To find the area enclosed by,
y = 2x2 ...(i)
And y = x2 + 4 ...(ii)
On solving the equation (i) and (ii),
2x2 = x2 + 4
Or x2 = 4
Or x = \(\pm 2\)
\(\therefore\) y = 8
Equation (1) represents a parabola with vertex (0, 0) and axis as y - axis.
Equation (2) represents a parabola with vertex (0,4) and axis as the y - axis.
Points of intersection of parabolas are A(2, 8) and B(– 2, 8).
These are shown in the graph below: -
Required area = Region AOBCA
= 2(Region AOCA)
Hence, proved that the area common to the two parabolas y = 2x2 and y = x2 + 4 is 32/3 sq. Units.