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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Solution | Comment 8 of 10 |
(In reply to re(2): Solution by Bon)

Integration by parts is slightly different when evaluating definite integrals. In this case, instead of 1, we get (b-a) over that interval.

The problem is still with the constants of integration. Also, the integration of the constant 1 is nonunique - technically, x +- C derives to 1. In general, the problem int(d(x+1)) = int(d(x)) does not give you x+1 = x. It gives you x + 1 + C1 = x + C2. So you get C2 = C1 + 1, not 1 = 0.


  Posted by Eric on 2004-07-13 15:42:03
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