differentials for integration by parts

Question

Suppose I want to calculate $\int_0^1xe^xdx.$

I've seen this argument:

$d(uv)=(du)v+udv$

$\int_0^1d(uv)=\int_0^1(du)v+\int_0^1udv$

$uv|_0^1=\int_0^1(du)v+\int_0^1udv$

$u=x, du=dx, v=e^x, dv=e^xdx$

$xe^x|_0^1=\int_0^1dxe^x+\int_0^1xe^xdx$

$ \begin{equation} \begin{split} \int_0^1xe^xdx&= xe^x|_0^1-\int_0^1dxe^x\\ &= (1\cdot e^1-0\cdot e^0)-(e^1-e^0)\\ &= (e^1)-(e^1-1)\\ &= 1\\ \end{split} \end{equation} $

How do I explain this proof to someone? Is there a way to define a differential so that this proof is meaningful? What are the objects we are talking about that allow us to calculate this way? Is there a simple and well defined way for me to understand what this means? If there is a reference on differentials where this analysis is well defined, please share that.

Answer 1

If $uv$ is the function defined by $(uv)(x) = u(x)v(x)$, then the product rule says that $$ (uv)'(x) = u'(x)v(x)+u(x)v'(x). $$ Therefore, $$ (uv)(x) \bigg|_0^1 = \int_0^1 (uv)'(x) dx = \int_0^1 u'(x)v(x) dx + \int_0^1 u(x)v'(x) dx, $$ and then $$ \int_0^1 u(x)v'(x) dx = (uv)(x) \bigg|_0^1 - \int_0^1 u'(x)v(x) dx. $$ Now put $u(x) = x$ and $v'(x) = e^x$, so that $u'(x) = 1$ and $v(x)=e^x$.