let give f(x)=(√(x+y)) +1 and D={(x,y)∈R^2 / 0≤x≤1 and −1≤y≤1} find the value of ∫∫ f(x,y)dxdy .


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Question Number 28038 by abdo imad last updated on 19/Jan/18
$let give f(x)=(√(x+y)) +1 and D={(x,y)∈R^2 / 0≤x≤1 and −1≤y≤1} find the value of ∫∫ f(x,y)dxdy .$
$l e t g i v e f (x) = \sqrt{x + y} + 1 a n d D = {(x, y) \in R^{2} / 0 ⩽ x ⩽ 1$ $a n d - 1 ⩽ y ⩽ 1} f i n d t h e v a l u e o f \int \int f (x, y) d x d y .$
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