Change of variable in 1 dimension

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 anti-derivatives, 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 $$\begin{eqnarray*} \int_0^{10} e^{-x/5} dx & = & \int_0^2 5 e^{-u} du \\ & = & 5(1-e^{-2}). \end{eqnarray*}$$