Discussion
The value of \( \displaystyle\int_{0}^{1}\!\!\int_{0}^{2} xy\,dy\,dx \) is
Answer: D
Integrate inside first, treating \( x \) as constant:
\[ \int_{0}^{2} xy\,dy \;=\; x\left[\frac{y^{2}}{2}\right]_{0}^{2} \;=\; 2x. \]
Then integrate outside:
\[ \int_{0}^{1} 2x\,dx \;=\; \left[x^{2}\right]_{0}^{1} \;=\; 1. \]
The limits are constant and the integrand separates, so the product \( \left(\int_{0}^{1}x\,dx\right)\left(\int_{0}^{2}y\,dy\right) = (1/2)(2) \) gives the same value
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