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    I had this question which asked me to integrate something using the substitution:

     u^2 = 2x -1

    The current x limits were 5 and 1, so the u limits would be  \frac {+}{} 3 , \frac {+}{} 1 .

    The question "worked", and I got the right answer, using 3 and 1, so why can you just discount the negative 3 and 1?

    Thanks.
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    It's just sloppy notation really. What you're really doing is substituting u=\sqrt{2x-1} (which is valid since 2x-1 \ge 0 between 1 and 5). This is the positive root, hence why you use 1 and 3; then we just use the fact that u^2=2x-1 to make the substitution simpler.
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    (Original post by nuodai)
    It's just sloppy notation really. What you're really doing is substituting u=\sqrt{2x-1} (which is valid since 2x-1 \ge 0 between 1 and 5). This is the positive root, hence why you use 1 and 3; then we just use the fact that u^2=2x-1 to make the substitution simpler.
    Why is it  u =\sqrt{2x-1} and not  u = \frac {+}{} \sqrt {2x-1} ? Why does it have to be the positive root we are taking?
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    (Original post by H3rrW4rum)
    Why is it  u =\sqrt{2x-1} and not  u = \frac {+}{} \sqrt {2x-1} ? Why does it have to be the positive root we are taking?
    It doesn't, but if you let u=-\sqrt{2x-1} then you'd get some other minus signs popping up that end up cancelling out, giving you the same answer.

    As a simple example, take \displaystyle \int_1^5 \sqrt{2x+1}\, dx. The substitution u=\sqrt{2x+1} gives \displaystyle \int_1^3 u . 2u\, du = \int_1^3 2u^2\, du. However, the substitution u=-\sqrt{2x+1} gives \displaystyle \int_{-1}^{-3} (-u) . 2u\, du = -\int_{-1}^{-3} 2u^2\, du = \int_{-3}^{-1} 2u^2\, du. These both give you the same thing; in fact, in the latter case you can make a further substitution u=-v and you end up with \displaystyle \int_1^3 2v^2\, dv, which is exactly the same as before but with u relabelled as v.
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    (Original post by nuodai)
    It doesn't, but if you let u=-\sqrt{2x-1} then you'd get some other minus signs popping up that end up cancelling out, giving you the same answer.

    As a simple example, take \displaystyle \int_1^5 \sqrt{2x+1}\, dx. The substitution u=\sqrt{2x+1} gives \displaystyle \int_1^3 u . 2u\, du = \int_1^3 2u^2\, du. However, the substitution u=-\sqrt{2x+1} gives \displaystyle \int_{-1}^{-3} (-u) . 2u\, du = -\int_{-1}^{-3} 2u^2\, du = \int_{-3}^{-1} 2u^2\, du. These both give you the same thing; in fact, in the latter case you can make a further substitution u=-v and you end up with \displaystyle \int_1^3 2v^2\, dv, which is exactly the same as before but with u relabelled as v.
    Ok so if they gave you it in the sloppy notation, how would you know whether to take the +ve or -ve? I am fairly certain in C4 it will always simply be the +ve, but it just annoyed me :P I guess you could just change it from u^2 to u...
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    (Original post by H3rrW4rum)
    Ok so if they gave you it in the sloppy notation, how would you know whether to take the +ve or -ve? I am fairly certain in C4 it will always simply be the +ve, but it just annoyed me :P I guess you could just change it from u^2 to u...
    As I just showed you, it doesn't matter whether you take the positive or negative root for the substitution, as long as the sign you pick for the limits matches the sign you pick for the substitution. It just strikes me as making more work for yourself if you pick the negative root, unless your function had lots of minus signs in the right places.
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    (Original post by nuodai)
    As I just showed you, it doesn't matter whether you take the positive or negative root for the substitution, as long as the sign you pick for the limits matches the sign you pick for the substitution. It just strikes me as making more work for yourself if you pick the negative root, unless your function had lots of minus signs in the right places.
    Thanks Nuodai,

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    Ah just went back to the original question and the denominator of the integral was

     \sqrt(2x - 1)

    So they obviously want us to take the positive root, and simply change this to "u".
 
 
 
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