Let's Prove that 1 = 0 1=0

Calculus Level 5

Read through all the steps below carefully:

  1. We know from the product rule that d d x ( u v ) = d v d x u + d u d x v \frac{d}{dx}(uv)=\frac{dv}{dx}u+\frac{du}{dx}v .
  2. Integrate both sides with respect to x x to get u v = ∫ u d v + ∫ v d u ⟹ ∫ u d v = u v − ∫ v d u uv=\int u\ dv+\int\ v\ du\implies \int u\ dv=uv-\int v\ du .
  3. Consider ∫ 1 2 d x x \int_1^2 \frac{dx}{x} .
  4. Define u = 1 x u=\frac{1}{x} and d v = d x dv=dx .
  5. Then, d u = − d x x 2 du=-\frac{dx}{x^2} and v = x v=x .
  6. Substitute the variables into the equation in step 2 to get ∫ 1 2 d x x = 1 + ∫ 1 2 d x x \int_1^2 \frac{dx}{x}=1+\int_1^2 \frac{dx}{x} .
  7. Subtract both sides of the equation in step 6 by ∫ 1 2 d x x \int_1^2 \frac{dx}{x} to get 0 = 1 0=1 .

In which of these steps did I first make a mistake by using flawed logic?

If you enjoyed this problem, you may want to consider trying this problem .

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 There is no error

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Nick Turtle by Nick Turtle , 21, USA

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2 solutions

Nick Turtle
Jan 30, 2018

Note that the moderators have deemed my original intended answer as incorrect. The below was my original answer and reasoning.

No error occurred at step 7. It is allowed to subtract both sides by the same number ( ∫ 1 2 d x x \int_1^2 \frac{dx}{x} , as a definite integral, evaluates to a number and there is no constant of integration involved).

Also, some may argue that in step 5, d v = d x dv=dx implies v = x + C v=x+C , where C C is an arbitrary constant, instead of just v = x v=x . However, precisely because it is an arbitrary constant, we can set C C to any value, i.e. 0 0 . Indeed, if you substitute v = x + C v=x+C into the equation, you will find that the arbitrary constant cancels itself out. Thus, though steps were missed, no error occurred in this step that would have resulted in the erroneous conclusion of 1 = 0 1=0 .

So, where did the problem occur? At step 6, when we substituted the variables in, we forgot to include the boundaries of the integral. The correct equation should be ∫ 1 2 d x x = [ 1 ] x = 1 x = 2 + ∫ 1 2 d x x \int_1^2 \frac{dx}{x}=[1]_{x=1}^{x=2}+\int_1^2 \frac{dx}{x}

Now, [ 1 ] x = 1 x = 2 = 0 [1]_{x=1}^{x=2}=0 . If we had done this, we would have got the correct result 0 = 0 0=0 at the end.

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It might not matter, but the statement is still wrong, so step 5 is indeed incorrect.

Sharky Kesa - 3 years, 3 months ago
Raven Herd
Feb 4, 2018

You cannot subtract an integral as it is indefinite and indeterminate.Hence step 6 is wrong!

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