Answer
$A = \frac{{23}}{6}$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}} \cr
& y = {x^2} + 2,{\text{ }}y = - x - 1,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& {\text{Let }} \cr
& f\left( x \right) = {x^2} + 1,{\text{ }}g\left( x \right) = - x - 1 \cr
& x = a = 0{\text{ and }}x = b = 1 \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_0^1 {\left[ {\left( {{x^2} + 2} \right) - \left( { - x - 1} \right)} \right]} dx \cr
& A = \int_0^1 {\left( {{x^2} + 2 + x + 1} \right)} dx \cr
& A = \int_0^1 {\left( {{x^2} + x + 3} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + 3x} \right]_0^1 \cr
& A = \left[ {\frac{{{{\left( 1 \right)}^3}}}{3} + \frac{{{{\left( 1 \right)}^2}}}{2} + 3\left( 1 \right)} \right] - \left[ {\frac{{{{\left( 0 \right)}^3}}}{3} + \frac{{{{\left( 0 \right)}^2}}}{2} + 2\left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{1}{3} + \frac{1}{2} + 3 \cr
& A = \frac{{23}}{6} \cr} $$
