Question

Solve:
${\int }_0^{\left[x\right]}{4}^{x-\left[x\right]}dx$

Answer 1

Let $y={\int }_0^{\left(x\right)}{4}^{x{-}^{\left(x\right)}}dx$

Taking log on both side

$\Rightarrow \log y=4{\int }_0^{\left(x\right)}$

let $y={\int }_0^{\left(x\right)}{4}^{x-\left(x\right)}dx$

For periodic function $f(0+x)=f\left(x\right)$

${\int }_0^{na}f\left(x\right)dx=n{\int }_0^{a}f\left(x\right)dx$

$a-\left[x\right]$ is periodic with period $1$

$y=\left[x\right]{\int }_0^1{4}^{x-\left[x\right]}dx$

$\left[x\right]=0\forall x\in (0,1)$

$\Rightarrow y=\left[x\right]{\int }_0^1{4}^{x}dx$

$=\left[x\right]{\left(\frac{4x}{\ln 4}\right)}_0^1$

$y=\left[x\right]\frac{3}{\ln 4}$