Question
Find the area of the region
{(x, y): 0 ≤ y ≤ x2 + 1 , 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}.
Solution
The given region is
{( x, y): 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Thus region is the intersection of the following regions:
R1 = {(x, y) : 0 ≤ y ≤ x2 + 1}
R2 = { (x, y) : 0 ≤ y ≤ x + 1}
R3 = {(x, y) : 0 ≤ x ≤ 2}
The function with graph in the figure is
Consider the equations
y = x2 + 1 ...(1)
and y = x + 1 ...(2)
Putting y = x + 1 in (1), we get
x + 1 = x2 + 1, or x = x2 ⇒ x2 - x = 0 ⇒ x(x - 1) = 0
∴ x = 0, 1
∴ from (2), y = 1, 2
∴ curve (1) and (2) intersect in the points P (0, 1) and Q (1, 2).
The region considered is bounded by
y = f(x),
y = 0
x = 0
and x = 2
required area = 
