Answer
It can be rewritten as
$$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx$$.
Work Step by Step
In the problem, it is asked only to rewrite the integral as the difference of two integrals, not to evaluate it. By the difference rule given in table 5.4 we have
$$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx.$$
