Mathematics - Area Under Curves Question with Solution | TestHub

Inside the square, $[\sqrt{x}]=[\sqrt{y}]=1$, so the curves are $y^2=4x$ and $x^2=4y$. Within the square, $C_1$ runs from $(1,2)$ to $(4,4)$ and $C_2$ from $(2,1)$ to (4, 4). Area $={\int }_1^2(2\sqrt{x}-1)dx+{\int }_2^4\left(2\sqrt{x}-\frac{x^2}{4}\right)dx=$ $\left(\frac{8\sqrt{2}}{3}-\frac{7}{3}\right)+\left(6-\frac{8\sqrt{2}}{3}\right)=6-\frac{7}{3}=\frac{11}{3}$. ZZZ DUMMY - DO NOT IMPORT \end{document}