- #1
MIB
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why substitution method for integration always work ?
Why can we completely treat dx and du known in substitution method completely like differentials even if we don't have ∫f(g(x))g'(x) dx , i.e : why we can substitute x in terms of u and dx in terms of du .
thanks
Why can we completely treat dx and du known in substitution method completely like differentials even if we don't have ∫f(g(x))g'(x) dx , i.e : why we can substitute x in terms of u and dx in terms of du .
thanks
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