The area (in square units) of the region bounded by the curves y + 2 x 2 = 0 and y + 3 x 2 = 1 , is equal to

To find point of intersection of two given curves

Put $y=-2x^2$ in $y+3x^2=1$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

Let the given equations be $y_1=1-3x^2$ and $y_2=-2x^2$.

The desired area would be,

${\int }_{-1}^1\left(y_1-y_2\right)dx ={\int }_{-1}^1\left(\left(1-3x^2\right)-\left(-2x^2\right)\right)dx$

$={\int }_{-1}^1\left(1-x^2\right)dx$

$={\left(x- \frac{x^3}{3}\right)}_{-1}^1= \left(\left(1-\frac{1}{3}\right)- \left(-1+\frac{1}{3}\right)\right)$

$=\frac{2}{3}- \left(\frac{-2}{3}\right)$

$=\frac{4}{3} \mathrm{sq}. \mathrm{units}$ .