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Find the area enclosed by the curve x = t^2 – 2t , y=sqrt(t), and the y-axis?

Find the area enclosed by the curve x = t^2 – 2t , y=sqrt(t), and the y-axis.
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Dr. Juanita Wolff

Feb 20, 2025

First find the the value of t where the curve intersects the Y-axis. This is when x = 0.x = t^2 – 2t = 0 = t(t – 2)So t= 0 and t = 2dA = (0 – x)*dy …. Since the curve has negative x in this regiony = SQRT(t) and dy = [(1/2)/SQRT(t)]dtdA = [2t – t^2][(1/2)/SQRT(t)]dtdA = [t^(1/2) – (1/2)t^(3/2)]dtIntegrate to get: A = (2/3)t^(3/2) – (1/5)t^(5/2)Now evaluate from t= 0 to t = 2.Area = [(2/3)2^(3/2) – (1/5)2^(5/2)] – [0]Area = SQRT(2)[4/3 – 4/5]Area = SQRT(2)[8/15) = 0.754... Show More
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