Area bounded by the curve $y=log x,$ x-axis , x=1 and x=2 in square units is :

log 4 $log (\frac{4}{e})$ $2log \, 2+1$ $log\, 4-2$

$log (\frac{4}{e})$

The correct answer is Option (2) → $\log (\frac{4}{e})$ $⇒\int\limits_1^2ydx$ $=\int\limits_1^2\log xdx$ $=[x\log x]_1^2-\int 1dx$ Using $\int uv\,dx=\int u\int v\,dx-\int u'\int v\,dx\,dx$ $=[x\log x-x]_1^2$ $=2\log 2-2+1$ $=\log 4-\log e⇒\log (\frac{4}{e})$