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Asked in JEE Main 9th Jan 1st Shift 2020 · Integrating a position-dependent force
W=∫⃗F· d⃗r=∫(-x dx+y dy).
Along the segment x runs from 1 to 0 and y from 0 to 1; each term integrates on its own variable.
∫₁⁰(-x) dx=[-(x²)/2]₁⁰=1/2 and ∫₀¹y dy=1/2.
W=1/2+1/2=1 J.
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