Discussion
Integrating \( xy \) over the triangle bounded by \( x = 0 \), \( y = 0 \) and \( x + y = 1 \) yields
Answer: D
For each \( x \) the variable \( y \) runs up to \( 1 - x \):
\[ \int_{0}^{1}\!\!\int_{0}^{1-x} xy\,dy\,dx \;=\; \int_{0}^{1} x\,\frac{(1-x)^{2}}{2}\,dx. \]
Expanding gives \( \tfrac{1}{2}\int_{0}^{1}\left(x - 2x^{2} + x^{3}\right)dx = \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{2}{3} + \tfrac{1}{4}\right) = \tfrac{1}{24} \). Over the whole unit square the same integrand would give \( 1/4 \)
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