Answer:
Step-by-step explanation:
integrate with respect to x
Area is symmetric so will be twice the integral from zero to 1
( I cannot seem to figure out how to make the lower limit negative in this editor)
[tex]A = 2\int\limits^1_0 {(x^2 + 3) - 4x^2} \, dx[/tex])
[tex]A = 2\int\limits^1_0 {-3x^2 + 3} \, dx[/tex]
[tex]A = 6\int\limits^1_0 {-x^2 + 1} \, dx[/tex]
A = 6(-⅓x³ + x)[tex]\left \| {{x=1} \atop {x=0}} \right.[/tex]
A = 6((-⅓(1)³ + (1)) - (-⅓(0)³ + (0)))
A = 6(⅔)
A = 4 units²
