Integrating positive function on interval from -1 to 1 but result is negative I have the following function: 
$$\int_{-1}^1 \frac1 {x^4} dx$$ 
and my result seems to be:
$$\frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?

Since
$\dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$\int_{-1}^1 \frac1 {x^4} dx
$
is
$\lim_{c \to 0^+} (\int_{-1}^{-c} \frac1 {x^4} dx+\int_{c}^1 \frac1 {x^4} dx)
$.
We have
$\int \dfrac{dx}{x^4}
=\int x^{-4} dx
=\dfrac{x^{-3}}{-3}
=-\dfrac{1}{3x^3}
$
so
$\begin{array}\\
\int_{-1}^{-c} \frac1 {x^4} dx
&=-\dfrac{1}{3x^3}|_{-1}^{-c}\\
&=-(\dfrac{1}{3(-c)^3}-\dfrac{1}{3(-1)^3})\\
&=-(-\dfrac{1}{3c^3}+\dfrac{1}{3})\\
&=\dfrac{1}{3c^3}-\dfrac{1}{3}\\
\end{array}
$
and
$\begin{array}\\
\int_{c}^{1} \frac1 {x^4} dx
&=-\dfrac{1}{3x^3}|_{c}^{1}\\
&=-(\dfrac{1}{3}-\dfrac{1}{3c^3})\\
&=\dfrac{1}{3c^3}-\dfrac{1}{3}\\
\end{array}
$
Adding these we get
$\dfrac{2}{3c^3}-\dfrac{2}{3}
$,
and the
$\dfrac{2}{3c^3}
$
overwhelms the
$-\dfrac{2}{3}
$.
