Practicing Success
$\int\limits_0^{[x] / 3} \frac{8^x}{2^{[3 x]}} d x$ where [.] denotes the greatest integer function, is equal to |
$\frac{[x]}{\ln 2}$ $\frac{[x]}{\ln 4}$ $\frac{2[x]}{\ln 2}$ $\frac{[x]}{\ln 8}$ |
$\frac{[x]}{\ln 8}$ |
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] $\Rightarrow I=[x] \int\limits_0^{1 / 3} 2^{3 x} d x=[x]\left[\frac{2^{3 x}}{3 \log 2}\right]_0^{1 / 3}=\frac{[x]}{\log 8}$ |