$

$$ \therefore {(\alpha  + \beta  + 2\gamma )^2}$$

$$ = {({2^2} - \sqrt 2  - 2 - 2)^2}$$

$$ = {(\sqrt 2 )^2} = 2$$

Solutions · 1