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

[VietDao29]

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) &gt; 0, \forall x \in \mathbb{R} and -1 < 1, but why \int_{-1}^{1}f(x)dx &lt; 0

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

 

[iNCREDiBLE]

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) &gt; 0, \forall x \in \mathbb{R}
\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2
Why this happened? It's obvious that f(x) &gt; 0, \forall x \in \mathbb{R} and -1 < 1, but why \int_{-1}^{1}f(x)dx &lt; 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.

 

[Fermat]

...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.

 

[lurflurf]

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.

 

[VietDao29]

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