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The fundamental theorem of calculus connects the Riemann integral with differentiation. Its first part says that for continuous $h:[a,b]\to \mathbb{R}$, the accumulation function $H(x)={\int }_a^{x}h(t)dt$ is continuous on $[a,b]$ and differentiable inside it, withTo prove this, boundedness of $h$ gives $|H(y)-H(x)|\le \Vert h{\Vert }_{\infty }|y-x|$, proving continuity. For an interior point and a sufficiently small nonzero increment $u$,Its absolute value is at most ${\sup }_{|t-x|\le |u|}|h(t)-h(x)|$, which tends to zero by continuity of $h$ at $x$. This proves the derivative formula. The same estimate gives the appropriate one-sided endpoint derivatives.

The second part says that if $Q$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $Q^{\prime }=h$ with $h$ continuous, thenIndeed $(Q-H{)}^{\prime }=0$, so the mean value theorem makes $Q-H$ constant, and evaluating at the endpoints proves the formula. A useful stronger version only requires $Q^{\prime }$ to be Riemann integrable. For a partition $a=x_0<\cdots <x_N=b$, the mean value theorem on each subinterval provides ${\xi }_j\in (x_{j-1},x_j)$ such thatAs the mesh tends to zero, these tagged Riemann sums converge to ${\int }_a^b{Q}^{\prime }$, proving that version too. Its integrability hypothesis must not be omitted.

An integrable integrand need not give an everywhere differentiable accumulation function. For a counterexample, let $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. This bounded step function is Riemann integrable, butIts left and right derivatives at $1/2$ are zero and one. Thus $F$ need not be differentiable at every interior point. The first part of the fundamental theorem of calculus does apply at every point where the integrand is continuous.

A derivative need not have a Riemann integral. Define on all of $\mathbb{R}$At zero $f(h)/h=h\sin (1/h^2)\to 0$, so $f^{\prime }(0)=0$; elsewhere the product rule and chain rule giveAt $x_n=(2\pi n{)}^{-1/2}$, this equals $-2/x_n\to -\infty$. Hence $f^{\prime }$ is unbounded on $[0,1]$, whereas a properly Riemann integrable function on a compact interval must be bounded. The requested Riemann integral therefore need not exist, even though $f$ is differentiable everywhere. This example is an unbounded derivative obstruction to Riemann integrability; it does not contradict the theorem's version that explicitly assumes integrability of the derivative.