Q23·Section A (Section A)·23 of 60Limits after reversing a double integralIf ∫y=01∫x=y2−1−(y−1)2f(x,y) dx dy=∫x=01∫y=0α(x)f(x,y) dy dx+∫x=12∫y=0β(x)f(x,y) dy dx,\int_{y=0}^1\int_{x=y}^{2-\sqrt{1-(y-1)^2}}f(x,y)\,dx\,dy=\int_{x=0}^1\int_{y=0}^{\alpha(x)}f(x,y)\,dy\,dx+\int_{x=1}^2\int_{y=0}^{\beta(x)}f(x,y)\,dy\,dx,∫y=01∫x=y2−1−(y−1)2f(x,y)dxdy=∫x=01∫y=0α(x)f(x,y)dydx+∫x=12∫y=0β(x)f(x,y)dydx, then α(x)\alpha(x)α(x) and β(x)\beta(x)β(x) are OptionsAα(x)=x,β(x)=1+1−(x−2)2\alpha(x)=x, \beta(x)=1+\sqrt{1-(x-2)^2}α(x)=x,β(x)=1+1−(x−2)2Bα(x)=x,β(x)=1−1−(x−2)2\alpha(x)=x, \beta(x)=1-\sqrt{1-(x-2)^2}α(x)=x,β(x)=1−1−(x−2)2Cα(x)=1+1−(x−2)2,β(x)=x\alpha(x)=1+\sqrt{1-(x-2)^2}, \beta(x)=xα(x)=1+1−(x−2)2,β(x)=xDα(x)=1−1−(x−2)2,β(x)=x\alpha(x)=1-\sqrt{1-(x-2)^2}, \beta(x)=xα(x)=1−1−(x−2)2,β(x)=xPick an option to check your answer.Show answerShow solution Report a problem with this question← Q22All questionsQ24 →Keyboard: A–D answer · S solution · ←/→ previous/next