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Q.
The area of the plane region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is equal to
Application of Integrals
Solution:
$x+2y^{2} = 0 \Rightarrow y^{2} = -\frac{x}{2}$
[Left handed parabola with vertex at (0, 0)]
$ x +3y^{2} = 1 \Rightarrow y^{2} = -\frac{1}{3} \left(x-1\right) $
[Left handed parabola with vertex at (1, 0)]
Solving the two equations we get the points of intersection as (-2, 1), (-2, -1)
The required area is ACBDA, given by
$= \left|\int\limits^{1}_{-1} \left( 1-3y^{2} - 2y^{2}\right)dy\right| = \left|y - \frac{5y^{3} }{3}\right|^{1}_{-1} $
$= \left|\left(1- \frac{5}{3}\right) - \left(-1 + \frac{5}{3}\right) \right| = 2 \times\frac{2}{3} = \frac{4}{3} $ sq. units.
