Integral dupla
Enviado: Ter 04 Dez, 2018 23:14
calcule a integral, sendo B={(x,y)∈R^2 |x≥0, x^5-x≤y≤0}
[tex3]\int\limits_{}^{}\int\limits_{B}^{} y dxdy[/tex3]
[tex3]\int\limits_{}^{}\int\limits_{B}^{} y dxdy[/tex3]
Solução:
x [tex3]^{5}[/tex3] - x = 0
x.( x⁴ - 1 ) = 0
x = 0 ou x⁴ - 1 = 0 → x⁴ = 1 → x = 1 , x = - 1 , x = i ou x = - i
Graficamente
- 15441172192736704773080512356970.jpg (42.2 KiB) Exibido 516 vezes
Analisando o gráfico acima, resulta que ;
B = { ( x , y )∈R² / 0 ≤ x ≤ 1 , x [tex3]^{5}[/tex3] - x ≤ y ≤ 0 }
Então;
[tex3]I=\int\limits_{0}^{1}\int\limits_{x^5-x}^{0} y \ dydx[/tex3]
[tex3]I=\int\limits_{0}^{1}[ \frac{y^2}{2}]_{x^5-x}^{0}\ dx[/tex3]
[tex3]I=\int\limits_{0}^{1}[ \frac{0^2}{2}-\left(\frac{(x^5-x)^2}{2}\right)]\ dx[/tex3]
[tex3]I=\int\limits_{0}^{1} -\left(\frac{(x^{10}-2x^6+x^2)}{2}\right)\ dx[/tex3]
[tex3]I=-\frac{1}{2}.\int\limits_{0}^{1} (x^{10}-2x^6+x^2)\ dx[/tex3]
[tex3]I=-\frac{1}{2}.[\frac{x^{11}}{11}-\frac{2x^7}{7}+\frac{x^3}{3}]_{0}^{1}[/tex3]
[tex3]I=-\frac{1}{2}.[\frac{1^{11}}{11}-\frac{2.1^7}{7}+\frac{1^3}{3} - (\frac{0^{11}}{11}-\frac{2.0^7}{7}+\frac{0^3}{3})][/tex3]
[tex3]I=-\frac{1}{2}.(\frac{1}{11}-\frac{2}{7}+\frac{1}{3})[/tex3]
[tex3]I=-\frac{1}{2}.(\frac{21-66+77}{231})[/tex3]
[tex3]I=-\frac{1}{2}.(\frac{32}{231})=-\frac{16}{231}[/tex3]
Portanto, [tex3]I=\int\limits_{0}^{1}\int\limits_{x^5-x}^{0} y \ dydx=-\frac{16}{231}[/tex3]
Bons estudos!