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An antiderivative recovers a function from its local rate of change. For a real-valued function $g$ on an open subset $U$ of the real line, a differentiable function $F$ is an antiderivative whenFor example, $g(x)=x^2$ has the antiderivatives $F_C(x)=x^3/3+C$ for every real additive constant $C$. Their derivatives agree, while their values can differ. On an interval, the constant of integration describes the entire freedom in the answer. ^[ref]

The indefinite-integral notation $\int g(x)dx=F(x)+C$ records this family on an interval. For the illustrated family, the extra condition $F(1)=2$ selects $C=5/3$. Thus reversing differentiation requires both a rate and a reference value if a unique function is wanted.

The fundamental theorem of calculus supplies a construction when $g$ is continuous on an interval $I$. Choose $x_0\in I$ and set $F(x)={\int }_{x_0}^{x}g(t)dt$; then $F^{\prime }=g$. Integrability alone does not guarantee a classical antiderivative. The Heaviside step function is Riemann integrable on bounded intervals, but cannot be a derivative throughout an interval containing its jump: Darboux theorem for derivatives requires every intermediate value. ^[ref]

A basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

For fixed temperatures and stress-free impermeable boundaries, linear marginal Rayleigh-Bénard convection with dimensionless horizontal wavenumber $q$ obeys $\mathrm{Ra}(q)=(q^2+{\pi }^2{)}^3/q^2$. Differentiation gives $q_c=\pi /\sqrt{2}$ and the displayed minimum. The threshold depends on the mechanical boundary conditions: two rigid no-slip boundaries instead give approximately $1708$.

The Dawson function obeys $F_D^{\prime }+2z{F}_D=1$ by differentiation. Its large-real-argument asymptotic expansion begins $F_D(z)=1/(2z)+1/(4z^3)+O(z^{-5})$. This inverse-power behavior contrasts with the separate exponentially growing factors in its definition.

Let $f$ be a real function with $\Vert f{\Vert }_w^2={\int }_a^{b}f(x{)}^{2}w(x)dx<\infty$, and let $V_m$ be the vector space of real polynomials of degree at most $m$. Its orthogonal projection is the least-squares polynomial in an orthogonal-polynomial basisEach coefficient depends on just one weighted inner product with $f$ and one squared norm. In particular, the coefficients $c_0,\dots ,c_m$ stay the same when the allowed polynomial degree increases. This statement assumes a fixed weight, interval and normalization of the orthogonal polynomials.

To see why the least-squares normal equations decouple, first write a trial polynomial as $q={\sum }_{j=0}^m{a}_j{\varphi }_j$ in any basis of $V_m$. The residual must be orthogonal to each permitted direction, givingThe Gram matrix usually couples the unknown coefficients. With ${\varphi }_j=p_j$, it becomes $G_{ij}=h_i{\delta }_{ij}$, leaving the separate equations $h_i{a}_i=b_i$. For the monomial basis on $[-1,1]$ with $w=1$, by contrast, $G_{ij}$ is zero when $i+j$ is odd and $2/(i+j+1)$ when $i+j$ is even. Distinct even powers, or distinct odd powers, remain coupled.

There is also a proof of the minimum that uses no differentiation. The defining coefficients give $⟨f-q_m,p_n{⟩}_w=0$ for $0\le n\le m$. Thus $f-q_m$ is orthogonal to all of $V_m$. For every $v\in V_m$, the Pythagorean theorem in an inner-product space givesThe second term is nonnegative and vanishes exactly when $v=q_m$, proving that the least-squares solution is unique. This is an integrated squared-error optimum; it does not assert a smallest error at every point or a smallest uniform norm.

For $N>m$, the nested orthogonal projections satisfy the exact residual budgetIndeed, $q_N-q_m$ is a linear combination of mutually orthogonal polynomials, and $f-q_N$ is orthogonal to it. Every newly retained component removes exactly $c_n^2{h}_n$ from the squared error. No infinite expansion or completeness assumption is needed for this finite identity. The Legendre least-squares projections of a cubic make every term explicit.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

Set $v=x^n{e}^{-x}$. Direct differentiation gives $x{v}^{\prime }=(n-x)v$. Apply the Leibniz rule to differentiate $n+1$ times:Hence $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. By Rodrigues' formula, $v^{(n)}=n!e^{-x}{L}_n$, so the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substitution gives the displayed Laguerre differential operator on polynomials eigenvalue equation. This derivation also covers $n=0$.

Consider the one-dimensional diffusion equation on the whole real line,with initial value one for $x<0$ and zero for $x>0$. This is an idealized Heaviside step function profile, not a measured material or an apparatus specification. The heat kernel gives a convolution with the initial profile:The change of variables $s=(x-y)/(2\sqrt{Dt})$ turns the convolution into a Gaussian integral tail. The resulting similarity variable $\eta =x/(2\sqrt{Dt})$ shows that the transition width grows like $\sqrt{Dt}$: multiplying elapsed time by four doubles every fixed-profile distance from the original interface.

At every positive time, $u(0,t)=1/2$, while $u(x,t)\to 1$ as $x\to -\infty$ and $u(x,t)\to 0$ as $x\to +\infty$. For every fixed $x\ne 0$, the limit as $t\to 0^{+}$ returns the specified initial value. The value assigned at the single initial discontinuity does not change this convolution. The error function therefore explains smoothing without pretending that a discontinuous initial profile remains sharp.

Direct differentiation checks the partial differential equation. Writing $\eta =x/(2\sqrt{Dt})$ givesThe model assumes constant diffusion coefficient, no advection, and no finite-domain boundary effects. For an interpretation as heat conduction, the heat equation applies to a normalized temperature difference under its own material assumptions.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

Use the Malthusian delay differential equation with time measured in units of its constant delay:Here $a$ is a real constant. On the first interval $y(t-1)=1$, hence $y(t)=1+at$. On the next interval the delayed input is $1+a(t-1)$, already known, so the method of steps for a delay differential equation givesEach polynomial integrates the preceding interval's delayed polynomial, and its integration constant ensures continuity. At $a=1$, the step endpoints are $y(1)=2$, $y(2)=7/2$, and $y(3)=37/6$. At $t=3/2$, the value is $21/8$, but its derivative is the earlier value $y(1/2)=3/2$. This is the geometric distinction from the ordinary differential equation $y^{\prime }=y$, whose derivative would equal its current value.

More generally, for any integer $n\ge 0$,At an integer join the newly introduced term vanishes. Inside a step, differentiation lowers each power by one and gives exactly $a$ times the preceding step evaluated at $t-1$. Together with $y(0)=1$, this verifies the formula by mathematical induction; it is an exact construction, not a truncated power series.

The history and the new equation need not prescribe matching derivatives at the starting time; their mismatch can reappear later in higher derivatives.^[ref] In this example, $y^{\prime }(0^{-})=0$ and $y^{\prime }(0^{+})=a$. For $a\ne 0$, $y^{\prime }$ is continuous at $t=1$ but $y^{\prime \prime }$ jumps there by $a^2$; at $t=2$, the first jump is in $y^{\prime \prime \prime }$, of size $a^3$. The solution of a differential equation is therefore continuously differentiable for positive time, although it need not have continuous derivatives of every order at step boundaries.

For $A_T=-\overline{A}B/2$ and $B_T=-A^2/4$, write $A=R{e}^{i\theta }$, $B=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. The polar equations areWhere the polar phases are defined, direct differentiation gives the first integrals $R^2-2P^2$ and $R^{2}P\sin \delta$. In particular $R^2{\theta }_T$ and $P^2{\varphi }_T$ are constant. If $\delta =\pi$ and $R^2-2P^2=C>0$, then $P_T=P^2/2+C/4$, a Riccati equation whose positive solution develops a finite-time pole. This is finite-time blowup of the reduced amplitude system; it does not establish blowup of the full wave equation beyond the domain of the weakly nonlinear expansion.

The real exponential function $\exp (x)=e^x$ converts addition into multiplication:Its values are positive, and it defines a bijection from the real numbers to the positive real numbers. The natural logarithm is its inverse function. These properties connect arithmetic with calculus: adding an amount to the input multiplies the output by a fixed factor, and the local rate of change equals the current value. The exponential series gives one construction; an equivalent characterization is the unique solution of a differential equation $y^{\prime }=y$, $y(0)=1$.

For a concrete example, $y(t)=10e^{0.2t}$ starts at $10$ and satisfies $y^{\prime }=0.2y$. The absolute derivative increases with $y$, while the relative rate $y^{\prime }/y$ stays $0.2$. At $t=5$, $y=10e\approx 27.1828$ and $y^{\prime }\approx 5.43656$. The exponential function therefore models accumulation proportional to what is already present, rather than addition of a fixed amount at every step. A positive constant rate gives exponential growth; a negative one gives exponential decay.

Every positive real base $b\ne 1$ fits the same construction through $b^x=e^{x\ln b}$. Thus its derivative is $(\ln b)b^x$: base $e$ is special because $\ln e=1$. The exponential function is also unchanged in form by differentiation and integration, helping explain its repeated appearance in linear ordinary differential equations. Its real definition, inverse relationship, and derivative rules are developed in OpenStax, integrals, exponential functions, and logarithms.

For a Schwartz function, with the angular-frequency convention,More generally, the identity holds when integration by parts is justified and the boundary term vanishes. Repeated differentiation gives $\widehat{f^{(m)}}(k)=(ik{)}^m\hat{f}(k)$. The short proof transfers the derivative from $f$ to $e^{-ikx}$ by integration by parts. High frequencies receive larger multipliers, so rapid ripples contribute more strongly to a derivative. For the heat equation, the second derivative becomes multiplication by $-k^2$, directly explaining the faster decay of fine spatial structure.

For electrons directed normally toward an ideal planar collector, a retarding electric potential difference $V_r>0$ imposes the barrier $e{V}_r$. Write its transfer as $T(E)=H(E-e{V}_r)$, where $H$ is the Heaviside step function and $e$ the positive elementary charge. Thus a perfectly sharp loss need not produce a sharp drop in electric current: the drop also depends on where the outgoing energy lies relative to the barrier. For an oblique incident electron at angle $\theta$ to the normal, the same planar model instead tests $E{\cos }^2\theta >e{V}_r$; angular scattering can change collection even at unchanged total kinetic energy.

For example, choose a monoenergetic incoming beam with $u=5.6\mathrm{eV}$, $p=0.75$ and $\Delta =4.9\mathrm{eV}$. After one encounter, $25\%$ have $5.6\mathrm{eV}$ and $75\%$ have $0.7\mathrm{eV}$. A $1.0\mathrm{V}$ retarder collects only $0.25I_0$, whereas a $0.5\mathrm{V}$ retarder collects $I_0$; here $I_0$ is the incident electric current with fixed geometrical acceptance and no background. The mean outgoing kinetic energy is $1.925\mathrm{eV}$, independently checked by $5.6-0.75(4.9)$. These are chosen inputs and calculated fractions, not historical measurements.

More generally, let $f(E;V)$ be the normalized outgoing probability density function at a fixed accelerating electric potential difference $V$, for the normally incident idealization. A retarding scan givesThe second equation follows by differentiating the lower integration limit, with Dirac delta functions understood for sharp beams. It explains why an integrated collector readout can reveal an energy distribution. With angles present it probes the distribution of normal kinetic energy; finite instrumental resolution smooths it, and numerical differentiation amplifies noise. Keep the incoming beam fixed during this scan. A scan of $V$ changes $f$ itself and cannot be differentiated by the same inversion formula.

Consider a ball moving horizontally with no net horizontal force in the laboratory. Treat its horizontal motion as that of a free particle: ${\mathbf{r}}_{\mathrm{lab}}=(ut,0)$ with constant $u>0$. Axes rotating counterclockwise at constant angular velocity $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ about the launch point assign it the positionThis is an exact rotation matrix transformation, with the axes coinciding at $t=0$. Taking $u=0.25\mathrm{m}{\mathrm{s}}^{-1}$, $\Omega =0.8\mathrm{rad}{\mathrm{s}}^{-1}$, and $0\le t\le 3\mathrm{s}$ gives the two views below. At $t=3\mathrm{s}$, the laboratory position is $(0.75,0)\mathrm{m}$, but the rotating position is approximately $(-0.5530,-0.5066)\mathrm{m}$. Both assign the same distance $0.75\mathrm{m}$ from the origin.

The entire rotating acceleration must include the centrifugal acceleration as well as the Coriolis acceleration. To see the balance explicitly, write $c=\cos \Omega t$ and $s=\sin \Omega t$. Direct differentiation givesThe two apparent contributions areAt launch $\mathbf{r}=0$, so only the Coriolis acceleration contributes to the initial bending. Later the centrifugal acceleration also contributes; the full curved trajectory cannot be explained by keeping only the Coriolis force. In particular, $|{\mathbf{v}}_{\mathrm{rot}}{|}^2=u^2(1+{\Omega }^2{t}^2)$ grows even though the Coriolis force does no work: the rotating-frame kinetic energy increase is supplied by work from the centrifugal force. The laboratory kinetic energy remains $m{u}^2/2$.

Rotate an ellipse with positive semiaxes $a,b$ about an axis a distance $r>a$ from its centre. For the parametrized surface $\sigma (u,v)=((r+a\sin u)\cos v,(r+a\sin u)\sin v,b\cos u)$, put $R=r+a\sin u$ and $D^2=a^2{\cos }^{2}u+b^2{\sin }^{2}u$. An outward unit normal is $N=(b\sin u\cos v,b\sin u\sin v,a\cos u)/D$. Direct differentiation gives the first fundamental form and second fundamental form, with the convention $I{I}_{ij}=⟨{\sigma }_{ij},N⟩$:Reversing the unit normal reverses $II$, but the Gaussian curvature $K=a{b}^2\sin u/(R{D}^4)$ is unchanged.

The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]