Question 293 of 416MediumMCQ
The value of the integral ∫410[x2−28x+196]+[x2][x2]dx, where [x] denotes the greatest integer less than or equal to x, is: Solutions · 1
Official solution
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 =4∫10dx=[x]410 Answer the question to unlock