Question

If [.] denotes the greatest integer function, then ${\int }_0^{\infty }\left[2e^{-x}\right]dx$=

• A. $0$
• B. $e^2$
• C. $2/e$
• D. $\log 2$

Answer 1

Answer: (D)

$\log 2$

Put $e^{-x}=y\therefore -e^{-x}dx=dy\Rightarrow dx=\frac{dy}{-y}$ and limits are adjusted as 1 to 0.
$\therefore I={\int }_1^0\left[2y\right]\frac{dy}{-y}={\int }_0^1\frac{\left[2y\right]}{y}dy$
As y
varies from 0 to 1, 2y varies from 0 to 2
$\therefore \left[2y\right]=0or1...\left[A\right]$

$\therefore I={\int }_0^{1/2}\frac{1}{y}\left[2y\right]dy+{\int }_{1/2}^1\frac{1}{y}\left[2y\right]dy={\int }_0^{1/2}\frac{1}{y}.0dy+{\int }_{1/2}^1.1dy\text{by (A)}=(\log y{)}_{1/2}^1=0-\log \frac{1}{2}=\log 2$