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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: Solution | Comment 3 of 10 |
(In reply to Solution by np_rt)

Integration by parts, which is int(u dv) = uv - int(v du), can work in reverse, even with a constant. The integral of 1 is indeed x.

I still say the problem is in the omission of constants. The indefinite integral of (1/x) is ln x + C. I can't stress the + C enough. Indefinite integrals do not equal each other because of these constants. The algebra fails:

int(1/x) dx = 1 + int(1/x) dx

ln x + C1 = 1 + ln x + C2

Because C1 and C2 can be anything, the problem is irrelevant. In fact, solving any indefinite integral in this problem yields an additional constant. The initial (1/x), for example, can have another constant.

Edited on July 8, 2004, 4:34 pm
  Posted by Eric on 2004-07-08 16:31:09

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