Question

The area of the region S = {(x, y): 3x² ≤ 4y ≤ 6x + 24} is ______.

पर्याय

• 25

• 26

• 27

• 28

Answer 1

The area of the region S = {(x, y): 3x² ≤ 4y ≤ 6x + 24} is 27.

Explanation:

S = {(x, y): 3x² ≤ 4y ≤ 6x + 24} 

Here, y = `(3x^2)/4`, is the parabala and 4y = 6x + 24 is the line.

Area of the region A = `int_(-2)^4 ((6x + 24)/4, (3x^2)/4)dx = [3/4x^2 + 6x - x^3/4]_(-2)^4`

⇒ A = `3/4(16 - 4) + 6(4 + 2) - 1/4(64 + 8)`

⇒ A = 9 + 36 – 18

⇒ A = 27 square unit