Question

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

विकल्प

• 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 = ${\displaystyle \frac{3x^2}{4}}$, is the parabala and 4y = 6x + 24 is the line.

Area of the region A = ${\displaystyle {\int }_{-2}^4(\frac{6x+24}{4},\frac{3x^2}{4})dx={[\frac{3}{4}{x}^2+6x-\frac{x^3}{4}]}_{-2}^4}$

⇒ A = ${\displaystyle \frac{3}{4}(16-4)+6(4+2)-\frac{1}{4}(64+8)}$

⇒ A = 9 + 36 – 18

⇒ A = 27 square unit