a função f (x) = 5x/(x²+3) representa o comportamento de um escoamento unidimensional de um fluido ideal, num intervalo -1, 1
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Boa noite Rafa!
Solução!
Escoamento ideal ou escoamento sem atrito, é aquele no qual não existem tensões de escoamento satisfazendo as certa condições, é chamado de fluido ideal.
Vamos usar a substituição para resolvermos a integral!
![\displaystyle \int_{-1}^{1} \frac{5x}{ x^{2}+3} dx \\\\\\\\\\\
5\displaystyle \int_{-1}^{1} \frac{x}{ x^{2}+3} dx \\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint_%7B-1%7D%5E%7B1%7D+%5Cfrac%7B5x%7D%7B+x%5E%7B2%7D%2B3%7D+dx+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A5%5Cdisplaystyle+%5Cint_%7B-1%7D%5E%7B1%7D+%5Cfrac%7Bx%7D%7B+x%5E%7B2%7D%2B3%7D+dx+%5C%5C%5C%5C%5C%5C%5C+ "\displaystyle \int_{-1}^{1} \frac{5x}{ x^{2}+3} dx \\\\\\\\\\\
5\displaystyle \int_{-1}^{1} \frac{x}{ x^{2}+3} dx \\\\\\\")
![\displaystyle \int \frac{1}{2u}du\\\\\\\\\
\frac{1}{2}\displaystyle \int \frac{1}{u}du\\\\\\\\
Lembrando~~de~~ \frac{1}{x}=lnx\\\\\\\
\frac{1}{2}\displaystyle \int \frac{ln u}{2} \\\\\\\\\
Substituindo!\\\\\\\
5I= \frac{ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\
I= \frac{5ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint+%5Cfrac%7B1%7D%7B2u%7Ddu%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cfrac%7B1%7D%7B2%7D%5Cdisplaystyle+%5Cint+%5Cfrac%7B1%7D%7Bu%7Ddu%5C%5C%5C%5C%5C%5C%5C%5C%0ALembrando%7E%7Ede%7E%7E+%5Cfrac%7B1%7D%7Bx%7D%3Dlnx%5C%5C%5C%5C%5C%5C%5C%0A+%5Cfrac%7B1%7D%7B2%7D%5Cdisplaystyle+%5Cint++%5Cfrac%7Bln+u%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%0ASubstituindo%21%5C%5C%5C%5C%5C%5C%5C%0A5I%3D+%5Cfrac%7Bln+%28+x%5E%7B2%7D+%2B3%29%7D%7B2%7D%5Cbigg%7C_%7B-1%7D%5E%7B1%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cfrac%7B5ln+%28+x%5E%7B2%7D+%2B3%29%7D%7B2%7D%5Cbigg%7C_%7B-1%7D%5E%7B1%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0A%0A "\displaystyle \int \frac{1}{2u}du\\\\\\\\\
\frac{1}{2}\displaystyle \int \frac{1}{u}du\\\\\\\\
Lembrando~~de~~ \frac{1}{x}=lnx\\\\\\\
\frac{1}{2}\displaystyle \int \frac{ln u}{2} \\\\\\\\\
Substituindo!\\\\\\\
5I= \frac{ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\
I= \frac{5ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\")
![I= \dfrac{5ln ( x^{2} +3)}{2}-\dfrac{5ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\
I= \dfrac{5ln ( 1^{2} +3)}{2}-\dfrac{5ln ( (-1)^{2} +3)}{2}\\\\\\\\
I= \dfrac{5ln ( 1 +3)}{2}-\dfrac{5ln ( 1 +3)}{2}\\\\\\\
I= \dfrac{5ln ( 4)}{2}-\dfrac{5ln ( 4)}{2}\\\\\\\
I= 5ln2-5ln2\\\\\\\
\boxed{I=0}](https://tex.z-dn.net/?f=I%3D+%5Cdfrac%7B5ln+%28+x%5E%7B2%7D+%2B3%29%7D%7B2%7D-%5Cdfrac%7B5ln+%28+x%5E%7B2%7D+%2B3%29%7D%7B2%7D%5Cbigg%7C_%7B-1%7D%5E%7B1%7D+%5C%5C%5C%5C%5C%5C%5C%5C+%0AI%3D+%5Cdfrac%7B5ln+%28+1%5E%7B2%7D+%2B3%29%7D%7B2%7D-%5Cdfrac%7B5ln+%28+%28-1%29%5E%7B2%7D+%2B3%29%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cdfrac%7B5ln+%28+1+%2B3%29%7D%7B2%7D-%5Cdfrac%7B5ln+%28+1+%2B3%29%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0AI%3D+%5Cdfrac%7B5ln+%28+4%29%7D%7B2%7D-%5Cdfrac%7B5ln+%28+4%29%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0AI%3D+5ln2-5ln2%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BI%3D0%7D%0A%0A%0A "I= \dfrac{5ln ( x^{2} +3)}{2}-\dfrac{5ln ( x^{2} +3)}{2}\bigg|_{-1}^{1} \\\\\\\\
I= \dfrac{5ln ( 1^{2} +3)}{2}-\dfrac{5ln ( (-1)^{2} +3)}{2}\\\\\\\\
I= \dfrac{5ln ( 1 +3)}{2}-\dfrac{5ln ( 1 +3)}{2}\\\\\\\
I= \dfrac{5ln ( 4)}{2}-\dfrac{5ln ( 4)}{2}\\\\\\\
I= 5ln2-5ln2\\\\\\\
\boxed{I=0}")
Justificativa:
Como o escoamento do liquido ideal logo seu resultado nesse intervalo é zero pois não apresenta tensão atrito ou viscosidade.
Boa noite!
Bons estudos!
Solução!
Escoamento ideal ou escoamento sem atrito, é aquele no qual não existem tensões de escoamento satisfazendo as certa condições, é chamado de fluido ideal.
Vamos usar a substituição para resolvermos a integral!
Justificativa:
Como o escoamento do liquido ideal logo seu resultado nesse intervalo é zero pois não apresenta tensão atrito ou viscosidade.
Boa noite!
Bons estudos!
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