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Q.
If $\int\limits_0^{11} \frac{11^x}{11^{[x]}} d x=\frac{k}{\log 11}$, (where [ ] denotes greatest integer function) then value of $k$ is
Integrals
Solution:
$I=\int\limits_0^{11} 11^{(x)} dx =\int\limits_0^{11 \times 1} 11^{(x)} dx =11 \int\limits_0^1 11^{\{x\}} dx \{\because \{ x \}$ is periodic with period 1$\}$
$=11 \int\limits_0^1 11^x dx =11\left[\frac{11^x}{\ell n 11}\right]_0^1=11\left[\frac{11}{\ell n 11}-\frac{1}{\ell n 11}\right]$
$=\frac{110}{\ell n 11}=\frac{ k }{\ell n 11} \Rightarrow k =110$