Oxford University MAT 2018 Solved Papers

The curves $\mathit{y}=\sqrt{\mathit{x}}$ and $\mathit{y}=\mathit{x}-2$ intersect when $\mathit{x}=4$. We want the area of region $\mathit{R}$ in the sketch below. We will integrate $\sqrt{\mathit{x}}$ from 0 to 4 and subtract the area of triangle $\mathit{T}$. This gives the required area as ${\underset{0}{\int }}^4\sqrt{\mathit{x}}\mathit{d}\mathit{x}-\frac{1}{2}\cdot 2\cdot 2={\left[\frac{2}{3}{\mathit{x}}^{3∕2}\right]}_0^4-2=\frac{10}{3}$. The answer is (d).