Question

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

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

Answer 1

Answer: (B)

$I=\underset{4}{\overset{10}{\int }}\frac{\left[x^2\right]dx}{\left[x^2-28x+196\right]+\left[x^2\right]}\dots \dots .$ (i)
Use property $\underset{a}{\overset{b}{\int }}f(a+b-x)dx=\underset{a}{\overset{b}{\int }}f(x)dx$
$\Rightarrow I=\underset{4}{\overset{10}{\int }}\frac{\left[x^2-28x+196\right]dx}{\left[x^2\right]+\left[x^2-28x+196\right]}\dots \dots .$ (ii)
by (i) and (ii)
$2I=\underset{4}{\overset{10}{\int }}dx=10-4=6$
$I=3$