Calculate the area of the figure bounded by the lines y = 4-x ^ 2 y = (x-2) ^ 2 y = 0.

Let’s simplify the second quadratic function:

y = (x – 2) ² = x² – 4 * x + 4.

We find the intersection points of the graphs, we get:

4 – x² = x² – 4 * x + 4,

2 * x² – 4 * x = 0,

2 * x * (x – 2) = 0,

x = 0,

x = 2.

Having built the graphs of both functions, we find out that the required area is bounded by two parabolas, and the graph of the function y = 4 – x² is higher than the graph of the function y = x² – 4 * x + 4, so the area is the integral of the difference between these functions:

s = integral (0 to 2) (4 – x² – x² + 4 * x – 4) dx = integral (0 to 2) (-2 * x² + 4 * x) dx = -2 * x³ / 3 + 2 * x² (from 0 to 2) = -16 / 3 + 8 = 8/3 units².