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प्रश्न
The area of the region S = {(x, y): 3x2 ≤ 4y ≤ 6x + 24} is ______.
विकल्प
25
26
27
28
MCQ
रिक्त स्थान भरें
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उत्तर
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
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