Discussion
The line integral of \( \mathbf{F} = y\mathbf{i} + x\mathbf{j} \) along the straight path from \( (0,0) \) to \( (1,1) \) is
Answer: B
On the path \( y = x \), so with \( x = t \) and \( y = t \) for \( t \) from zero to one,
\[ \int_{C}\left(y\,dx + x\,dy\right) \;=\; \int_{0}^{1}\left(t + t\right)dt \;=\; \left[t^{2}\right]_{0}^{1} \;=\; 1. \]
The field is the gradient of \( xy \), so the value is just \( xy \) evaluated between the ends, \( 1 - 0 \), whatever path is taken
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