JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of the region \(\left\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R}^{2} | 4 \mathrm{x}^{2} \leq \mathrm{y} \leq 8 \mathrm{x}+12\right)\) is
- A \(\frac{127}{3}\)
- B \(\frac{125}{3}\)
- C \(\frac{124}{3}\)
- D \(\frac{128}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{128}{3}\)
Step-by-step Solution
Detailed explanation
\(4 x^{2}-y \leq 0\) and \(8 x-y+12 \geq 0\) On solving \(y=4 x^{2}\) and \(\quad y=8 x+12\) We get \(A(-1,4)\;and\; B(3,36)\) Required area \(=\) area of the shaded region \(=\int_{-1}^{3}\left(8 x+12-4 x^{2}\right) d x=\frac{128}{3}\)
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