calculate ∫∫_([1,3]^2 ) e^(−x−y) ln(2x+y)dxdy


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 70872 by Abdo msup. last updated on 09/Oct/19

$c a l c u l a t e \int \int_{{[1, 3]}^{2}} e^{- x - y} l n (2 x + y) d x d y$
Commented by Abdo msup. last updated on 11/Oct/19

$l e t I = \int \int_{{[1, 3]}^{2}} e^{- (x + y)} l n (2 x + y) d x d y l e t$ $c o n s i d e r t h e d i f f e o m o r p h i s m$ $(u, v) \to (x, y) = (φ_{1} (u, v), φ_{2} (u, v)) / x + y = u a n d$ $2 x + y = v \Rightarrow x = - u + v a n d y = 2 u - v \Rightarrow$ $φ_{1 (u, v)} = - u + v a n d φ_{2} (u, v) = 2 u - v$ $M_{j} (φ) = (\begin{matrix} \frac{\partial φ_{1}}{\partial u} \frac{\partial φ_{1}}{\partial v} \\ \frac{\partial φ_{2}}{\partial u} \frac{\partial φ_{2}}{\partial v} \end{matrix})$ $= (\begin{matrix} - 1 1 \\ 2 - 1 \end{matrix}) a n d {d e t M}_{j} = - 1 \Rightarrow$ $w e h a v e 1 ⩽ x ⩽ 3 a n d 1 ⩽ y ⩽ 3 \Rightarrow 2 ⩽ x + y ⩽ 6 \Rightarrow$ $2 ⩽ u ⩽ 6$ $2 ⩽ 2 x ⩽ 6 \Rightarrow 3 ⩽ 2 x + y ⩽ 9 \Rightarrow 3 ⩽ v ⩽ 9 \Rightarrow$ $I = \int \int_{2 ⩽ u ⩽ 6 a n d 3 ⩽ v ⩽ 9} e^{- u} l n v d u d v$ $= \int_{2}^{6} e^{- u} d u . \int_{3}^{9} l n v d v$ $= {[- e^{- u}]}_{2}^{6} . {[v l n v - v]}_{3}^{9}$ $= (e^{- 2} - e^{- 6}) (9 l n (9) - 9 - 3 l n (3) + 3)$ $= (e^{- 2} - e^{- 6}) (15 l n (3) - 6)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum