Calculate the value of
[tex]\int (x^3-3x)dx[/tex][tex]\int (x^3-3x)dx=\frac{x^4}{4}-\frac{3x^2}{2}[/tex]
Apply the limits implies,
[tex]\begin{gathered} \int ^{\frac{1}{2}}_{-1}(x^3-3x)dx=\lbrack\frac{x^4}{4}-\frac{3x^2}{2}\rbrack^{\frac{1}{2}}_{-1} \\ =\frac{1}{64}-\frac{3}{8}-\frac{1}{4}+\frac{3}{2} \\ =\frac{57}{64} \end{gathered}[/tex]
Therefore, the answer is 0.89.
Nearest integer is 1.
