Question 876816
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


*[tex \large \int_{0}^{1} x(x-1)^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}]