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

The curves are $x=-2y^2$ and $x=1-3y^2$.Find the points of intersection by setting $x=x$:
$$-2y^2=1-3y^2⟹y^2=1⟹y=\pm 1$$
The area is given by integrating with respect to $y$:
$$A={\int }_{-1}^1(x_{\text{right}}-x_{\text{left}})dy={\int }_{-1}^1[(1-3y^2)-(-2y^2)]dy$$
$$A={\int }_{-1}^1(1-y^2)dy={\left[y-\frac{y^3}{3}\right]}_{-1}^1$$
$$A=\left(1-\frac{1}{3}\right)-\left(-1+\frac{1}{3}\right)=\frac{2}{3}-\left(-\frac{2}{3}\right)=\frac{4}{3}$$