JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Area Under Curves

The region lies above the $x$-axis and below both $y=x^2+1$ and $y=x+1$, which cross at $x=0$ and $x=1$. On $\frac{1}{2}\le x\le 1$ the parabola is the lower of the two and so caps the region, while on $1\le x\le 2$ the line is lower and caps it. The area is ${\int }_{1/2}^1(x^2+1)dx+{\int }_1^2(x+1)dx=\frac{19}{24}+\frac{5}{2}=\frac{79}{24}$.