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Exercise 5.1 · Q10

Q.Find the area of the region included between: y2=2xy^2 = 2x, line y=2xy = 2x

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Substituting y=2xy = 2x into y2=2xy^2 = 2x: 4x2=2x⇒2x(2x−1)=0⇒x=04x^2 = 2x \Rightarrow 2x(2x-1) = 0 \Rightarrow x = 0 or x=12x = \frac{1}{2}. The curves meet at (0,0)(0,0) and (12,1)\left(\frac{1}{2}, 1\right). Between these, the parabola y=2xy = \sqrt{2x} lies above the line y=2xy = 2x (checking at x=0.25x = 0.25: 0.5≈0.707>0.5\sqrt{0.5} \approx 0.707 > 0.5). So …

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