Notez integrala de calculat cu I. Avem
[tex]I=\int_0^3(9x^2-6x^3+x^4)dx=\left(\dfrac{9x^3}{3}-\dfrac{6x^4}{4}+\dfrac{x^5}{5}\right)|_0^3=[/tex]
[tex]=x^3\left(3-\dfrac{3x}{2}+\dfrac{x^2}{5}\right)\Displaystyle|_0^3=27\left(3-\dfrac92+\dfrac95\right)=32,4.[/tex]
