- #1
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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]
Why this happened? 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]
I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
What have I done wrong?
Viet Dao,
[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]
Why this happened? 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]
I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
What have I done wrong?
Viet Dao,