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Question Number 172944 by flor last updated on 03/Jul/22

$$\int {\int }_{d}dxdyx=y^2-1x=1-y$$$$D=?$$

Commented by kaivan.ahmadi last updated on 04/Jul/22

$$x=y^2-1\Rightarrow y^2=x+1\Rightarrow y=\pm \sqrt{x+1}$$$$D_1:$$$$-1⩽x⩽0,-\sqrt{x+1}⩽y⩽\sqrt{x+1}$$$$-------------$$$$y=1-x\Rightarrow y=1-y^2+1\Rightarrow y^2-y-2=0$$$$(y-2)(y+1)=0\Rightarrow y=2,-1$$$$ify=-1\Rightarrow x=1-y=1+1=2$$$$D_2:$$$$0⩽x⩽2,-\sqrt{x+1}⩽y⩽1-x$$$$-------------$$$$D=D_1\cup D_2$$$$$$

Answered by kaivan.ahmadi last updated on 04/Jul/22

Commented by kaivan.ahmadi last updated on 04/Jul/22

$$redline:y=1-x$$

Commented by CElcedricjunior last updated on 05/Jul/22

$$\mathit{A}={\int }_{-1}^1{\int }_{\mathit{y}-1}^{{\mathbf{y}}^2-1}\mathbf{dxdy}$$$$={\int }_{-1}^1{\left[\mathbf{x}\right]}_{\mathbf{y}-1}^{y^2-1}\mathbf{dy}$$$$={\int }_{-1}^1({\mathbf{y}}^2-\mathbf{y})\mathbf{dy}$$$$={[\frac{{\mathbf{y}}^3}{3}+\frac{{\mathbf{y}}^2}{2}+2\mathbf{y}]}_{-1}^1$$$$=\frac{2}{3}$$$$\mathcal{A}=\frac{2}{3}$$$$$$$$$$$$.........\mathcal{L}ec\acute{el}\stackrel{`}{ebre}cedricjunior.........$$$$$$

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