$\int\limits_0^{[x] / 3} \frac{8^x}{2^{[3 x]}} d x$ where [.] denotes the greatest integer function, is equal to
Answer & explanation
Correct answer: option 4
Let $I=\int\limits_0^{[x] / 3} \frac{8^x}{2^{[3 x]}} d x$. Then,
$I =\int\limits_0^{[x] / 3} 2^{3 x-[3 x]} d x$
$\Rightarrow I =[x] \int\limits_0^{1 / 3} 2^{3 x-[3 x]} d x$
[∵ $2^{3 x-[3 x]}$ is periodic with period 1/3]
$\Rightarrow I=[x] \int\limits_0^{1 / 3} 2^{3 x} d x=[x]\left[\frac{2^{3 x}}{3 \log 2}\right]_0^{1 / 3}=\frac{[x]}{\log 8}$