Mathematics - Area Under Curves Question with Solution | TestHub

If $1/2<x<1$, then $0<{\log }_{y}x<1$; with $0<y<1$ this gives $0<y<x$, area ${\int }_{1/2}^{1}xdx=3/8$. If $1<x<2$, then ${\log }_{y}x>1$; with $y>1$ this gives $1<y<x$, area ${\int }_1^2(x-1)dx=1/2$. The total is 7/8.