Discussion

Evaluating \( \displaystyle\int_{0}^{1}\!\!\int_{0}^{1}\left(x + y\right)dy\,dx \) gives

Answer: A

Do the inner integral with \( x \) held fixed:

\[ \int_{0}^{1}\left(x + y\right)dy \;=\; x + \frac{1}{2}. \]

Then the outer one:

\[ \int_{0}^{1}\left(x + \frac{1}{2}\right)dx \;=\; \frac{1}{2} + \frac{1}{2} \;=\; 1. \]

By symmetry each variable contributes \( 1/2 \), so the halves add rather than averaging to one of them

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