calculate ∫∫_(0≤x≤1 ,0≤y≤1) (x+2y)e^(2x−y) dxdy


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Question Number 43539 by abdo.msup.com last updated on 11/Sep/18

$c a l c u l a t e \int \int_{0 ⩽ x ⩽ 1, 0 ⩽ y ⩽ 1} (x + 2 y) e^{2 x - y} d x d y$
Commented by maxmathsup by imad last updated on 16/Sep/18
$let consider the diffeomorphisme (u,v) →ϕ(u,v)=(ϕ_1 (u,v),ϕ_2 (u,v))= (x,y) /u=x+2y and v=2x−y ⇒ { ((x+2y=u)),((2x−y=v ⇒ { ((2x+4y=2u)),((2x−y=v)) :})) :} ⇒ { ((5y=2u−v)),((x=u−2(((2u−v)/5)) ⇒ { ((y=(2/5)u−(1/5)v)),((x=(1/5)u +(2/5)v)) :})) :} ⇒ϕ(u,v)=(ϕ_1 ,ϕ_2 )=(x,y)=((1/5)u+(2/5)v,(2/5)u−(1/5)v)⇒ M_j (ϕ) = ((((∂ϕ_1 /∂u) (∂ϕ_1 /∂v))),(((∂ϕ_2 /∂u) (∂ϕ_2 /∂v))) ) = ((( (1/5) (2/5))),(((2/5) −(1/5))) ) and detM_j =−(1/(25)) −(4/(25)) =−(1/(25)) we have 0≤x≤1 and 0≤2y≤2 ⇒0≤x+2y≤3 ⇒0≤u≤3 0≤2x≤2 and −1≤−y≤0 ⇒−1≤2x−y≤ 2 ⇒−1≤v≤2 ⇒ ∫∫f(x,y)dxdy = ∫∫_w foϕ(u,v) ∣detM_j (ϕ)∣ du dv =∫∫_(0≤u≤3 and −1≤v≤2) u e^v (1/(25)) du dv =(1/(25)) ∫_0 ^3 udu .∫_(−1) ^2 e^v dv =(1/(25))[ (u^2 /2)]_0 ^3 .[ e^v ]_(−1) ^2 =(1/(50))(9)( e^2 −e^(−1) ) =(9/(50)){ e^2 −(1/e)} .$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m e (u, v) \to φ (u, v) = (φ_{1} (u, v), φ_{2} (u, v)) =$ $(x, y) / u = x + 2 y a n d v = 2 x - y \Rightarrow {\begin{cases} x + 2 y = u \\ 2 x - y = v \Rightarrow {\begin{cases} 2 x + 4 y = 2 u \\ 2 x - y = v \end{cases} \end{cases}$ $\Rightarrow {\begin{cases} 5 y = 2 u - v \\ x = u - 2 (\frac{2 u - v}{5}) \Rightarrow {\begin{cases} y = \frac{2}{5} u - \frac{1}{5} v \\ x = \frac{1}{5} u + \frac{2}{5} v \end{cases} \end{cases}$ $\Rightarrow φ (u, v) = (φ_{1}, φ_{2}) = (x, y) = (\frac{1}{5} u + \frac{2}{5} v, \frac{2}{5} u - \frac{1}{5} v) \Rightarrow$ $M_{j} (φ) = (\begin{matrix} \frac{\partial φ_{1}}{\partial u} \frac{\partial φ_{1}}{\partial v} \\ \frac{\partial φ_{2}}{\partial u} \frac{\partial φ_{2}}{\partial v} \end{matrix}) = (\begin{matrix} \frac{1}{5} \frac{2}{5} \\ \frac{2}{5} - \frac{1}{5} \end{matrix}) a n d {d e t M}_{j} = - \frac{1}{25} - \frac{4}{25} = - \frac{1}{25}$ $w e h a v e 0 ⩽ x ⩽ 1 a n d 0 ⩽ 2 y ⩽ 2 \Rightarrow 0 ⩽ x + 2 y ⩽ 3 \Rightarrow 0 ⩽ u ⩽ 3$ $0 ⩽ 2 x ⩽ 2 a n d - 1 ⩽ - y ⩽ 0 \Rightarrow - 1 ⩽ 2 x - y ⩽ 2 \Rightarrow - 1 ⩽ v ⩽ 2 \Rightarrow$ $\int \int f (x, y) d x d y = \int \int_{w} f o φ (u, v) ∣ {d e t M}_{j} (φ) ∣ d u d v$ $= \int \int_{0 ⩽ u ⩽ 3 a n d - 1 ⩽ v ⩽ 2} u e^{v} \frac{1}{25} d u d v = \frac{1}{25} \int_{0}^{3} u d u . \int_{- 1}^{2} e^{v} d v$ $= \frac{1}{25} {[\frac{u^{2}}{2}]}_{0}^{3} . {[e^{v}]}_{- 1}^{2} = \frac{1}{50} (9) (e^{2} - e^{- 1}) = \frac{9}{50} {e^{2} - \frac{1}{e}} .$
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