<p>Evaluate \(\displaystyle\int_0^\infty[2e^{-x}]\,dx\) where \([\cdot]\) is GIF. [JEE Advanced 2014]</p>
Step-by-Step Solution
Key Concept: 2e^(-x)=1 when x=ln2; =2 when x=0. GIF=1 on [0,ln2), GIF=0 on [ln2,\infty). Integral = 1 \cdot ln2.
<div class='solution'>
<p>$2e^{-x}=1\Rightarrow x=\ln 2$. $2e^{-x}=2\Rightarrow x=0$.</p>
<ul>
$x\in[0,\ln 2)$: $2e^{-x}\in(1,2]$, so $[2e^{-x}]=1$
$x=\ln 2$: $2e^{-x}=1$, $[1]=1$
$x>\ln 2$: $2e^{-x}<1$, so $[2e^{-x}]=0$
</ul>
<p>$$I=\int_0^{\ln 2}1\,dx+\int_{\ln 2}^\infty 0\,dx=\ln 2$$</p>
Correct Answer: A