Find the area of the region enclosed by the parabolas y=x^2 and y+2x-x^2
a = ∫[0,1] (2x-x^2)-(x^2) dx
To find the area of the region enclosed by the parabolas y = x^2 and y + 2x - x^2, we need to find the points of intersection and integrate the difference of the functions over the interval between the intersection points.
First, let's find the points where the two parabolas intersect by setting them equal to each other:
x^2 = y + 2x - x^2
Rearranging the equation, we get:
2x^2 - 2x - y = 0
We can rewrite the equation as:
y = 2x^2 - 2x
Now let's find the points of intersection by setting the two parabolas equal to each other:
x^2 = 2x^2 - 2x
Rearranging the equation, we get:
x^2 - 2x^2 + 2x = 0
Simplifying, we have:
-x^2 + 2x = 0
Factoring out an x, we get:
x(x - 2) = 0
From this, we see that x = 0 or x = 2.
Now we need to find the y-coordinates of the points of intersection by substituting the x-values back into either of the original equations. Let's use y = x^2:
For x = 0:
y = (0)^2 = 0
So one point of intersection is (0, 0).
For x = 2:
y = (2)^2 = 4
So the other point of intersection is (2, 4).
Now we integrate the difference of the functions y = x^2 and y + 2x - x^2 over the interval between the x-values of the points of intersection, which are 0 and 2:
Area = ∫[0, 2] (x^2 - (2x^2 - 2x))dx
Simplifying the integrand:
Area = ∫[0, 2] (3x^2 - 2x)dx
To integrate, we will use the power rule for integration:
∫ x^n dx = (x^(n+1))/(n+1)
Applying the power rule, we get:
Area = [(3/3)x^3 - (2/2)x^2] | from 0 to 2
Simplifying further:
Area = (x^3 - x^2) | from 0 to 2
Now, let's plug in the upper and lower limits of integration:
Area = (2^3 - 2^2) - (0^3 - 0^2)
Area = (8 - 4) - (0 - 0)
Area = 4 - 0
Therefore, the area of the region enclosed by the parabolas y = x^2 and y + 2x - x^2 is 4 square units.