[m]=(\frac{x^3}{3})|^{5} _{1}-6\cdot (\frac{x^2}{2})|^{5} _{1}+5(x)|^{5} _{1}=\frac{5^3}{3}-\frac{1^3}{3}-6\cdot ( \frac{5^2}{2}-\frac{1^2}{2})+5\cdot (5-1)=\frac{124}{3}-6\cdot 12+5\cdot 4=-\frac{32}{3}[/m]
Второй способ:
[m] ∫^{5} _{1}((x-3)^2-4)dx=∫^{5} _{1}(x-3)^2dx-4∫^{5} _{1}dx=(\frac{(x-3)^3}{3})|^{5} _{1}-4(x)|^{5} _{1}=\frac{(5-3)^3}{3}-\frac{(1-3)^3}{3}-4(5-1)=\frac{8}{3}-(-\frac{8}{3})-4\cdot 4=\frac{16}{3}-16=\frac{32}{3}[/m]