[tex] \displaystyle I =\int\limits_{-1}^1(1+2x^{2015})e^{-|x|}\,dx \\ \\ \text{Primul pas este s\u{a} o imp\u{a}r\c{t}im in dou\u{a}:} \\ \\ I = \int\limits_{-1}^1 e^{-|x|}\, dx + \int \limits_{-1}^1{2x^{2015}e^{-|x|}}\,dx\\ \\ [/tex]
[tex] \text{Observ\u{a}m c\u{a} aceast\u{a} expresie este o sum\u{a} de o func\c{t}ie par\u{a} }(e^{-|x|}) \\ \text{\c{s}i o func\c{t}ie impar\u{a} } (2x^{2015}e^{-|x|}).[/tex]
[tex]\text{Noi avem:}\\ \\ \displaystyle I=\int\limits_{-L}^L\underset{functie~ para}{\underbrace{f(x)}}\,dx + \int\limits_{-L}^L \underset{functie~ impara}{\underbrace{g(x)}}\,dx\\ \\ \text{\c{S}tim c\u{a}:} \\ \\ \bullet ~ \text{Pentru o func\c{t}ie par\u{a}:} \\ \int\limits_{-L}^Lf(x) \, dx = 2\int\limits_{0}^{L}f(x) \, dx, \quad \text{iar in cazul nostru:} \\ \\ \Rightarrow \int\limits_{-1}^{1}e^{-|x|}\, dx = 2\int\limits_{0}^{1} e^{-|x|}\, = 2\int\limits_{0}^{1} e^{-x}\,dx, ~~\text{am sc\u{a}pat de modul deoarece} [/tex]
[tex] \displaystyle\text{intervalul de integrare este pozitiv: }[0,1]. \\ = 2\int\limits_{0}^1}(-e^{-x})'\,dx = 2\cdot (-e^{-x})\Big|_{0}^1 = 2\Big(1-\dfrac{1}{e}\Big) \\ \\ \bullet \text{Pentru o functie impar\u{a}:} \\ \int\limits_{-L}^L g(x) \, dx = 0, \quad \text{iar in cazul nostru:} \\ \\ \Rightarrow \int\limits_{-1}^{1} 2x^{2015}e^{-|x|} \, dx = 0 \\ \\ \\ \Rightarrow I = \int\limits_{-1}^1 {e^{-|x|}}\, dx + \int\limits_{-1}^1 {2x^{2015}e^{-|x|}}\, dx [/tex]
[tex] \Rightarrow I = 2\cdot \Big(1-\dfrac{1}{e}\Big) + 0\\ \\ \Rightarrow \boxed{I = \dfrac{2(e-1)}{e}} [/tex]
