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Yet Another 0=1 (Posted on 2004-07-08) Difficulty: 2 of 5
Find the error in this proof of 0=1:

∫(1/x) dx
= ∫(1/x)*(1) dx (Mult. Identity)
= (1/x) x - ∫(-1/x^2)*x dx (Integ by Parts)
= 1 + ∫(1/x) dx (Simplify)

Hence, ∫(1/x) dx = 1 + ∫(1/x) dx, therefore 0 = 1.

See The Solution Submitted by Brian Smith    
Rating: 2.0000 (6 votes)

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Solution | Comment 2 of 10 |

I don't quite agree with Eric. The reason doesn't have to do with the integrals being indefinite.

The reason it doesn't work has to do with the basis of integration by parts. Integration by parts comes from the product rule for derivates, (uv)'=uv'+u'v, leading to d(uv)=udv+vdu.

If uv=constant, then integration by parts fails because the left hand side would be 0. It's absurd to say the integral of the differential of a constant because the differential of the constant is 0.

The above proof relies on the faulty principle that the integral of the differential of a constant is equal to the constant.


  Posted by np_rt on 2004-07-08 16:11:44
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