The area (in square units) of the region bounded by the curves y +...

To find point of intersection of two given curves

Put $y = - 2{x}^{2}$ in $y + 3{x}^{2} = 1$

$⇒ {x}^{2} = 1$

$⇒ x = \pm 1$

Let the given equations be ${y}_{1} = 1 - 3{x}^{2}$ and ${y}_{2} = - 2{x}^{2}$.

The desired area would be,

${∫}_{- 1}^{1}\left({y}_{1} - {y}_{2}\right)dx = {∫}_{- 1}^{1}\left(\left(1 - 3{x}^{2}\right) - \left(- 2{x}^{2}\right)\right)dx$

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

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

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

$= \dfrac{4}{3}sq . units$ .