# Integration by substitutionWatch

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Thread starter 4 years ago
#1
When using a substitution such as u^2=x+1 (basically anything to the positive power), when you substitute the limits for x with the limits for u, how do you choose whether to take the positive or negative root (when there are no restrictions on u)?
It can't be an arbitrary choice because say your integral was u^2 + u, if you chose the negative root you'd get a difference of 2u, so it's a numerical difference rather than just a change in sign.
I've always taken the positive root, and I always get the right answer, I'm just not sure why.
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4 years ago
#2
(Original post by hhattiecc)
When using a substitution such as u^2=x+1 (basically anything to the positive power), when you substitute the limits for x with the limits for u, how do you choose whether to take the positive or negative root (when there are no restrictions on u)?
It can't be an arbitrary choice because say your integral was u^2 + u, if you chose the negative root you'd get a difference of 2u, so it's a numerical difference rather than just a change in sign.
I've always taken the positive root, and I always get the right answer, I'm just not sure why.

Now which would you pick?
0
4 years ago
#3
(Original post by hhattiecc)
When using a substitution such as u^2=x+1 (basically anything to the positive power), when you substitute the limits for x with the limits for u, how do you choose whether to take the positive or negative root (when there are no restrictions on u)?
It can't be an arbitrary choice because say your integral was u^2 + u, if you chose the negative root you'd get a difference of 2u, so it's a numerical difference rather than just a change in sign.
I've always taken the positive root, and I always get the right answer, I'm just not sure why.
You must always choose an invertible (1-1) function over the integration range.

You can set and work with that, or you can set and work with that, but you must choose one or the other and be consistent.
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Thread starter 4 years ago
#4
(Original post by Phichi)

Now which would you pick?
The positive root, but I'm still not sure why you'd reject the negative root since the relation still holds.
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Thread starter 4 years ago
#5
(Original post by davros)
You must always choose an invertible (1-1) function over the integration range.

You can set and work with that, or you can set and work with that, but you must choose one or the other and be consistent.
I guess having to use a 1-1 function provides a reason for rejecting the negative root, but is there a solid mathematical argument as to why you can't use a function that involves multiple values of u for a given value of x?

I just can't understand why, if the relation still holds and at those particular values of u you will still get that particular value of x, you can't use a combination of a positive and negative root.
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4 years ago
#6
(Original post by hhattiecc)
I guess having to use a 1-1 function provides a reason for rejecting the negative root, but is there a solid mathematical argument as to why you can't use a function that involves multiple values of u for a given value of x?

I just can't understand why, if the relation still holds and at those particular values of u you will still get that particular value of x, you can't use a combination of a positive and negative root.
This is NOT what I'm saying. You can choose the negative root but you have to do that everywhere!

You can choose the positive root OR the negative root, but not both.

If you want a reason for needing an invertible function. think about a definite integral with x-limits a and b. What happens if you set u = (x-a)(x-b)? Both limits transform to 0, which means that the integral has value 0. Therefore, all definite integrals are 0
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