calculate ∫∫_D (x+y)e^(x+y) dxdy with D = {(x,y)∈R^2 / 0<x<2 and 1<y<2 }


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Question Number 36190 by prof Abdo imad last updated on 30/May/18
$calculate ∫∫_D (x+y)e^(x+y) dxdy with D = {(x,y)∈R^2 / 0<x<2 and 1<y<2 }$
$c a l c u l a t e \int \int_{D} (x + y) e^{x + y} d x d y w i t h$ $D = {(x, y) \in R^{2} / 0 < x < 2 a n d 1 < y < 2}$
Commented bymaxmathsup by imad last updated on 20/Aug/18
$I =∫_1 ^2 (∫_0 ^2 (x+y)e^(x+y) dx)dy =∫_1 ^2 A(y)dy with A(y)=∫_0 ^2 (x+y)e^(x+y) dx A(y) = e^(y ) ∫_0 ^2 (x+y)e^x dx =e^y ∫_0 ^2 (xe^x +y e^x )ex =e^y { ∫_0 ^2 xe^x dx +y ∫_0 ^2 e^x dx} but ∫_0 ^2 e^x dx =[e^x ]_0 ^2 =e^2 −1 and by parts ∫_0 ^2 x e^x dx =[x e^x ]_0 ^2 −∫_0 ^2 e^x dx =2e^2 −e^2 +1 =e^2 +1 ⇒A(y)=e^y (e^2 +1+y(e^2 −1)) =(e^2 +1)e^y +(e^2 −1)y e^y ⇒ I = ∫_1 ^2 { (e^2 +1)e^y +(e^2 −1)ye^y )dy =(e^2 +1)∫_1 ^2 e^y dy +(e^2 −1) ∫_1 ^2 y e^y dy but ∫_1 ^2 e^y dy =e^2 −e and ∫_1 ^2 y e^y dy =[y e^y ]_1 ^2 −∫_1 ^2 e^y dy =2e^2 −e −e^2 +e =e^2 ⇒ I =(e^2 +1)(e^2 −e) +(e^2 −1)e^2 I =e^4 −e^3 +e^2 −e +e^4 −e^2 ⇒ I =2e^(4 ) −e^3 .$
$I = \int_{1}^{2} (\int_{0}^{2} (x + y) e^{x + y} d x) d y = \int_{1}^{2} A (y) d y w i t h A (y) = \int_{0}^{2} (x + y) e^{x + y} d x$ $A (y) = e^{y} \int_{0}^{2} (x + y) e^{x} d x = e^{y} \int_{0}^{2} ({x e}^{x} + y e^{x}) e x$ $= e^{y} {\int_{0}^{2} {x e}^{x} d x + y \int_{0}^{2} e^{x} d x} b u t$ $\int_{0}^{2} e^{x} d x = {[e^{x}]}_{0}^{2} = e^{2} - 1 a n d b y p a r t s$ $\int_{0}^{2} x e^{x} d x = {[x e^{x}]}_{0}^{2} - \int_{0}^{2} e^{x} d x = 2 e^{2} - e^{2} + 1 = e^{2} + 1 \Rightarrow A (y) = e^{y} (e^{2} + 1 + y (e^{2} - 1))$ $= (e^{2} + 1) e^{y} + (e^{2} - 1) y e^{y} \Rightarrow$ $I = \int_{1}^{2} {(e^{2} + 1) e^{y} + (e^{2} - 1) {y e}^{y}) d y$ $= (e^{2} + 1) \int_{1}^{2} e^{y} d y + (e^{2} - 1) \int_{1}^{2} y e^{y} d y b u t$ $\int_{1}^{2} e^{y} d y = e^{2} - e a n d \int_{1}^{2} y e^{y} d y = {[y e^{y}]}_{1}^{2} - \int_{1}^{2} e^{y} d y$ $= 2 e^{2} - e - e^{2} + e = e^{2} \Rightarrow I = (e^{2} + 1) (e^{2} - e) + (e^{2} - 1) e^{2}$ $I = e^{4} - e^{3} + e^{2} - e + e^{4} - e^{2} \Rightarrow I = 2 e^{4} - e^{3} .$
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