Answer :
Answer:[tex]A=\frac{14}{9}\cdot \sqrt{2}\ unit^2[/tex]
Step-by-step explanation:
Given
Two curves are given
[tex]y=10 x^2[/tex]
and [tex]y=x^2+2[/tex]
the two curves intersect at
[tex]10x^2=x^2+2[/tex]
[tex]x=\pm \frac{\sqrt{2}}{3}[/tex]
to get the we need to integrate the curves over x axis
[tex]A=\int_{-\frac{\sqrt{2}}{3}}^{\frac{\sqrt{2}}{3}}\left ( x^2+3-10x^2\right )dx[/tex]
[tex]A=2\int_{0}^{\frac{\sqrt{2}}{3}}\left ( 3-9x^2\right )dx[/tex]
[tex]A=2\left [ 3\left ( \frac{\sqrt{2}}{3}\right )-\frac{9}{3}\left ( \frac{\sqrt{2}}{3}\right )^3\right ][/tex]
[tex]A=2\sqrt{2}\left [ 1-\frac{2}{9}\right ][/tex]
[tex]A=\frac{14}{9}\cdot \sqrt{2}\ unit^2[/tex]
