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

5/3

1/3

2/3

4/3

Solution

4/3

Solving the equations we get the points of intersection (–2, 1) and (–2, 1)The bounded region is shown as shaded region

$The space required space area space equals space 2 integral subscript 0 superscript 1 space left parenthesis 1 minus 3 straight y squared right parenthesis space minus space left parenthesis negative 2 straight y squared right parenthesis space equals space 2 integral subscript 0 superscript 1 space left parenthesis 1 minus straight y squared right parenthesis space dy space equals space 2 space open square brackets straight y space minus space straight y cubed over 3 close square brackets subscript 0 superscript 1 space equals space 2 space straight x space 2 over 3 space equals space 4 over 3$