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Simple integral - but a question of methodology

  1. Apr 4, 2007 #1
    simple integral -- but a question of methodology

    [tex]\int \frac{(x-1)}{2} dx[/tex]

    First, try substitution. Let [tex]u = x - 1[/tex], and [tex]du = dx[/tex]


    [tex]\int \frac{u}{2} du[/tex]

    [tex]\frac{1}{2} \int u \ du[/tex]

    [tex]\frac{1}{2} \times \frac{u^{2}}{2}[/tex]

    [tex]\frac{1}{2} \times \frac{(x-1)^2}{2}[/tex]

    [tex]\frac{(x-1)^{2}}{4} + C[/tex]

    Or we could do this:

    [tex]\int \frac{(x-1)}{2} dx[/tex]

    [tex]\int \frac{1}{2}(x-1)dx[/tex]

    [tex]\frac{1}{2} \int(x-1)dx[/tex]

    [tex]\frac{1}{2} (\frac{x^2}{2} - x) + C[/tex]

    [tex]\frac{1}{4} (x^2 - 2x) + C[/tex]

    Why should one of them be right?
     
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2
    They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.
     
  4. Apr 4, 2007 #3

    VietDao29

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    The last line is wrong. You've pulled out the factor 1 / 4 incorrectly.
    It should read:
    [tex]\frac{1}{4} \left( x ^ 2 - 2x \right)[/tex]
    Both are correct, the two results differ from each other by a constant 1/4.
    You should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant.
    Proof: Differentiate the LHS with respect to x, we have:
    (F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0)
    = F'(x) = f(x)
    Can you get this? :)
     
  5. Apr 4, 2007 #4
    Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.

    Thoughts?
     
  6. Apr 4, 2007 #5
    Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.
     
  7. Apr 4, 2007 #6

    VietDao29

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    Nope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have:
    [tex]\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)[/tex]
    [tex]\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)[/tex], they are still the same.
    The constant will automatically vanish when you subtract the two terms. :)
     
  8. Apr 4, 2007 #7
    Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
     
    Last edited: Apr 4, 2007
  9. Apr 4, 2007 #8
    I'm sorry, I should have been clearer. Message edited accordingly.
     
  10. Apr 4, 2007 #9

    VietDao29

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    Uhm, well, no, they are not different.
    Well, here we go:
    [tex]\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \left( \frac{(x - 1) ^ 2}{4} \right) \right|_0 ^ 1 = \frac{(1 - 1) ^ 2}{4} - \frac{(0 - 1) ^ 2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}[/tex]

    [tex]\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \frac{1}{4} \left( x ^ 2 - 2x \right) \right|_0 ^ 1 = \frac{1}{4} \left( (1 ^ 2 - 2 \times 1) - (0 ^ 2 - 0) \right) = \frac{1}{4} (-1 - 0) = -\frac{1}{4}[/tex]
    They are the same. :)
     
  11. Apr 4, 2007 #10
    Wow. I'm embarrassed to say the least. You, sir, rock my world.
     
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