FALSE: if $u$ and $v$ are differentiable on $[a,b]$ then ${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v$

False claim: if $u,v:[a,b]\to \mathbb{R}$ are differentiable at every point of $[a,b]$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set), then

${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v\mathrm{.}$

That is If $u,v$ are differentiable on $[a,b]$ with $u^{\mathrm{\prime }},v^{\mathrm{\prime }}$ integrable, then ${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v$ with the hypothesis "$u^{\mathrm{\prime }}$ and $v^{\mathrm{\prime }}$ are integrable" deleted, and it is false.

The falsity is undefinedness, not a wrong number. Take $[a,b]=[0,1]$, let $u:=G$ be the everywhere-differentiable function of A function differentiable on $[0,1]$ whose derivative is unbounded, hence not Riemann integrable, and let $v(x):=x$. Then $u$ and $v$ are differentiable at every point of $[0,1]$, and $u{v}^{\mathrm{\prime }}=G$ is continuous hence integrable, so the left-hand side exists. But $u^{\mathrm{\prime }}v$ is the function $x\mapsto x{G}^{\mathrm{\prime }}(x)$, which is unbounded on $[0,1]$, hence has no Darboux sums at all (For bounded $f$ on $[a,b]$ and a partition $P$: the infimum $m_i$ and supremum $M_i$ of $f$ on the $i$-th subinterval, and the lower and upper Darboux sums $L(f,P)={\sum }_i{m}_i{\mathrm{\Delta }}_i$ and $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$) and is not Riemann integrable: the symbol ${\int }_0^1{u}^{\mathrm{\prime }}v$ on the right-hand side does not denote. An equation one of whose sides is undefined is not a true equation.

The correct hypothesis, and when it is automatic. If $u,v$ are differentiable on $[a,b]$ with $u^{\mathrm{\prime }},v^{\mathrm{\prime }}$ integrable, then ${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v$ asks that $u^{\mathrm{\prime }}$ and $v^{\mathrm{\prime }}$ be integrable, which is what makes $(uv{)}^{\mathrm{\prime }}=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$ integrable and lets the second fundamental theorem be applied to $uv$. It holds automatically when $u$ and $v$ are continuously differentiable, since a continuous function on $[a,b]$ is integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

• A second, even simpler witness. Taking $v\equiv 1$ instead, so $v^{\mathrm{\prime }}=0$, makes the left-hand side ${\int }_0^{1}0=0$ and the right-hand side $G(1)\cdot 1-G(0)\cdot 1-{\int }_0^1{G}^{\mathrm{\prime }}$, whose last term is undefined for the same reason. That version is the observation that the second fundamental theorem itself needs its integrability hypothesis; the witness in the refutation is given instead because it keeps both integrands genuinely non-constant.

• This is the same defect as in the fundamental theorem, seen through a product. If $u,v$ are differentiable on $[a,b]$ with $u^{\mathrm{\prime }},v^{\mathrm{\prime }}$ integrable, then ${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v$ is proved by applying The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$ to $uv$, and that theorem needs $(uv{)}^{\mathrm{\prime }}$ integrable. Deleting the hypothesis here deletes it there.

• Nothing is claimed about the identity holding whenever both sides happen to exist. If $u^{\mathrm{\prime }}$ and $v^{\mathrm{\prime }}$ are integrable the identity is If $u,v$ are differentiable on $[a,b]$ with $u^{\mathrm{\prime }},v^{\mathrm{\prime }}$ integrable, then ${\int }_a^{b}u{v}^{\mathrm{\prime }}=u(b)v(b)-u(a)v(a)-{\int }_a^b{u}^{\mathrm{\prime }}v$; what happens when $u^{\mathrm{\prime }}v$ and $u{v}^{\mathrm{\prime }}$ are integrable without $u^{\mathrm{\prime }}$ and $v^{\mathrm{\prime }}$ being so is not addressed anywhere on this page.