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Question: The area of the plane region bounded by the curves x + 2y<sup>2</sup> = 0 and x + 3y<sup>2</sup> = 1...

The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to-

A

1/3

B

2/3

C

4/3

D

5/3

Answer

4/3

Explanation

Solution

Area = 2 ∫01(1−3y2)−(−2y2)dy\int _ { 0 } ^ { 1 } \left( 1 - 3 y ^ { 2 } \right) - \left( - 2 y ^ { 2 } \right) d y

= 2

= 2. = 2. (1−13)\left( 1 - \frac { 1 } { 3 } \right)= 43\frac { 4 } { 3 }