Question

If $\underset{0}{\overset{4}{\int }}\sqrt{\left\{x\right\}}dx=\alpha$, then the value of $3\alpha$ is
(where $\{\cdot \}$ denotes fractional part function)

• A. $8$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is A $8$

$\underset{0}{\overset{4}{\int }}\sqrt{\left\{x\right\}}dx=\underset{0}{\overset{4}{\int }}\sqrt{x-\left[x\right]}dx=\underset{0}{\overset{1}{\int }}\sqrt{x}dx+\underset{1}{\overset{2}{\int }}\sqrt{x-1}dx+\underset{2}{\overset{3}{\int }}\sqrt{x-2}dx+\underset{3}{\overset{4}{\int }}\sqrt{x-3}dx=\frac{2}{3}[1+1+1+1]=\frac{8}{3}=\alpha$
$\therefore 3\alpha =8$