Question

Por favor, alguém pode me ajudar????
O resultado da integral dupla ∫_1^2 ∫_0^1▒(3x –y^2) dx dy

Answer 1

Olá



Regra de integração de polinômios:


$\displaystyle\mathsf{\int x^pdx~=~ \frac{x^{p+1}}{p+1} +C}$

$\displaystyle\mathsf{ \int\limits^2_1 \, \int\limits^0_1 {(3x-y^2)} \, dxdy }$



A ordem de integração está em dx, então as outras variáveis, diferentes de 'x', se tornam constantes.


$\displaystyle\mathsf{ \int\limits^2_1 \, \left[ \int\limits^1_0 {(3x-y^2)} \, dx\right] dy }\\\\\\\\\mathsf{ \int\limits^2_1 \, \left[ \left(\frac{3x^2}{2} -y^2x\right)\bigg|^1_0 \, \right] dy }\\\\\\\\\mathsf{ \int\limits^2_1 \, \left[ \left(\frac{3(1)^2}{2} -y^2(1)\right)~-~\left(\frac{3(0)^2}{2} -y^2(0)\right) \, \right] dy }\\\\\\\\\mathsf{ \int\limits^2_1 { \left(\frac{3}{2} -y^2}\right) \, dy }$



Integrando em dy


$\displaystyle \mathsf{ \int\limits^2_1 { \left(\frac{3}{2} -y^2}\right) \, dy }}\\\\\\\\\mathsf{ \left(\frac{3y}{2}- \frac{y^3}{3}\right)\bigg|^2_1 }\\\\\\\\\mathsf{ \underbrace{\left(\frac{3(2)}{2}- \frac{(2)^3}{3}\right)}_{= \frac{1}{3} } ~-~\underbrace{\left(\frac{3(1)}{2}- \frac{(1)^3}{3}\right)}_{= \frac{7}{6} } }\\\\\\\\\mathsf{ \frac{1}{3}~-~ \frac{7}{6} }\\\\\\\\\boxed{\mathsf{ -\frac{5}{6} }}$