To find this area, it is necessary to solve an integral, actually the sum of 2 integrals
![\int (x^2+2x+2)dx+\int (3x+4)dx](https://img.qammunity.org/2023/formulas/mathematics/college/z1ieut3eaa3xylrzvf3dss26ggvtgnmb92.png)
The first one must be evaluated from -3 to 2 and the second one from 2 to 3
![\begin{gathered} \int (x^2+2x+2)dx+\int (3x+4)dx \\ ((x^3)/(3)+x^2+2x)+((3x^2)/(2)+4x) \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ui9f1j3lqmswwr356o30ove6vk6ra4ixtw.png)
Evaluate the first integral
![\begin{gathered} (x^3)/(3)+x^2+2x\text{ (From -3 to 2)} \\ ((2^3)/(3)+2^2+2\cdot2)-(((-3)^3)/(3)+(-3)^2+(2\cdot-3)) \\ (8)/(3)+4+4-(-(27)/(3)+9-6) \\ (35)/(3)+5=(50)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nml3hey90xqe2izntk1zdftyfa3irsue6s.png)
Evaluate the second integral
![\begin{gathered} (3x^2)/(2)+4x\text{ (From 2 to 3)} \\ ((3\cdot(3^2))/(2)+4\cdot3)-((3\cdot(2^2))/(2)+4\cdot2) \\ ((27)/(2)+12)-((12)/(2)+8) \\ (15)/(2)+4=(23)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dx7645ruubgd8y5j55g51ugmucwjaydpe1.png)
Now, solve the sum
![\begin{gathered} (50)/(3)+(23)/(2) \\ (100+69)/(6)=(169)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8mw8urw8y8acsjmu9f4rn8kuts0oqrc16v.png)
The area is 169/6