SOLUTION: Find the area enclosed between the curve y = x(x - 1)2 and the axis y = 0,
establishing first where they intersect.
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Question 876816: Find the area enclosed between the curve y = x(x - 1)2 and the axis y = 0,
establishing first where they intersect.
Found 2 solutions by Fombitz, richard1234:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
From the graph, it looks like the limits of integration are and
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
The curve and the line y = 0 intersect at x = 0 and x = 1. Since x(x-1)^2 is positive along (0,1), the area enclosed is
^2 \, dx = \int_{0}^{1} x^3 - 2x^2 + x \, dx = \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} \, \, |^{1}_{0} = \frac{1}{12})
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