Simple integral - but a question of methodology

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  • #1
ianb
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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?
 
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Answers and Replies

  • #2
ZioX
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They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.
 
  • #3
VietDao29
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...
[tex]\frac{1}{4} (x^2 - \frac{x}{2}) + C[/tex]

Why should one of them be right?
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? :)
 
  • #4
ianb
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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?
 
  • #5
ianb
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Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.
 
  • #6
VietDao29
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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?
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. :)
 
  • #7
ianb
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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?
 
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  • #8
ianb
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I'm sorry, I should have been clearer. Message edited accordingly.
 
  • #9
VietDao29
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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. 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?
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. :)
 
  • #10
ianb
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Wow. I'm embarrassed to say the least. You, sir, rock my world.
 

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