The given integral is
[tex]\int ^3_0(x^3-6x)dx[/tex]
Recall the fundamental theorem of calculus,
[tex]\int ^b_af(x)dx=F(b)-F(a)[/tex]
The given intergral is
[tex]\int ^b_af(x)dx=\int ^3_0(x^3-6x)dx[/tex][tex]F(x)=\int f(x)dx=\int (x^3-6x)dx[/tex]
Integrating with respect to x, we get
[tex]F(x)=\frac{x^4}{4}-\frac{6x^2}{2}[/tex]
The antiderivative of f is
[tex]F(x)=\frac{x^4}{4}-3x^2[/tex]
Replace x=3 in F(x) to compute F(3).
[tex]F(3)=\frac{(3)^4}{4}-3(3)^2=20.25-27=6.75[/tex]
Replace x=0 in F(x) to compute F(0).
[tex]F(0)=\frac{(0)^4}{4}-3(0)^2=0[/tex]
The given integral is
[tex]\int ^3_0(x^3-6x)dx=F(3)-F(0)[/tex]
Substitute F(3)=6.75 and F(0)=0, we get
[tex]\int ^3_0(x^3-6x)dx=6.75-0[/tex]
Hence the answer is
[tex]\int ^3_0(x^3-6x)dx=6.75=6\frac{3}{4}[/tex]
