Let the area enclosed by the lines x + y = 2 ,   y = 0 , x = 0 and the curve f ( x ) = min x 2 + 3 4 ,…

Given,

The lines $x+y=2, y=0$, $x=0$ and the curve $f\left(x\right)=\min \left\{x^2+\frac{3}{4},1+\left[x\right]\right\}$ where $\left[x\right]$ denotes the greatest integer $\le x$,

Now plotting the diagram of given function we get,

 

Now from above diagram, the area enclosed is given by,

$A=\left[{\int }_0^{\frac{1}{2}}\left(x^2+\frac{3}{4}\right)dx\right]+\frac{1}{2}\left(\frac{1}{2}+\frac{3}{2}\right)\times 1=\frac{5}{12}+1$

$\Rightarrow 12A=17$