Question

The value of the integral $\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]}{[x^2-28x+196]+\left[x^2\right]}dx,$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$, is

Answer 1

$I=\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]}{[x^2-28x+196]+\left[x^2\right]}dx\Rightarrow I=\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]}{\left[\right(x-14{)}^2]+\left[x^2\right]}dx$
Putting $x\to 14-x$, we get
$\Rightarrow I=\underset{4}{\overset{10}{\int }}\frac{\left[\right(x-14{)}^2]}{\left[x^2\right]+\left[\right(x-14{)}^2]}dx$
Adding both, we get
$\Rightarrow 2I=\underset{4}{\overset{10}{\int }}dx=6\therefore I=3$