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$\int_{0}^{[x]} \frac{4^{x}}{4^{[x]}} dx$ equals, where [.] denotes greatest integer functions

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$\frac{3 [x]}{2 \log 2}$

$6 [x] \log 2$

$\frac{[x]}{2 \log 2}$

$3 [x] \log 2$

answer is A.

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Detailed Solution

$\int_{0}^{[x]} 4^{x - [x]} dx = [x] \int_{0}^{1} 4^{x - [x]} dx$
x-[x] is periodic function with period 1 unit
$= [x] \int_{0}^{1} 4^{x} dx = \frac{[x]}{\log 4} (4 - 1) = \frac{3 [x]}{2 \log 2}$

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