Question

Cine ma ajuta cu integrala asta ?

Answer 1

$\displaystyle I =\int_{-2019}^{2019} \dfrac{2019^x-2019^{-x}}{1+ x^{2020}}\, dx \\ \\ \\ \text{Proprietate a integralei de}\text{finite:} \\ \\ \int_{a}^bf(x) \,dx = \int_{a}^b f(a+b-x)\, dx \\ \\\\\Rightarrow I = \int_{-2019}^{2019}\dfrac{2019^{-2019+2019 - x}-2019^{-(-2019+2019-x)}}{1+(-2019+2019-x)^{2020}}\, dx \\ \\ \\\Rightarrow I =\int_{-2019}^{2019} \dfrac{2019^{-x}-2019^{x}}{1+ x^{2020}}\, dx \\ \\ \\ \text{Dar }I =\int_{-2019}^{2019} \dfrac{2019^x-2019^{-x}}{1+ x^{2020}}\, dx$

$\\ \\ \Rightarrow I = -I \\\\ \Rightarrow 2I = 0 \\ \\ \Rightarrow \boxed{I = 0}$