Question

The value of ${\int }_0^6(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}})dx$ is equal to

• A. $6\sqrt{3}$
• B. $4\sqrt{3}$
• C. $12\sqrt{3}$
• D. none of these

Answer 1

The correct option is A $6\sqrt{3}$

$x+\sqrt{12x-36}=(x-3)+3+2\sqrt{3}\sqrt{x-3}={|\sqrt{x-3}-3|}^2$

$\therefore I={\int }_0^6(\sqrt{x+}\sqrt{12x-36}+\sqrt{x-\sqrt{12x-36}})dx={\int }_0^6|\sqrt{x-3}-\sqrt{3}|dx+{\int }_0^6|\sqrt{x-3}+\sqrt{3}|dx\because \sqrt{x-3}>\sqrt{3}\Rightarrow x-3>3\Rightarrow x>6\therefore I={\int }_0^6\sqrt{3}dx=6\sqrt{3}$