Consider the attached figure (note that the axis are not in a 1:1 scale).
We want the (finite) area between the two parabolas, where [tex] y = 6x^2 [/tex] is the green one, and [tex] y = x^2+10 [/tex] is the red one.
We need the following result: given a function f, the integral
[tex] \displaystyle \int_a^b f(x)\; dx [/tex]
returns the area between the function and the x axis.
So, if we subtract the area below the green function between points A and B from the area below the red function between points A and B, we have exactly the area between the two. So, we need the x coordinates of points A and B.
Since they are the points of interception between the two parabolas, we can compute them by solving the equation
[tex] 6x^2 = x^2+10 \implies 5x^2=10\implies x^2=2 \implies x=\pm\sqrt{2} [/tex]
So, we need
[tex] \displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (x^2+10)-6x^2 = \int_{-\sqrt{2}}^{\sqrt{2}} -5x^2+10 [/tex]
The primitive of this function is
[tex] F(x) = -\cfrac{5x^3}{3} + 10x + C [/tex]
And we need to compute it at the two endpoints: the area is given by
[tex] A = F(\sqrt{2}) - F(-\sqrt{2}) = \cfrac{20\sqrt{2}}{3} - (- \cfrac{20\sqrt{2}}{3}) = \cfrac{40\sqrt{2}}{3} [/tex]
