Official solution
\[\int\limits_0^x \, \]f(t) dt = x2 + \[\int\limits_x^1 \, \] t2f(t) dt f '\[\left( {{1 \over 2}} \right)\] = ?
Differentiate w.r.t. 'x'
f(x) = 2x + 0 - x2 f(x)
f(x) = \[{{2x} \over {1 + {x^2}}}\] \[ \Rightarrow \] f '(x) = \[{{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}\]
f '(x) = \[{{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}\]
f '\[\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}\]