Question

Evaluate ${\int }_1^{a}x\cdot a^{-\left[lo{g}_{a}x\right]}dx,(a>1)$.

Answer 1

Given : ${\int }_1^{a}x.a^{-\left[{\log }_{a}x\right]}dx$

We know ,

${\log }_{a}1=0{\log }_{a}a=1\left[{\log }_{a}x\right]=0for1<x<a$

${\int }_1^{9}x.a^{-\left[{\log }_{a}x\right]}dx={\int }_1^{9}xdx=\frac{1}{2}{\left|x^2\right|}_1^a\frac{1}{2}(a^2-1)$

Hence the correct answer is $\frac{1}{2}(a^2-1)$