Answer:
[tex]A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36[/tex]
Step-by-step explanation:
The equations are:
[tex]y = x^{2} + 2x + 3[/tex]
[tex]y = 2x + 12[/tex]
The two graphs intersect when:
[tex]x^{2} + 2x + 3 = 2x + 12[/tex]
[tex]x^{2} = 0[/tex]
[tex]x_{1} = 3\\x_{2} = -3[/tex]
To find the area under the curve for the first equation:
[tex]A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx[/tex]
To find the area under the curve for the second equation:
[tex]A_{2} = \int\limits^3__-3}{2x + 12} \, dx[/tex]
To find the total area:
[tex]A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx[/tex]
Simplifying the equation:
[tex]A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx[/tex]
Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).
