Question

calcular a integral dupla 2x-y no intervalo 0e1 e1e2

Answer 1

Olá
resolvendo:

$\int\limits^1_0 \int\limits^2_1 {(2x-y)} \, dy {} \, dx \\ \\ integrando (y) temos: \\ \\ \int\limits^1_0 {(2xy- \frac{ y^{2} }{2}) } \left[\begin{array}{ccc}2\\\\1\end{array}\right \, dx \\ \\ substituindo..em..(y)..temos: \\ \\ \int\limits^1_0 {(2x.2- \frac{ 2^{2} }{2}-(2x.1- \frac{ 1^{2} }{2} )) } \, dx \\ \\ \int\limits^1_0 {(4x-2-2x+ \frac{1}{2} )} \, dx \\ \\ \int\limits^1_0 {(2x- \frac{3}{2}) } \, dx$
$( \frac{2 x^{2} }{2}- \frac{3}{2} x) \left[\begin{array}{ccc}1&&\\&&\\0\end{array}\right \\ \\ ( x^{2} - \frac{3}{2} x ) \left[\begin{array}{ccc}1&&\\&&\\0&&\end{array}\right \\ \\ substituindo(x) \\ \\ \left[\begic}\\\\ 1^{2}- \frac{3}{2}.1-( 0^{2} - \frac{3}{2} .0) }\right] \\ \\ 1- \frac{3}{2} \\ \\ - \frac{1}{2} ...Resposta$

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