Is 0=1? Then find the mistake

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

We know that I=logx I= \log x

But, I=1x(1)dx=(x)1x(1x2)xdx\displaystyle I =\int \dfrac{1}{x}(1)dx =(x)\dfrac{1}{x}-\int \bigg(-\dfrac{1}{x^2} \bigg) xdx

1+I=I1=01+I=I \Rightarrow 1=0

What is the mistake?

I do not know the mistake

Note by Sparsh Sarode
2 years, 9 months ago

No vote yet
1 vote

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you actually got 1x1x=1\int \frac{1}{x} -\int \frac{1}{x}=1 . There is absolutely nothing wrong with that. Those are indefinite integrals.

Kushal Patankar - 2 years, 9 months ago

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

Sparsh Sarode - 2 years, 9 months ago

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I should have added limits.

Sparsh Sarode - 2 years, 9 months ago

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

Kushal Patankar - 2 years, 9 months ago

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@Kushal Patankar I didn't quite get u, u meant xx=baba1 \dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1

Sparsh Sarode - 2 years, 9 months ago

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

Kushal Patankar - 2 years, 9 months ago

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@Kushal Patankar Correct me if I am wrong, isn't 2121=11=1 \dfrac{2-1}{2-1}=\dfrac{1}{1}=1 ?

Sparsh Sarode - 2 years, 9 months ago

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@Sparsh Sarode its bbaa\frac{b}{b}-\frac{a}{a} and still 1ab=1a1b=11=01|_a^b = 1_a-1_b=1-1=0.

Kushal Patankar - 2 years, 9 months ago

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

Sparsh Sarode - 2 years, 9 months ago

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@Sparsh Sarode Sure... Glad to help.

Kushal Patankar - 2 years, 9 months ago

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

Calvin Lin Staff - 2 years, 9 months ago

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@Calvin Lin Ok, surely

Sparsh Sarode - 2 years, 9 months ago

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

Rishabh Jain - 2 years, 9 months ago

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How is LHS not equal to 0

Sparsh Sarode - 2 years, 9 months ago

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Integration constants need not be equal.

Kushal Patankar - 2 years, 9 months ago

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@Kushal Patankar K.. I agree on that

Sparsh Sarode - 2 years, 9 months ago

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

Hjalmar Orellana Soto - 2 years, 9 months ago

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

Sparsh Sarode - 2 years, 9 months ago

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Oh, right, thank you

Hjalmar Orellana Soto - 2 years, 9 months ago

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@Hjalmar Orellana Soto Welcome.. :)

Sparsh Sarode - 2 years, 9 months ago

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You cannot cancel the integrals on both sides because each function has infinitely many anti-derivatives off by a constant. That's why, for good reason, you should always include "+C" in your answers.

Math Nerd 1729 - 1 year, 10 months ago

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