Question on integration of a gradient.

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Why

[tex]\int_a^b \nabla T\; d\vec l \;=\; T(b)-T(a)[/tex]

Why integration of a gradient is always path independent?
 
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yungman said:
Why

[tex]\int_a^b \nabla T\; d\vec l \;=\; T(b)-T(a)[/tex]

Why integration of a gradient is always path independent?
This is the essence of the gradient theorem, which is a generalisation of the fundamental theorem of calculus. One can prove the gradient theorem with a simple application of Stokes' theorem.