JEE Mains · Maths · STD 12 - 8. Application and integration
The area enclosed by \(y ^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to
- A \(\frac{16 \sqrt{2}}{6}\)
- B \(\frac{11 \sqrt{2}}{6}\)
- C \(\frac{13 \sqrt{2}}{6}\)
- D \(\frac{5 \sqrt{2}}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{13 \sqrt{2}}{6}\)
Step-by-step Solution
Detailed explanation
Area of \(\Delta ABC =\frac{1}{2}(\sqrt{2}) \cdot 1=\frac{\sqrt{2}}{2}\) So required Area \(=\int_{0}^{4}(\sqrt{8 x}-\sqrt{2} x) d x-\frac{\sqrt{2}}{2}\) \(=\frac{32 \sqrt{2}}{3}-8 \sqrt{2}-\frac{\sqrt{2}}{2}=\frac{13 \sqrt{2}}{6}\)
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