Application of Integrals · Question 28

Let the area of the region \(x, y) : 2y x2 + 3, y + |x| 3, y |x - 1|\ be A. Then 6A is equal to :

1. 14
2. 18
3. 16
4. 12

Answer

(1) 14

Solution

Boundaries: y = x2 + 32 (parabola), y = 3 - |x| and y = |x - 1|

y |x - 1| and y 3 - |x| -1 x 2

For x 0: x2 + 32 3 - x x2 + 2x - 3 0 x 1

For x < 0: x2 + 32 3 + x (x + 1)(x - 3) 0, true for -1 x < 0

upper boundary is the parabola on [-1, 1] and y = 3 - x on [1, 2]

[figure]

A = -11 ( x2 + 32 - (1 - x) )dx + 12 ((3 - x) - (x - 1) )dx

-11 x2 + 32 dx = 12 ( 23 + 6 ) = 103

-11(1 - x) dx = 2

12(4 - 2x) dx = [4x - x2 ]12 = 4 - 3 = 1

A = 103 - 2 + 1 = 73

6A = 14

Answer: (1)