To find this area, it is necessary to solve an integral, actually the sum of 2 integrals
[tex]\int (x^2+2x+2)dx+\int (3x+4)dx[/tex]
The first one must be evaluated from -3 to 2 and the second one from 2 to 3
[tex]\begin{gathered} \int (x^2+2x+2)dx+\int (3x+4)dx \\ (\frac{x^3}{3}+x^2+2x)+(\frac{3x^2}{2}+4x) \\ \end{gathered}[/tex]
Evaluate the first integral
[tex]\begin{gathered} \frac{x^3}{3}+x^2+2x\text{ (From -3 to 2)} \\ (\frac{2^3}{3}+2^2+2\cdot2)-(\frac{(-3)^3}{3}+(-3)^2+(2\cdot-3)) \\ \frac{8}{3}+4+4-(-\frac{27}{3}+9-6) \\ \frac{35}{3}+5=\frac{50}{3} \end{gathered}[/tex]
Evaluate the second integral
[tex]\begin{gathered} \frac{3x^2}{2}+4x\text{ (From 2 to 3)} \\ (\frac{3\cdot(3^2)}{2}+4\cdot3)-(\frac{3\cdot(2^2)}{2}+4\cdot2) \\ (\frac{27}{2}+12)-(\frac{12}{2}+8) \\ \frac{15}{2}+4=\frac{23}{2} \end{gathered}[/tex]
Now, solve the sum
[tex]\begin{gathered} \frac{50}{3}+\frac{23}{2} \\ \frac{100+69}{6}=\frac{169}{6} \end{gathered}[/tex]
The area is 169/6
