If $$I = \int_0^1 {\frac{{x\,dx}}{{8 + {x^3}}}} $$    then the smallest interval in which $$I$$ lies is

Solution :
$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{x}{{8 + {x^3}}} \cr & {\text{Then }}f'\left( x \right) = \frac{{1.\left( {8 + {x^3}} \right) - x\left( {3{x^2}} \right)}}{{{{\left( {8 + {x^3}} \right)}^2}}} = \frac{{8 - 2{x^3}}}{{{{\left( {8 + {x^3}} \right)}^2}}} > 0{\text{ for }}x\, \in \left[ {0,\,1} \right] \cr & \therefore f\left( x \right)\,{\text{is m}}{\text{.i}}{\text{. in }}\left[ {0,\,1} \right].{\text{ So, }}f\left( 0 \right) \leqslant f\left( x \right) \leqslant f\left( 1 \right) \cr & \therefore 0 \leqslant f\left( x \right) \leqslant \frac{1}{{8 + 1}} \cr & \therefore \int_0^1 {0\,dx < } \int_0^1 {f\left( x \right)} dx < \frac{1}{9}\int_0^1 {1\,dx} \cr & \therefore 0 < \int_0^1 {f\left( x \right)} dx < \frac{1}{9} \cr} $$