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Is 0=1? Then find the mistake

Consider the integral, \(\displaystyle I =\int \dfrac{1}{x}dx \)

We know that \( I= \log x \)

But, \(\displaystyle I =\int \dfrac{1}{x}(1)dx =(x)\dfrac{1}{x}-\int \bigg(-\dfrac{1}{x^2} \bigg) xdx\)

\(1+I=I \Rightarrow 1=0\)

What is the mistake?

I do not know the mistake

Note by Sparsh Sarode
10 months, 2 weeks ago

No vote yet
1 vote

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you actually got \(\int \frac{1}{x} -\int \frac{1}{x}=1 \). There is absolutely nothing wrong with that. Those are indefinite integrals.

Kushal Patankar - 10 months, 2 weeks ago

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How is LHS not equal to 0

Sparsh Sarode - 10 months, 2 weeks ago

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Integration constants need not be equal.

Kushal Patankar - 10 months, 2 weeks ago

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@Kushal Patankar K.. I agree on that

Sparsh Sarode - 10 months, 2 weeks ago

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I should have added limits.

Sparsh Sarode - 10 months, 2 weeks ago

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It is \(\frac{x}{x} |_{a}^{b}\). That yeilds zero. 1 there has limits.

Kushal Patankar - 10 months, 2 weeks ago

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@Kushal Patankar Exactly... Then he should get correctly.

Rishabh Cool - 10 months, 2 weeks ago

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@Kushal Patankar I didn't quite get u, u meant \( \dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1\)

Sparsh Sarode - 10 months, 2 weeks ago

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@Sparsh Sarode I can't believe you just stated that...

Kushal Patankar - 10 months, 2 weeks ago

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@Kushal Patankar Correct me if I am wrong, isn't \( \dfrac{2-1}{2-1}=\dfrac{1}{1}=1\) ?

Sparsh Sarode - 10 months, 2 weeks ago

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@Sparsh Sarode its \(\frac{b}{b}-\frac{a}{a}\) and still \(1|_a^b = 1_a-1_b=1-1=0\).

Kushal Patankar - 10 months, 2 weeks ago

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@Kushal Patankar Oops! I can't believe I made such a mistake... Anyways thx..

Sparsh Sarode - 10 months, 2 weeks ago

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@Sparsh Sarode Can you post this as a problem so others can learn from this misconception?

Calvin Lin Staff - 10 months, 2 weeks ago

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@Calvin Lin Ok, surely

Sparsh Sarode - 10 months, 2 weeks ago

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@Sparsh Sarode Sure... Glad to help.

Kushal Patankar - 10 months, 2 weeks ago

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What if I add limits from 1 to 2..you still get 1=0

Sparsh Sarode - 10 months, 2 weeks ago

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And what happens without limits? \[\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}\]

Hjalmar Orellana Soto - 10 months, 2 weeks ago

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It's still correct.. Integration constants need not be same

Sparsh Sarode - 10 months, 2 weeks ago

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Oh, right, thank you

Hjalmar Orellana Soto - 10 months, 2 weeks ago

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@Hjalmar Orellana Soto Welcome.. :)

Sparsh Sarode - 10 months, 2 weeks ago

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