Aplicind formula lui newton L . calculati .: clasa 12 Rog mult Multumesc

Răspuns:

[tex]\int\limits^2_ {-2} \(x-2)(x+1), dx[/tex]=

[tex]\int\limits^2_ {-2} \,(x^2-x-2) dx[/tex]=║[tex]\int\limits^2_ {-2} x^2dx -\int\limits^2_ {-2} xdx+\int\limits^2_ {-2}-2 \, dx =[/tex]

[tex]\frac{x^3}{3}[/tex]║₋₂²-x²/2║₋₂²-2x║₋₂²=

2³/3-(-2)³/3-(2²/2-(-2)²/2-2(2-(-2))=

8/3+8/3-(4/2-4/2)-2*4=

16/3-0-8=

(16-24)/3= -8/3

Explicație pas cu pas:

[tex]\displaystyle \int_{-2}^2 (x-2)(x+1)\, dx= \int_{-2}^2(x^2-x-2)\, dx = \\ \\ = \int_{-2}^2 \left[\left(x-\frac{1}{2}\right)^2-\dfrac{9}{4}\right]dx = \int_{-2}^2\left(x-\frac{1}{2}\right)^2 dx-\int_{-2}^2 \dfrac{9}{4}\, dx=\\ \\ = \dfrac{\left(x-\frac{1}{2}\right)^{3}}{3}\Bigg|_{-2}^{2}-\dfrac{9}{4}x\Big|_{-2}^2 =\dfrac{(2-\frac{1}{2})^3}{3}-\dfrac{(-2-\frac{1}{2})^3}{3}-\dfrac{9}{4}\cdot 2+\dfrac{9}{4}\cdot (-2)=\\ \\ = \dfrac{27}{8\cdot 3}+\dfrac{125}{8\cdot 3}-9 = \dfrac{27+125-216}{24} = -\dfrac{64}{24} = \boxed{-\dfrac{8}{3}}[/tex]