Question

The value of the integral ${\int }_4^{10}\frac{\left[x^2\right]dx}{[x^2-28x+196]+\left[x^2\right]}$, where $\left[x\right]$ denotes the greatest integer less than or equal to $x$, is:

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

Answer 1

The correct option is D $3$

$I={\int }_4^{10}\frac{\left[x^2\right]}{[x^2-28x+196]+\left[x^2\right]}dx$ ..... (1)

Use ${\int }_a^{b}f(a+b-x)dx={\int }_a^{b}f\left(x\right)dx$

$I={\int }_4^{10}\frac{\left[\right(x-14{)}^2]}{\left[x^2\right]+\left[\right(x-14{)}^2]}dx$ ..... (2)

$\left(1\right)+\left(2\right)$

$2I={\int }_4^{10}\frac{\left[\right(x-14{)}^2]+[x^2]}{\left[x^2\right]+\left[\right(x-14{)}^2]}dx$

$2I={\int }_4^{10}dx$

$2I=6$

$I=3$