Answer:[tex]\frac{8}{3}\times \sqrt{\frac{2}{5}}[/tex]
Step-by-step explanation:
Given two upward facing parabolas with equations
[tex]y=6x^2 & y=x^2+2[/tex]
The two intersect at
[tex]6x^2=x^2+2[/tex]
[tex]5x^2=2[/tex]
[tex]x^2[/tex]=[tex]\frac{2}{5}[/tex]
x=[tex]\pm \sqrt{\frac{2}{5}}[/tex]
area enclosed by them is given by
A=[tex]\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx[/tex]
A=[tex]\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx[/tex]
A=[tex]4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ][/tex]
A=[tex]\frac{8}{3}\times \sqrt{\frac{2}{5}}[/tex]