Question

A integral dupla integral de 0 até 3 integral de1 até 2 xy/ 2 +y² dydx vale:

Answer 1

Resposta:

Ok, vamos lá:

$\int\limits^3_0\int\limits^2_1 {\frac{xy}{2} +y^2} dydx\\\int\limits^3_0( \int\limits^2_1{\frac{xy}{2} dy +\int\limits^2_1 y^2} dy ) dx\\\int\limits^3_0( {\frac{x}{2}\int\limits^2_1y \; dy +\int\limits^2_1 y^2} dy ) dx\\\int\limits^3_0( {\frac{x}{2}*\frac{y^2}{2}\int\limits^2_1 dy +\frac{y^3}{3}\int\limits^2_1} dy ) dx\\\int\limits^3_0( {\frac{x}{2}*\frac{3}{2} +\frac{7}{3} ) dx\\\int\limits^3_0 {\frac{3x}{4} +\frac{7}{3} dx$

Pronto, a segunda derivada fá foi resolvida, agora resta apenas a primeira:

$\int\limits^3_0 {\frac{3x}{4} +\frac{7}{3} dx\\\int\limits^3_0 {\frac{3x}{4} dx +\int\limits^3_0\frac{7}{3} dx$

$\frac{3}{4} \int\limits^3_0 x\;dx + \frac{7}{3}x\int\limits^3_0\\\frac{3}{4} * \frac{x^2}{2} \int\limits^3_0 + \;\frac{7}{3}x\int\limits^3_0\\=\frac{3}{4} * \frac{9}{2} + \frac{7}{3}*3\\\\=\frac{27}{8} + 7\\\\=\frac{83}{8}\\\\= 10.375$

Answer 2

Resposta:

$1,55958...$

Explicação passo-a-passo:

$\int _0^3\int _1^2\frac{xy}{2+y^2}dydx$

$=\int _0^3x\frac{1}{2}\left(\ln \left(6\right)-\ln \left(3\right)\right)dx$

$=\frac{9}{4}\ln \left(2\right)$