Question

The value of ${\int }_0^{\left[x\right]}\frac{2^x}{2^{\left[x\right]}}dx$, is equal to (where $[.]$ denotes the greatest integer function)

• A. $\frac{\left[x\right]}{\log 2}$
• B. $\frac{\left[x\right]}{2\log 2}$
• C. $\frac{\left[x\right]}{4\log 2}$
• D. none of these

Answer 1

The correct option is A $\frac{\left[x\right]}{\log 2}$

${\int }_0^{\left[x\right]}\frac{2^x}{2^{\left[x\right]}}dx={\int }_0^{\left[x\right]}{2}^{x-\left[x\right]}dx=\left[x\right]{\int }_0^1{2}^{x-\left[x\right]}dx$
$\because x-\left[x\right]$ is periodic with period 1
$2^{x-\left[x\right]}$ is also periodic with 1
$=\left[x\right]{\int }_0^{\left[x\right]}{2}^{x}dx=\left[x\right]{\left(\frac{2^x}{\log 2}\right)}_0^1=\frac{\left[x\right]}{\log 2}$