Question:
What is the area enclosed by the parabola \(4y = 3x^2\) and the line \(2y = 3x + 12\)?
परवलय \(4y = 3x^2\) और रेखा \(2y = 3x + 12\) द्वारा परिबद्ध क्षेत्रफल क्या है?
Options:
- 27 Square unit
- 36 Square unit
- 48 Square unit
- 54 Square unit
- 27 वर्ग इकाई
- 36 वर्ग इकाई
- 48 वर्ग इकाई
- 54 वर्ग इकाई
Correct Answer: a)
Explanation:
The area enclosed between the intersection points \(x = -2\) and \(x = 4\) is:
\[A = \int_{-2}^4 \left(y_{\text{line}} – y_{\text{parabola}}\right) dx = \int_{-2}^4 \left(\frac{3x + 12}{2} – \frac{3x^2}{4}\right) dx\]
\[A = \left[ \frac{3x^2}{4} + 6x – \frac{x^3}{4} \right]_{-2}^4\]
At \(x = 4\): \(\frac{3(16)}{4} + 6(4) – \frac{64}{4} = 12 + 24 – 16 = 20\).
At \(x = -2\): \(\frac{3(4)}{4} + 6(-2) – \frac{-8}{4} = 3 – 12 + 2 = -7\).
\[A = 20 – (-7) = 27\text{ square units}\]
Hence, the correct option is (a).
परिबद्ध क्षेत्रफल:
\[A = \int_{-2}^4 \left(\frac{3x + 12}{2} – \frac{3x^2}{4}\right) dx = \left[ \frac{3x^2}{4} + 6x – \frac{x^3}{4} \right]_{-2}^4\]
सीमाएँ रखने पर:
\[A = 20 – (-7) = 27\text{ वर्ग इकाई}\]
अतः सही विकल्प (a) है।