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Extremely Weird Integration.

(y^2 - 1)^2 = y^4 - 2y^2 + 1 . Aren't both equation the same? Why after integration , the answers become different?

Can anyone explain it?

Note by 柯 南
1 year, 8 months ago

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The integration of \((y^2-1)^2\) is incorrect. You seem to have used something similar to the chain rule of differentiation, which is not applicable to integration. The correct way to do it would be to expand and continue as you did on the left hand side. Raj Magesh · 1 year, 8 months ago

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@Raj Magesh (I'm a student in malaysia) but the formula is in d text book ? Are there ways to do that without expanding ? 柯 南 · 1 year, 8 months ago

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@柯 南 Could you please post a picture of the formula in your textbook? And no, there's no quicker way. However, if you write the integral as the limit of an infinite summation, you will obtain the correct answer but after a lot of tedious calculation. Raj Magesh · 1 year, 8 months ago

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@Raj Magesh Sorry , not text book , it's reference book. Maybe they wrote it wrong.

Thank you so much :D , i wondered about it for so many days 柯 南 · 1 year, 8 months ago

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@柯 南 No problem! Useful tip: check your integration by differentiating your answer. Raj Magesh · 1 year, 8 months ago

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The integration of (y^2 - 1)^2 is incorrect. therefore you got different computation Rusli Azis · 1 year, 8 months ago

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You can not divide by 2y, that only works when the derivative of the inside is a constant. In your case it was 2y which is not a constant. Hope that helps Gaurav Agarwal · 1 year, 8 months ago

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@Gaurav Agarwal Thank you. That was a clear explanation !

(btw , why 2y is not constant ? Isit because y^2 can be ± ? 柯 南 · 1 year, 8 months ago

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I check both so many times , can't find anything wrong with it. 柯 南 · 1 year, 8 months ago

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You have performed the integral of \((y^2-1)^2\) incorrectly. Samuel Jones · 1 year, 8 months ago

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@Samuel Jones What is the correct steps? 柯 南 · 1 year, 8 months ago

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