Respuesta :

[tex]\bf \displaystyle \int\limits_{-4}^4\ \left[ f(x)= \begin{cases} 2&-4\le x\le 0\\ 0&\ \textless \ x\le 4 \end{cases} \right] \cdot dx\\\\ -------------------------------\\\\[/tex]

[tex]\bf \displaystyle \int\limits_{-4}^0\ 2\cdot dx+\int\limits_{0}^4\ (5-x^2)\cdot dx \\\\\\ \displaystyle \int\limits_{-4}^0\ 2\cdot dx+\int\limits_{0}^4\ 5\cdot dx-\int\limits_{0}^4\ x^2\cdot dx\implies \left. 2x \cfrac{}{} \right]_{-4}^0 +\quad \left. 5x \cfrac{}{}\right]_{0}^4-\quad \left. \cfrac{x^3}{3}\right]_{0}^4 \\\\\\ 8+20-\cfrac{64}{3}\implies \cfrac{20}{3}[/tex]

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