Question

Find the area of the region bounded by the curves $x=\frac{1}{2},x=2,y=logx$ and $y=2^x$

• A. $\frac{4-\sqrt{2}}{\log 2}-\frac{5}{2}\log 2+\frac{3}{2}sq.units$
• B. $\frac{4+\sqrt{2}}{\log 2}-\frac{3}{2}\log 2+\frac{5}{2}sq.units$
• C. $\frac{4-\sqrt{2}}{\log 2}-\frac{3}{2}\log 2+\frac{5}{2}sq.units$
• D. $\frac{4+\sqrt{2}}{\log 2}-\frac{5}{2}\log 2+\frac{3}{2}sq.units$

Answer 1

The correct option is A $\frac{4-\sqrt{2}}{\log 2}-\frac{5}{2}\log 2+\frac{3}{2}sq.units$

Therefore the required area
$={\int }_{\frac{1}{2}}^2(2^x-\log x)dx$
$={[\frac{2^x}{\log 2}-(x\log x-x)]}_{\frac{1}{2}}^2$
$=\frac{4-\sqrt{2}}{\log 2}-\frac{5}{2}\log 2+\frac{3}{2}$