QJEEM16Ma664b: JEE Main 2016 Mathematics - Class 12 JEEM16Ma664b Solution: The value of the integral where [x] denotes the greatest…

Question

Verified Solution • 2025–26 Session moderate

The value of the integral

\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,

where [x] denotes the greatest integer less than or equal to x, is :

Accepted Answer

Let I =

\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}}

=

\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)

Using,

\int\limits_a^b {f\left( {a + b - x} \right)dx\,}

=

\,\,\int\limits_a^b {f(x)\,\,dx}

I =

\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)

Adding (1) and (2)

2I =

\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx

\Rightarrow

\,\,\,

2I =

\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}

= 6

= I = 3

Step-by-Step Verified Solution