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Miscellaneous Exercise 5(II) · Q46

Q.Find the area of the region lying between the parabolas: y2=xy^2 = x and x2=yx^2 = y

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Substituting y=x2y=x^2 into y2=xy^2=x: x4=x⇒x(x3−1)=0⇒x=0,1x^4=x\Rightarrow x(x^3-1)=0\Rightarrow x=0,1. Between them, y=xy=\sqrt x (from y2=xy^2=x) lies above y=x2y=x^2. So $A=\int_0^1(\sqrt x - x^2),dx = \left[\frac{2}{3}x^{3/2}-\frac{x^ …

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