
To find point of intersection of two given curves
Put y=−2x2 in y+3x2=1
⇒x2=1
⇒x=±1
Let the given equations be y1=1−3x2 and y2=−2x2.
The desired area would be,
∫−11(y1−y2)dx=∫−11((1−3x2)−(−2x2))dx
=∫−11(1−x2)dx
=(x−3x3)−11=((1−31)−(−1+31))
=32−(3−2)
=34sq.units .