Area bounded by larger part in I quadrant by x = 4y2 , x = 2 and y = x is A then 3A equals
\(6+\frac{1}{32}-2\sqrt2\)
\(2+\frac{1}{96}-\frac{2\sqrt2}{3}\)
\(\frac{2\sqrt2}{3}\)
- 96
The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(\frac{2\sqrt2}{3}\)
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