I would start by stating the Fundamental Theorem of Calculus which states that;
If a function f is continuous on a closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then
[tex]\int ^b_af(x)dx=F(b)-F(a)\text{ }[/tex]
Let
[tex]\begin{gathered} f(x)=x^3-6x^{} \\ F(x)=\int f(x)dx=\int (x^3-6x)dx \end{gathered}[/tex]
Recall that;
[tex]\int x^n=\frac{x^{n+1}}{n+1},n\ne-1[/tex]
That implies that,
[tex]F(x)=\int (x^3-6x)dx=\int x^3dx-\int 6xdx=\frac{x^4}{4}-6(\frac{x^2}{2})=\frac{x^4}{4}-3x^2+C[/tex]
Applying the Fundamental Theorem of Calculus, where a=0, b=3
[tex]\begin{gathered} \int ^3_0(x^3-6x)dx=F(3)-F(0) \\ F(3)=\frac{3^4}{4}-3(3)^2+C=\frac{81}{4}-27+C=-\frac{27}{4}+C \\ F(0)=\frac{0^4}{4}-3(0)^2+C=C \\ \Rightarrow\int ^3_0(x^3-6x)dx=-\frac{27}{4}+C-C=-\frac{27}{4} \end{gathered}[/tex]
So the answer is -27/4
