Answer:
D. 1/6
Explanation:
You want the area enclosed by the graph of y = x(1 -x) and the x-axis.
Integral
The area is the integral of the function value over the interval in which it is non-negative.
The zeros are at x=0 and x=1, so those are the limits of integration.
![\displaystyle \int_0^1{(x-x^2)}\,dx=\left[(x^2)/(2)-(x^3)/(3)\right]_0^1=(1)/(2)-(1)/(3)=(1)/(6)](https://img.qammunity.org/2024/formulas/mathematics/high-school/p3ddmzdfifcgapgowh8mtnu57tf5f4v0y6.png)
The enclosed area is 1/6 square units.
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