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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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re(2): Solution | Comment 4 of 10 |
(In reply to re: Solution by Bob)

Yes it is true that 1*dx, = x + C. But that it is not what we're dealing with here. We are dealing with d(1). You see the difference?

But after some more thought, I agree that there is a way to explain it using the fact that they are indefinite integrals. But you're not explaining it right.

Consider that there are limits to the integrals. That doesn't change anything about the integrals in the proof. What it does change is the 1. You still have to evaluate the difference. So if the limits are a to b. Then you'd have to say 1(b)-1(a), which is 1-1=0.

That's where the indefinite integral comes in. The only place in the proof that is affected by it being an indefinite integral is the uv part. But as I said in the earlier post, that only makes a difference when uv is a constant.


  Posted by np_rt on 2004-07-08 20:47:41
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