y=x/4=x^{3} where the line intersects the curve.

4x^{3}-x=0, x(4x^{2}-1)=0, so x=0 x=±½. y=±⅛.

We are only interested in the bound region which includes (½,⅛).

The region can be split into two, 0≤x≤½ and ½≤x≤9 where different integrals apply. As it happens the second integral is an area which can be calculated easily using geometry because the region is bounded here by straight lines. The first region is small and calculus is needed to evaluate its area.

x(4x^{2}-1)<0 between x=0 and x=½, so in this region the curve is closer to the x-axis, hence the integral ∫x^{3}dx=[x^{4}/4]_{0}^{½}=1/64 square units.

x(4x^{2}-1)>0 between x=½ and x=9, so in this region the line is closer to the x-axis, hence the integral ∫(x/4)dx=[x^{2}/8]_{½}^{9}=⅛(81-¼)=323/32 square units.

(Geometrically we have a long thin rectangle with dimension [⅛×(9-½)=17/16] + a triangle [½×(17/2)(9/4-1/8)=289/32]=323/32 square units.)

Now add the region areas together: 1/64+323/32=647/64 square units.