Concept history

Updated domain and linked exercises

Concept created

Let $u$ and $v$ be two real functions that are differentiable. Hence their product $uv$ is also differentiable and
$$
(u\cdot v)'=u'\cdot v+u\cdot v'.
$$
Assume now that at least two of these three functions admits an integral on the interval $[a,b], a<b$, then the third function also admits an integral on this interval by the last equation and one gets the equality
$$
\int_{a}^{b}(u\cdot v)'(x)\,\mathrm{d}x=\int_a^bu'(x)\cdot v(x)\,\mathrm{d}x+\int_a^bu(x)\cdot v'(x)\,\mathrm{d}x,

$$
which can also be written as
$$
u(b)\cdot v(b)-u(a)\cdot v(a)=\int_a^bu'(x)\cdot v(x)\,\mathrm{d}x+\int_a^bu(x)\cdot v'(x)\,\mathrm{d}x.

$$
And this method is called "integration by part".

**Remarks:**
$\bullet$ Most of the time, one starts by one of the two integrals on the right and the purpose of this method is to write the integral in another form, changing the integrand $u'\cdot v$ into $u\cdot v'$.
$\bullet$ This formula also work with the same hypothesis but with an open or a semi-open interval instead of a closed one. Also $a$ and $b$ can be infinite. One only needs to check whether two of the three integrals from the second equation converges before applying this method.