Answer:
D. 1/6
Step-by-step 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.
[tex]\displaystyle \int_0^1{(x-x^2)}\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}[/tex]
The enclosed area is 1/6 square units.
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