Discussion
Double integration over the region bounded by \( y = x^{2} \) and \( y = 2x \) gives an area of
Answer: A
The curves meet where \( x^{2} = 2x \), that is at \( x = 0 \) and \( x = 2 \), with the line above the parabola between them:
\[ A \;=\; \int_{0}^{2}\!\!\int_{x^{2}}^{2x} dy\,dx \;=\; \int_{0}^{2}\left(2x - x^{2}\right)dx \;=\; \left[x^{2} - \frac{x^{3}}{3}\right]_{0}^{2} \;=\; 4 - \frac{8}{3} \;=\; \frac{4}{3}. \]
The inner integral returns the vertical extent of the region at each \( x \), so the two areas four and \( 8/3 \) are the separate contributions rather than the answer
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