Find the area between the graphs of $y=9 x^{2}-x+4$ and $y=3 x^{2}-x+10$. Area $=\square$ (Type an integer or a simplified fraction.)

Step 1 ：Find the points of intersection by setting the two functions equal to each other: \(9x^2 - x + 4 = 3x^2 - x + 10\)

Step 2 ：Solve for x to find the intersection points: \(x = -1, x = 1\)

Step 3 ：Calculate the definite integral of the difference between the two functions from -1 to 1: \(\int_{-1}^{1} (9x^2 - x + 4) - (3x^2 - x + 10) \,dx\)

Step 4 ：The area is the absolute value of the result of the definite integral: \(|\int_{-1}^{1} (9x^2 - x + 4) - (3x^2 - x + 10) \,dx|\)

Step 5 ：The final answer for the area between the graphs is \(\boxed{8}\)