FALSE: if $u$ and $v$ are differentiable on $[a,b]$ then $\int_a^b uv' = u(b)v(b)-u(a)v(a)-\int_a^b u'v$

False claim: if $u, v : [a,b] \to \mathbb{R}$ are differentiable at every point of $[a,b]$ (The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ 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

$\int_a^b u\,v' \;=\; u(b)v(b) - u(a)v(a) \;-\; \int_a^b u'\,v .$

That is If $u,v$ are differentiable on $[a,b]$ with $u',v'$ integrable, then $\int_a^b u v' = u(b)v(b)-u(a)v(a) - \int_a^b u'v$ with the hypothesis "$u'$ and $v'$ 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 $uv' = G$ is continuous hence integrable, so the left-hand side exists. But $u'v$ is the function $x \mapsto x\,G'(x)$, 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$) and is not Riemann integrable: the symbol $\int_0^1 u'v$ 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 $u',v'$ integrable, then $\int_a^b u v' = u(b)v(b)-u(a)v(a) - \int_a^b u'v$ asks that $u'$ and $v'$ be integrable, which is what makes $(uv)' = u'v + uv'$ 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 $v' = 0$, makes the left-hand side $\int_0^1 0 = 0$ and the right-hand side $G(1)\cdot 1 - G(0)\cdot 1 - \int_0^1 G'$, 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 $u',v'$ integrable, then $\int_a^b u v' = u(b)v(b)-u(a)v(a) - \int_a^b u'v$ is proved by applying The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G' = f$ and $f$ is integrable, then $\int_a^b f = G(b)-G(a)$ to $uv$, and that theorem needs $(uv)'$ integrable. Deleting the hypothesis here deletes it there.

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