If $$\phi \left( x \right) = \int_x^{{x^2}} {\left( {t - 1} \right)dt,\,1 \leqslant x \leqslant 2,} $$        then the greatest value of $$\phi \left( x \right)$$  is :

Solution :
$$\eqalign{ & \phi \left( x \right) = \int_0^{{x^2}} {\left( {t - 1} \right)dt} - \int_0^x {\left( {t - 1} \right)dt} \cr & \therefore \phi '\left( x \right) = \left( {{x^2} - 1} \right).2x - \left( {x - 1} \right) = 2{x^3} - 3x + 1 \cr & \therefore \phi '\left( x \right) = \left( {x - 1} \right)\left( {2{x^2} + 2x - 1} \right) \cr & = 2\left( {x - 1} \right)\left( {{x^2} + x - \frac{1}{2}} \right) \cr & = 2\left( {x - 1} \right)\left\{ {{{\left( {x + \frac{1}{2}} \right)}^2} - \frac{3}{4}} \right\} \cr} $$
In $$1 \leqslant x \leqslant 2,\,\phi '\left( x \right) \geqslant 0.$$      So, $$\phi \left( x \right)$$  is m.i. in [1, 2]. Hence, $$\phi \left( 2 \right)$$  is the greatest value. Now, $$\phi \left( 2 \right) = \int_2^4 {\left( {t - 1} \right)dt} = \left[ {\frac{{{t^2}}}{2} - t} \right]_2^4 = \left( {8 - 4} \right) - \left( {2 - 2} \right) = 4\,$$