The area bounded by the curves , and is
- A.sq unit
- B.sq unit
- C.sq unit
- D.sq unit
Show answer and solution
Correct option: A
Solution
The three curves intersect to form a region. y = (x-1)² meets y = 1/4 at x = 1/2 and 3/2; y = (x+1)² meets y = 1/4 at x = -3/2 and -1/2. The two parabolas meet each other at x = 0, y = 1.
By symmetry, compute the left half (from x = -1/2 to 0) where the region is bounded above by y = (x+1)² and below by y = 1/4, then double it.
Area of one half = ∫ from -1/2 to 0 of [(x+1)² - 1/4] dx
= [((x+1)³)/3 - x/4] from -1/2 to 0
At x = 0: 1/3 - 0 = 1/3
At x = -1/2: ((1/2)³)/3 - (-1/2)/4 = 1/24 + 1/8 = 1/24 + 3/24 = 4/24 = 1/6
One half area = 1/3 - 1/6 = 1/6
Total area = 2 × 1/6 = 1/3 sq unit.
By symmetry, compute the left half (from x = -1/2 to 0) where the region is bounded above by y = (x+1)² and below by y = 1/4, then double it.
Area of one half = ∫ from -1/2 to 0 of [(x+1)² - 1/4] dx
= [((x+1)³)/3 - x/4] from -1/2 to 0
At x = 0: 1/3 - 0 = 1/3
At x = -1/2: ((1/2)³)/3 - (-1/2)/4 = 1/24 + 1/8 = 1/24 + 3/24 = 4/24 = 1/6
One half area = 1/3 - 1/6 = 1/6
Total area = 2 × 1/6 = 1/3 sq unit.
[AIEEE 2005]