Question:

Area bounded by larger part in I quadrant by x = 4y2 , x = 2 and y = x is A then 3A equals

Updated On: Apr 1, 2024
  • \(6+\frac{1}{32}-2\sqrt2\)

  • \(2+\frac{1}{96}-\frac{2\sqrt2}{3}\)

  • \(\frac{2\sqrt2}{3}\)

  • 96
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The Correct Option is A

Solution and Explanation

The correct answer is (A) : \(\frac{2\sqrt2}{3}\)

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