Question Number 100362 by Dara last updated on 26/Jun/20

∫_0 ^1 ∫_0 ^1 e^(2x+y) dydx

Answered by smridha last updated on 26/Jun/20

∫_0 ^1 dx.[e^(2x+y) ]_0 ^1   =∫_0 ^1 [e^(2x+1) −e^(2x) ]dx  =(e−1)[(e^(2x) /2)]_(0 ) ^1   =(1/2)(e−1)(e^2 −1)=(1/2)(e^3 −e^2 −e+1)

Answered by Ar Brandon last updated on 26/Jun/20

I=∫_0 ^1 ∫_0 ^1 e^(2x+y) dxdy=[∫_0 ^1 e^(2x) dx][∫_0 ^1 e^y dy]=[(e^(2x) /2)]_0 ^1 ∙[(e^y /1)]_0 ^1   ⇒I=(((e^2 −1)/2))(((e−1)/1))=(1/2)(e+1)(e−1)^2

Answered by mathmax by abdo last updated on 26/Jun/20

I =∫_0 ^1  ∫_0 ^1  e^(2x+y)  dy dx ⇒ I =∫_0 ^1  e^(2x)  dx.∫_0 ^1  e^y  dy  =[(1/2)e^(2x) ]_0 ^1 .[e^y ]_0 ^1  =(1/2)(e^2 −1)(e−1) =(1/2)(e^3 −e^2 −e+1)