The value of $\int\limits_0^{[x]} \frac{2^x}{2^{[x]}} d x$, is
Answer & explanation
Correct answer: option 2
We have,
$\int\limits_0^{[x]} \frac{2^x}{2^{[x]}} d x=\int\limits_0^{[x]} 2^{x-[x]} d x$
Since, $2^{x-[x]}$ is a periodic function with period one unit.
∴ $\int\limits_0^{[x]} 2^{x-[x]} d x=[x] \int\limits_0^1 2^{x-[x]} d x=[x] \int\limits_0^1 2^x d x$
$\Rightarrow \int\limits_0^{[x]} 2^{x-[x]} d x=\frac{[x]}{\log 2}\left[2^x\right]_0^1=\frac{[x]}{\log 2}(2-1)=\frac{[x]}{\log 2}$