Question

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

• A. $3$
• B. $7$
• C. $\frac{1}{3}$
• D. $6$

Answer 1

The correct option is A $3$

$I=\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]dx}{[x^2-28x+196]+\left[x^2\right]}=\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]dx}{\left[\right(14-x{)}^2]+[x^2]}\text{Using}\underset{a}{\overset{b}{\int }}f(a+b-x)=\underset{a}{\overset{b}{\int }}f\left(x\right)I=\underset{4}{\overset{10}{\int }}\frac{\left[\right(14-x{)}^2]dx}{\left[\right(14-x{)}^2]+[x^2]}\Rightarrow 2I=\underset{4}{\overset{10}{\int }}dx\Rightarrow 2I=10-4\Rightarrow I=3$