(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?

(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?

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TopNewestThe 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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– 柯 南 · 1 year, 8 months ago

(I'm a student in malaysia) but the formula is in d text book ? Are there ways to do that without expanding ?Log in to reply

– Raj Magesh · 1 year, 8 months ago

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.Log in to reply

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

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– Raj Magesh · 1 year, 8 months ago

No problem! Useful tip: check your integration by differentiating your answer.Log in to reply

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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(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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– 柯 南 · 1 year, 8 months ago

What is the correct steps?Log in to reply