Answer:
Step-by-step explanation:
Complete Question
The complete question is shown on the first uploaded image
Answer:
The area is [tex]A =8 sq\cdot unit[/tex]
Step-by-step explanation:
From the question we are told that
The first equation is [tex]f(x) = x^2 + x \ \ \ x< 1[/tex]
[tex]on[ -2 , 3 ][/tex]
The second equation is [tex]f(x) = 2 x \ \ \ x \ge 1[/tex]
This means that the limit of the area under the enclosed region is limited between -2 to 1 on the x- axis for first equation and 1 to 3 for second equation
Now the area under the region is evaluated as
[tex]A = \int\limits^1_{-2}{x^2 + x } \, dx + \int\limits^3_{1}{2x } \, dx[/tex]
[tex]A ={ \frac{x^3}{3} + \frac{x^2}{2} + c } | \left \ 1 } \atop {-2}} \right. + {\frac{2x^2}{2} }| \left \ 3} \atop {1}} \right.[/tex]
[tex]A =9 + c - 1 -c[/tex]
[tex]A =8 sq\cdot unit[/tex]
