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

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While integrating the function $$f(x) = \frac{1}{x ^ 2}$$, I came across something I don't understand:
$$\int \frac{1}{x ^ 2}dx = - \frac{1}{x} + C$$
Let $$f(x) := \frac{1}{x ^ 2}$$
$$f(x) > 0, \forall x \in \mathbb{R}$$
$$\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2$$
Why this happened? It's obvious that $$f(x) > 0, \forall x \in \mathbb{R}$$ and -1 < 1, but why $$\int_{-1}^{1}f(x)dx < 0$$

I think it should be : $$\int_{-1}^{1}f(x)dx = + \infty$$
What have I done wrong?
Viet Dao,

iNCREDiBLE
VietDao29 said:
While integrating the function $$f(x) = \frac{1}{x ^ 2}$$, I came across something I don't understand:
$$\int \frac{1}{x ^ 2}dx = - \frac{1}{x} + C$$
Let $$f(x) := \frac{1}{x ^ 2}$$
$$f(x) > 0, \forall x \in \mathbb{R}$$
$$\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2$$
Why this happened? It's obvious that $$f(x) > 0, \forall x \in \mathbb{R}$$ and -1 < 1, but why $$\int_{-1}^{1}f(x)dx < 0$$

I think it should be : $$\int_{-1}^{1}f(x)dx = + \infty$$
What have I done wrong?
Viet Dao,

It's an improper integral.

Last edited:
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VietDao29 said:
...I think it should be : $$\int_{-1}^{1}f(x)dx = + \infty$$
What have I done wrong?
Viet Dao,
There's a discontinuity in f(x) at x = 0.

You can't integrate over a range that includes a discontituity.

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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:
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Yup, thanks for the help. I see what I missed now.
Viet Dao,