Question

If $\left[x\right]$ denotes greatest integer function then,
${\int }_{10}^{11}\frac{\left[x^2\right]dx}{[x^2-42x+441]-\left[x^2\right]}$equals

• A. $\frac{1}{2}$
• B. $0$
• C. $\frac{21}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

$I={\int }_{10}^{11}\frac{\left[x^2\right]dx}{[x^2-42x+441]-\left[x^2\right]}={\int }_{10}^{11}\frac{\left[x^2\right]dx}{\left[{(x-21)}^2\right]-\left[x^2\right]}$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{10}^{11}\frac{\left[{(21-x)}^2\right]dx}{\left[x^2\right]-\left[{(21-x)}^2\right]}$
$\therefore 2I={\int }_{10}^{11}dx=-1\Rightarrow I=\frac{-1}{2}$