solve this: ∫∫_D x^2 e^(xy) dxdy D:{(x.y)∈R^2 /0≤x≤1. 0≤y≤2} ∫∫_D ((ydxdy)/((1+x^2 +y^2 )^(3/2) )). 0≤x≤1.0≤y≤1.


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Question Number 170501 by kndramaths last updated on 25/May/22
$solve this: ∫∫_D x^2 e^(xy) dxdy D:{(x.y)∈R^2 /0≤x≤1. 0≤y≤2} ∫∫_D ((ydxdy)/((1+x^2 +y^2 )^(3/2) )). 0≤x≤1.0≤y≤1.$
$s o l v e t h i s :$ $\int \int_{D} x^{2} e^{x y} d x d y$ $D : {(x . y) \in R^{2} / 0 ⩽ x ⩽ 1 . 0 ⩽ y ⩽ 2}$ $\int \int_{D} \frac{y d x d y}{{(1 + x^{2} + y^{2})}^{\frac{3}{2}}} . 0 ⩽ x ⩽ 1.0 ⩽ y ⩽ 1 .$
	Answered by FelipeLz last updated on 26/May/22

	$A = \int \int_{D} x^{2} e^{x y} d x d y = \underset{0}{\int^{2}} \underset{0}{\int^{1}} x^{2} e^{x y} d x d y = \underset{0}{\int^{1}} [\underset{0}{\int^{2}} x^{2} e^{x y} d y] d x$ $\underset{0}{\int^{2}} x^{2} e^{x y} d y = x^{2} \underset{0}{\int^{2}} e^{x y} d y = {[x^{2} \frac{e^{x y}}{x}]}_{y = 0}^{y = 2} = {[{x e}^{x y}]}_{y = 0}^{y = 2} = {x e}^{2 x} - x$ $A = \underset{0}{\int^{1}} ({x e}^{2 x} - x) d x = \underset{0}{\int^{1}} {x e}^{2 x} d x - \underset{0}{\int^{1}} x d x = {[\frac{1}{2} {x e}^{2 x} - \frac{1}{4} e^{2 x} - \frac{x^{2}}{2}]}_{0}^{1} = \frac{1}{2} e^{2} - \frac{1}{4} e^{2} - \frac{1}{2} - (0 - \frac{1}{4} - 0) = \frac{1}{4} (e^{2} - 1)$ $∙$ $B = \int \int_{D} \frac{y}{{(1 + x^{2} + y^{2})}^{3 / 2}} d x d y = \underset{0}{\int^{1}} {[\int \frac{y}{\sqrt{{(1 + x^{2} + y^{2})}^{3}}} d y]}_{y = 0}^{y = 1} d x$ $u = 1 + x^{2} + y^{2} \Rightarrow d u = 2 y d y$ $\int \frac{y}{\sqrt{{(1 + x^{2} + y^{2})}^{3}}} d y = \frac{1}{2} \int \frac{1}{\sqrt{u^{3}}} d u = - \frac{1}{\sqrt{u}} = - \frac{1}{\sqrt{1 + x^{2} + y^{2}}}$ $B = \underset{0}{\int^{1}} [- \frac{1}{\sqrt{2 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}}] d x = \underset{0}{\int^{1}} \frac{1}{\sqrt{1 + x^{2}}} d x - \underset{0}{\int^{1}} \frac{1}{\sqrt{2 + x^{2}}} d x = \underset{0}{\int^{1}} \frac{1}{\sqrt{1 + x^{2}}} d x - \frac{1}{\sqrt{2}} \underset{0}{\int^{1}} \frac{1}{\sqrt{1 + \frac{x^{2}}{2}}} d x$ $x = \tan (v) \Rightarrow d x = \sec^{2} (v) d v$ $\int \frac{1}{\sqrt{1 + x^{2}}} d x = \int \frac{\sec^{2} (v)}{\sqrt{1 + \tan^{2} (v)}} d v = \int \sec (v) d v = \ln ∣ \sec (v) + \tan (v) ∣ = \ln ∣ \sqrt{1 + x^{2}} + x ∣ = \sinh^{- 1} (x)$ $\frac{x}{\sqrt{2}} = \tan (t) \Rightarrow d x = \sqrt{2} \sec^{2} (t) d t$ $\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1 + \frac{x^{2}}{2}}} d x = \int \frac{\sec^{2} (t)}{\sqrt{1 + \tan^{2} (t)}} d t = \ln ∣ \sec (t) + \tan (t) ∣ = \ln ∣ \sqrt{1 + \frac{x^{2}}{2}} + \frac{x}{\sqrt{2}} ∣ = \sinh^{- 1} (\frac{x}{\sqrt{2}})$ $B = {[\sinh^{- 1} (x)]}_{0}^{1} - {[\sinh^{- 1} (\frac{x}{\sqrt{2}})]}_{0}^{1} = \sinh^{- 1} (1) - \sinh^{- 1} (\frac{1}{\sqrt{2}}) = \ln (\sqrt{2} + 1) - \ln (\frac{1 + \sqrt{3}}{\sqrt{2}}) = \ln (\frac{2 + \sqrt{2}}{1 + \sqrt{3}})$
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