Mudança de variável na integral
Enviado: 21 Dez 2021, 18:10
Suponha [tex3]f[/tex3]
contínua em [tex3][-1,1][/tex3]
. Calcule [tex3]\int\limits_{0}^{1}f(2x-1)dx[/tex3]
sabendo que [tex3]\int\limits_{-1}^{1}f(u)du=5[/tex3]
.
5/2
Resposta
5/2
[tex3]\mathsf{2 \cdot x \ - \ 1 \ = \ u \ \rightarrow \ 2 \ = \ \dfrac{du}{dx} \ \rightarrow \ du \ = \ \dfrac{dx}{2}}[/tex3]
[tex3]\mathsf{0 \ \leq \ x \ \leq \ 1 \ \therefore \ -1 \ \leq \ \underbrace{2\cdot x \ - \ 1}_{u} \ \leq \ 1}[/tex3]
[tex3]\mathsf{\int_{0}^{1} \ f(2\cdot x \ - 1) \ dx \ = \ \int_{-1}^{1} \ f(u) \ \dfrac{du}{2}}[/tex3]
[tex3]\mathsf{= \ \dfrac{\overbrace{\int_{-1}^{1} \ f(u) \ du}^{5}}{2} \ = \ \boxed{\mathsf{\dfrac{5}{2}}}}[/tex3]