JEE Main 201610 Apr, Morning shiftMathematicsMediumIntegral Calculus369× since 2002The value of the integral ∫410[x2]dx[x2−28x+196]+[x2],\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,4∫10[x2−28x+196]+[x2][x2]dx, where [x] denotes the greatest integer less than or equal to x, is :AOption A. 6BOption B. 3COption C. 7DOption D. 13{1 \over 3}31CheckSkip · show solution
