Question

puteti sami explicati integrala asta nu inteleg de ce la sfarsit apare doar -y si nu -(x+y)$\int\limits^1_0 ( \int\limits^2_1 {lnx+y} \, dy )dx= \int\limits^1_0 {(x+y)ln(x+y)-y} (pentru/y/ de/ la 1 la 2), dx$

Answer 1

$\int\limits^2_1 {\ln(x+y)} \, dy = \int\limits^2_1 {(x+y)'_y\cdot \ln(x+y)} \, dy = \\ \\ = \big[(x+y)\cdot \ln (x+y)\big]\Big|_1^2 - \int\limits^2_1 {(x+y)\cdot \big[\ln(x+y)\big]'_y} \, dy = \\ \\ =\big[(x+y)\cdot \ln (x+y)\big]\Big|_1^2 - \int\limits^2_1 {(x+y)\cdot \dfrac{(x+y)'_y}{x+y}} \, dy = \\ \\ =\big[(x+y)\cdot \ln (x+y)\big]\Big|_1^2 - \int\limits^2_1 {(x+y)\cdot \dfrac{(0+1)}{x+y}} \, dy =$ 
$=\big[(x+y)\cdot \ln (x+y)\big]\Big|_1^2-\int\limits^2_1 {(x+y)\cdot \dfrac{1}{x+y}} \, dy=$
$= \big[(x+y)\cdot \ln (x+y)\big]{\Big|_1^2} - \int\limits^2_1 {1}\, dy = \\ \\ = \big[(x+y)\cdot \ln (x+y)\big]\Big|_1^2-(y)\Big|_1^2 = \\ \\ = \Big[(x+y)\cdot \ln(x+y)-y\Big]\Big|_1^2 {} \quad_{(in~ functie~ de ~y)}$