∫ _ 0 ^ 2 ( |2 x ^ 2 - 3x| + [x - (1)/(2)])dx , where [t] is the...

${∫}_{0}^{2}\left(\left|2{x}^{2} - 3x\right| + \left[x - \dfrac{1}{2}\right]\right)dx$

$= {∫}_{0}^{2}\left|2{x}^{2} - 3x\right|dx + {∫}_{0}^{2}\left[x - \dfrac{1}{2}\right]dx$

Let ${I}_{1} = {∫}_{0}^{2}\left|2{x}^{2} - 3x\right|dx = {∫}_{0}^{\dfrac{3}{2}}\left(3x - 2{x}^{2}\right)dx + {∫}_{\dfrac{3}{2}}^{2}\left(2{x}^{2} - 3x\right)dx$

$= {\left(\dfrac{3{x}^{2}}{2} - \dfrac{2{x}^{3}}{3}\right)}_{0}^{\dfrac{3}{2}} + {\left(\dfrac{2{x}^{3}}{3} - \dfrac{3{x}^{2}}{2}\right)}_{\dfrac{3}{2}}^{2}$

$= \left(\dfrac{27}{8} - \dfrac{9}{4}\right) + \left(\dfrac{8}{3} - 6 - \dfrac{9}{4} + \dfrac{27}{8}\right) = \dfrac{9}{4} - \dfrac{10}{3} = \dfrac{19}{12}$

Let ${I}_{2} = {∫}_{0}^{2}\left[x - \dfrac{1}{2}\right]dx$

Assume $x - \dfrac{1}{2} = t$

i.e. ${I}_{2} = {∫}_{\dfrac{- 1}{2}}^{\dfrac{3}{2}}\left[t\right]dt$

$= {∫}_{- \dfrac{1}{2}}^{0}\left(- 1\right)dt + {∫}_{0}^{1}0 \cdot dt + {∫}_{1}^{\dfrac{3}{2}}1 \cdot dt$

$= {\left(- t\right)}_{- \dfrac{1}{2}}^{0} + {\left(t\right)}_{1}^{\dfrac{3}{2}} = - \dfrac{1}{2} + \left(\dfrac{3}{2} - 1\right) = 0$

Hence, ${∫}_{0}^{2}\left(\left|2{x}^{2} - 3x\right| + \left[x - \dfrac{1}{2}\right]\right)dx = \dfrac{19}{12}$