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Integrating the function 1/x^2. Something I don't understand!

  • Thread starter VietDao29
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VietDao29
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While integrating the function [tex]f(x) = \frac{1}{x ^ 2}[/tex], I came across something I don't understand:
[tex]\int \frac{1}{x ^ 2}dx = - \frac{1}{x} + C[/tex]
Let [tex]f(x) := \frac{1}{x ^ 2}[/tex]
[tex]f(x) > 0, \forall x \in \mathbb{R}[/tex]
[tex]\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2[/tex]:confused:
Why this happened??? :confused: It's obvious that [tex]f(x) > 0, \forall x \in \mathbb{R}[/tex] and -1 < 1, but why [tex]\int_{-1}^{1}f(x)dx < 0[/tex]
:confused:
I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
What have I done wrong???? :cry:
Viet Dao,
 

Answers and Replies

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VietDao29 said:
While integrating the function [tex]f(x) = \frac{1}{x ^ 2}[/tex], I came across something I don't understand:
[tex]\int \frac{1}{x ^ 2}dx = - \frac{1}{x} + C[/tex]
Let [tex]f(x) := \frac{1}{x ^ 2}[/tex]
[tex]f(x) > 0, \forall x \in \mathbb{R}[/tex]
[tex]\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2[/tex]:confused:
Why this happened??? :confused: It's obvious that [tex]f(x) > 0, \forall x \in \mathbb{R}[/tex] and -1 < 1, but why [tex]\int_{-1}^{1}f(x)dx < 0[/tex]
:confused:
I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
What have I done wrong???? :cry:
Viet Dao,
It's an improper integral.
 
Last edited:
Fermat
Homework Helper
872
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VietDao29 said:
...I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
What have I done wrong???? :cry:
Viet Dao,
There's a discontinuity in f(x) at x = 0.

You can't integrate over a range that includes a discontituity.
 
lurflurf
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2,418
122
The fundamental theorem does not hold as 1/x^2 is not continuous at x=0. It is also an improper integral that does not converge in the traditional sense.
 
Last edited:
VietDao29
Homework Helper
1,417
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Yup, thanks for the help. I see what I missed now. :smile:
Viet Dao,
 

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