The curves y = x and y = 1 - x^2 intersect at the points (-1, 0), (0, 0), and (1, 0). The region enclosed by these curves is a triangle with vertices at these points.
To find the area of this region without using the absolute value function, we can split the triangle into two parts: the part above the x-axis and the part below the x-axis.
For the part above the x-axis, the bounds of integration are x = 0 to x = 1. The integrand is y = x, the equation of the upper curve. The integral is:
∫[0,1] x dx = 1/2
For the part below the x-axis, the bounds of integration are x = -1 to x = 0. The integrand is y = 1 - x^2, the equation of the lower curve. However, since we cannot use the absolute value function, we need to split this integral into two parts as well. When x is between -1 and 0, the lower curve is y = 1 - x^2, and when x is between 0 and 1, the lower curve is y = x. Therefore, we have:
∫[-1,0] (1 - x^2) dx + ∫[0,1] x dx
Evaluating the first integral gives:
∫[-1,0] (1 - x^2) dx = x - (x^3/3) evaluated from -1 to 0 = 1/3
Therefore, the area of the region enclosed by the curves is:
1/2 + 1/3 = 5/6
So, the area of the region can be represented using two definite integrals.