Of all the techniques of integration that we have learned, the most
powerful, and the simplest, is $u$substitution. If $u = g(x)$, then
$$\int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du.$$
You probably learned this method in terms of antiderivatives, as a
way to turn the chain rule inside out. That sort of
reasoning doesn't generalize very well to multiple dimensions. Instead,
let's analyze $u$substitution from the definition of an integral.
By definition, $\int_a^b f(g(x)) g'(x) dx$ is the limit of a sum
$$\sum_{i=1}^N f(g(x_i^*)) g'(x_i^*) \Delta_i x,$$
where
we have broken the interval from $a$ to $b$ into $N$ pieces, $x_i^*$ is a
point in the $i$th piece, and $\Delta_i x$
is the length of the $i$th piece. If we
let $u_i^* = g(x_i^*)$, then $\Delta_i u \approx g'(x_i^*) \Delta_i x$, and
our sum is approximately
$$\sum_{i=1}^N f(u_i^*) \Delta_i u.$$
The limit of this sum is, by definition, the definite integral of $f(u) du$
from the starting value of $u$ (namely $g(a)$) to the ending value of $u$
($g(b)$).
We have converted an integral in $x$space into an integral in $u$space,
but the functions being integrated are not the same. In one case we integrate
$f(u)$. In the other case we integrate $f(g(x))$ times a distortion factor
$g'(x)$. This factor corrects for $\Delta u$ being a different size
than $\Delta x$.
We also learned about inverse $u$substitution, especially in the
context of trig substitutions. If $x=g(u)$ (rather than the other way
around), then
$$\int_a^b f(x) dx = \int_{\alpha}^{\beta} f(g(u)) g'(u) du,$$
where $a=g(\alpha)$ and $b=g(\beta)$. That's the same rule we had before,
only with the roles of $x$ and $u$ reversed. We say that $g$ is a
mapping from $u$space to $x$space, sending the interval
$[\alpha,\beta]$ to the interval $[a,b]$,
and the distortion factor $g'(u)$ is called the Jacobian of this
mapping. When we write the equation
$$dx = g'(u) du,$$
we are saying that the length of a short interval in $x$space is
$g'(u)$ longer than the interval of the corresponding interval in $u$space.
Example: Compute $\int_0^{10} e^{x/5} dx$.
Solution: We let $x=g(u)=5u$, so $e^{x/5}=e^{u}$ and $dx=5 du$.
The map $g$ sends the interval $[0,2]$ in $u$space to the interval
$[0,10]$ in $x$space, 
and $g'(u)=5$ tells us that the $x$interval is
5 times bigger than the $u$interval, so
\begin{eqnarray*} \int_0^{10} e^{x/5} dx & = & \int_0^2 5 e^{u} du \\
& = & 5(1e^{2}).
\end{eqnarray*}

