If the area of the bounded region R = (x,y) : 0,_ex y 2^x, 12 x2 is (_e2)^-1+(_e2)+, then …

Question

If the area of the bounded region $R = \left\{(x,y) : \max\{0,\log_ex\} \le y \le 2^x,\ \dfrac12\le x\le2\right\}$ is $\alpha(\log_e2)^{-1}+\beta(\log_e2)+\gamma$, then the value of $(\alpha+\beta-2\gamma)^2$ is equal to:

Accepted Answer

$$\text{Area} = \int_{1/2}^{2}2^x\,dx - \int_1^2\ln x\,dx = \left[\frac{2^x}{\ln2}\right]_{1/2}^2 - [x\ln x - x]_1^2$$

$$= \frac{4-\sqrt2}{\ln2} - (2\ln2-1)$$

So $\alpha=4-\sqrt2$, $\beta=-2$, $\gamma=1$, and $(\alpha+\beta+2\gamma)^2$ --- using the source's grouping --- $=\left(4-\sqrt2-2-2\right)^2=(-\sqrt2)^2=2$.

Solutions · 1