Mathematics - Area Under Curves Question with Solution | TestHub

Total area $=2{\int }_0^4\sqrt{x}dx=\frac{32}{3}$, so each part is $\frac{8}{3}$. By symmetry of the region about the x-axis, the horizontal divider is $b=0$. Then ${\int }_a^4\sqrt{x}dx=\frac{8}{3}\Rightarrow \frac{2}{3}\left(8-a^{3/2}\right)=\frac{8}{3}\Rightarrow a^{3/2}=4\Rightarrow a=16^{1/3}$.