Find the area of the region enclosed between and from to . Hint: Notice that this region consists of two parts.
To find the area of the region enclosed between the curves y = x^2 and y = 4 - x^2 from x = -1 to x = 1, we need to first find the points of intersection of the two curves.
Setting y = x^2 equal to y = 4 - x^2, we get:
x^2 = 4 - x^2
2x^2 = 4
x^2 = 2
x = ±√2
So, the points of intersection are at x = -√2 and x = √2.
Now, we can set up the integral to find the area of the region:
A = ∫[from -√2 to √2] (4 - x^2 - x^2) dx
A = ∫[from -√2 to √2] (4 - 2x^2) dx
A = [4x - (2/3)x^3] from -√2 to √2
A = (4√2 - (2/3)(2)^3) - (-4√2 - (2/3)(2)^3)
A = 8√2 - (16/3) + 8√2 - (16/3)
A = 16√2 - (32/3)
Therefore, the area of the region enclosed between the curves y = x^2 and y = 4 - x^2 from x = -1 to x = 1 is 16√2 - (32/3) square units.