Because the curve y = 3 − x^2 is always above y = 1 − x within the interval, you have to evaluate the
following integral:
[(3 − x^2 ) − (1 − x)] dx = (2 + x − x^2 ) dxTherefore, the area of the region is
PROBLEM 2. Find the area between the x-axis and the curve y = 2 − x^2 from x = 0 to x = 2.
Answer: First, sketch the graph over the interval.