Question

Ajutati-ma va rog ,repede

Answer 1

$\int\limits^1_{-1}{f(x)} \, dx = \int\limits^1_{-1} {(3x^{2} -x)} \, dx = 3\int\limits^1_{-1}{x^2} \, dx - \int\limits^1_{-1} {x} \, dx = 3*\frac{x^3}{3} -\frac{x^2}{2} = x^3 -\frac{x^2}{2} intre -1 si 1 = 1-1/2-(-1-1/2) = 2$

Answer 2

$\displaystyle f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=\left \{ {{3x^2-x,~~~~~x \in (-\infty,1]} \atop {\displaystyle 2+\frac{1}{x} }\cdot lnx,~x \in(1,+\infty)} \right. \\ \\ \int\limits_{-1}^1 f(x)dx=2 \\ \\ \int\limits _{-1}^1 f(x)dx=\int\limits_{-1}^1(3x^2-x)dx=\int\limits_{-1}^13x^2dx-\int\limits_{-1}^1xdx=3\int\limits_{-1}^1x^2dx-\int\limits_{-1}^1xdx=$

$\displaystyle =3 \cdot \frac{x^3}{3} \Bigg|_{-1}^1-\frac{x^2}{2} \Bigg|_{-1}^1=3\left(\frac{1^3}{3} -\frac{(-1)^3}{3} \right)-\left(\frac{1^2}{2} -\frac{(-1)^2}{2} \right)=\\ \\\\ =3 \cdot \frac{1+1}{3} -\frac{1-1}{2} =3 \cdot \frac{2}{3} -\frac{0}{2} =\frac{6}{3} =\mathbf{2}$