The area (in sq. units) of the region \ (x, y): 0 ≤ y ≤ x ^ 2 + 1,...

$0 \le y \le {x}^{2} + 1, 0 \le y \le x + 1, \dfrac{1}{2} \le x \le 2$

Drawing the graph according to the given information,

Area is, $A = {∫}_{\dfrac{1}{2}}^{1}\left({x}^{2} + 1\right) d x + {∫}_{1}^{2}\left(x + 1\right) d x$

$A = {\left[\dfrac{{x}^{3}}{3} + x\right]}_{\dfrac{1}{2}}^{1} + {\left[\dfrac{{x}^{2}}{2} + x\right]}_{1}^{2}$

$A = \left[\dfrac{1}{3} + 1\right] - \left[{\left(\dfrac{1}{2}\right)}^{3} \cdot \dfrac{1}{3} + \dfrac{1}{2}\right] + \left[\dfrac{4}{2} + 2\right] - \left[\dfrac{1}{2} + 1\right]$

$A = \dfrac{4}{3} - \left(\dfrac{1}{24} + \dfrac{1}{2}\right) + \left(2 + 2\right) - \dfrac{3}{2}$

$A = \dfrac{4}{3} - \dfrac{13}{24} + 4 - \dfrac{3}{2}$

$A = \dfrac{32 - 13 + 96 - 36}{24}$

$A = \dfrac{79}{24}$ sq. units.