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Question

Mathematics Question on Coordinate Geometry

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

A

6+132−226+\frac{1}{32}-2\sqrt2

B

2+196−2232+\frac{1}{96}-\frac{2\sqrt2}{3}

C

223\frac{2\sqrt2}{3}

D

96

Answer

6+132−226+\frac{1}{32}-2\sqrt2

Explanation

Solution

The correct answer is (A) : 223\frac{2\sqrt2}{3}