Discussion
The area enclosed between the straight line y = x and the parabola y = x2 in the x-y plane is
Answer: A
The enclosed area is 1/6.
The line and the parabola meet where \(x = x^{2}\), at x = 0 and x = 1. Between these points the line lies above the parabola, so:
\[ \begin{aligned} A &= \int_{0}^{1}(x-x^{2})\,dx \\ &= \left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} \\ &= \frac{1}{2}-\frac{1}{3} = \frac{1}{6} \end{aligned} \]
1/2 is the area under the line alone and 1/3 the area under the parabola alone; the region between them is their difference. 1/4 does not correspond to either.
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