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The signed accumulated rotation about a specified fixed axis. Unlike an angle specified only modulo $2\pi$, angular displacement counts complete revolutions as well as the final orientation. Its derivative with respect to time is angular velocity. The displayed formula therefore follows from the fundamental theorem of calculus.

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]

Write the magnetic flux linkage of the stationary pickup coil as $\lambda ={\sum }_{n=1}^N{\Phi }_n$, in SI units of weber. If every turn links the same magnetic flux $\Phi$, then $\lambda =N\Phi$; the turn count is dimensionless. With a fixed winding and terminal convention, Faraday's law of induction gives the induced electromotive forceThis is the fundamental theorem of calculus applied to magnetic flux linkage. Its units are $\mathrm{V}\mathrm{s}=\mathrm{Wb}$, not volts alone. Reversing the leads reverses the sign. In a high-input-electrical resistance voltage measurement the input approximately follows this electromotive force; coil loading and the frequency response of the electronics require separate calibration.

Separate direct drive linkage from material linkage: $\lambda ={\lambda }_{\mathrm{drive}}+{\lambda }_{\mathrm{material}}$. Even an empty pickup coil has $-d{\lambda }_{\mathrm{drive}}/dt$. In an ideal long, uniformly coupled specimen with negligible demagnetizing field correction and eddy current shielding, a useful material approximation isHere $A_s$ is specimen cross-sectional area, not the larger coil aperture; ${\mu }_0$ is vacuum permeability and $M$ the axial magnetization sampled by the coil. The units check directly: $(\mathrm{V}s/(\mathrm{A}\mathrm{m}))({\mathrm{m}}^2)(A/m)=\mathrm{V}\mathrm{s}$. Finite geometry and nonuniform magnetization generally require a calibrated coupling in place of ${\mu }_{0}N{A}_s$. The exact linkage identity survives that failure of the uniform approximation.

As an original illustrative calculation, stipulate $N=100$, $A_s=10^{-6}{\mathrm{m}}^2$, $\Delta M=10^{3}A/m$ and rounded ${\mu }_0=4\pi \times 10^{-7}\mathrm{V}s/(\mathrm{A}\mathrm{m})$. ThenA uniform change over $0.10\mathrm{ms}$ would give a rectangular pulse of $-1.25664\mathrm{mV}$; over $0.20\mathrm{ms}$ it gives $-0.628319\mathrm{mV}$ with the same area. Thus pulse height mixes change size with change speed. These numbers describe an ideal model, not observed pulses or material constants.

Calculus connects local change with accumulated change. A derivative asks how fast a function changes near one input; an integral adds its contributions across an interval. The fundamental theorem of calculus explains why these operations fit together: accumulating a continuous rate produces a quantity whose derivative is that rate.

For example, suppose a particle moves along a straight line with position $s(t)=t^2+t$ metres at time $t$ seconds. Its velocity is $s^{\prime }(t)=2t+1$ metres per second. Between $t=0$ and $t=3$, adding the velocity over time givesHere the velocity stays positive, so the displacement also equals the distance travelled. If the particle reverses direction, the integral of its signed velocity still gives displacement, while the integral of its speed gives total distance.

This is useful because local rules are often easier to specify than whole trajectories. A differential equation can prescribe a particle's velocity or a population's growth rate; integration then reconstructs its evolution, subject to an initial condition. In the other direction, a derivative extracts local information from a known function, and optimization uses that information to locate or approach promising inputs. The same two operations therefore connect measurement, prediction and design.

The visual connection is developed in derivative, integral and fundamental theorem of calculus. The definitions depend on limits, which turn progressively finer finite measurements into exact mathematical statements. A graph with a corner need not have a derivative there, and the conditions permitting integration or interchange of limits matter; calculus is more than a collection of symbolic rules.

Historically, Isaac Newton described changing quantities and their rates in the language of fluents and fluxions; his manuscript History of the Method of Fluxions records that approach.^[ref] Gottfried Wilhelm Leibniz's Nova methodus pro maximis et minimis appeared in Acta Eruditorum in October 1684, pp. 467–473; the original article is preserved in the HKUST Library collection.^[ref] The modern limit definitions used here should not be read back into those seventeenth-century formulations.

Online text: openstax.org/books/calculus-volume-3/pages/6-4-greens-theorem

This section belongs to the textbook whose senior contributing authors are Gilbert Strang and Edwin “Jed” Herman; no separate section byline is verified.

Cited passages: §6.4, Theorem 6.12 and equation (6.13): circulation form and hypotheses; the following proof: rectangle edge calculation and the fundamental theorem of calculus.

The displayed estimate, for all $x$ and real $t$, defines a coordinate Lipschitz constant $L_i$ for the partial derivative in coordinate $i$. It implies the coordinate descent lemmaby the fundamental theorem of calculus and integration of the Lipschitz continuity estimate. Subtracting an affine function leaves these constants unchanged.

The apparently weaker estimate along the singleton negative gradient direction is equivalent when every real step size is allowed: if the initial partial derivative is nonzero, any displacement can be written $-\alpha {\partial }_{i}f(x)e_i$; if it is zero but the final partial derivative is nonzero, reverse the endpoints; if both vanish the estimate is trivial.

Target: gradient-descent.bigb, under Lipschitz gradient.

For the cube $Q=(0,L{)}^n$ and $u\in H^1(Q)$,This follows from the pairwise-difference identityand a coordinate-by-coordinate path from $x$ to $y$. The Cauchy-Schwarz inequality bounds the squared sum of its $n$ increments by $n$ times the sum of their squares, while the one-dimensional fundamental theorem of calculus bounds each integrated increment by $L^{n+2}\Vert D_{i}u{\Vert }_2^2$. Division by $2|Q|=2L^n$ gives the stated constant.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

Suppose $h$ is continuously differentiable on $[0,L]$. For $0<x\le L$, the improper integral $g(x)={\int }_0^{x}h(t)(x-t{)}^{-1/2}dt$ exists, but substituting directly into the ordinary Leibniz integral rule produces an undefined endpoint term. First use integration by parts to obtainThe new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally integrable. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as ${\int }_0^x[{\int }_0^s{h}^{\prime }(t)(s-t{)}^{-1/2}dt]ds$ by Fubini's theorem; the inner integral is continuous for $s>0$, so the fundamental theorem of calculus applies. Moreover $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$: if $h(0)\ne 0$, the right difference quotient at zero diverges rather than defining a finite derivative.

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.

Suppose $f$ is a differentiable function on an open convex set $D\subseteq {\mathbb{R}}^n$ and has an $L$-Lipschitz gradient there, in the Euclidean norm. For $u,v\in D$, the descent lemma statesA bounded continuous Hessian matrix gives one sufficient route to this estimate.^[ref]

For a short proof, put $d=v-u$. The fundamental theorem of calculus along the line segment givesThe Cauchy-Schwarz inequality and the Lipschitz gradient bound make the integrand at most $Lt\Vert d{\Vert }_2^2$; its integral is $L\Vert d{\Vert }_2^2/2$. This proof needs no assumption that $f$ is a convex function.

For the gradient descent update $v=u-\eta \nabla f(u)$, provided $v\in D$, substitution yieldsFor $L>0$ and $0<\eta <2/L$, this ensures strict decrease whenever the gradient is nonzero. At $\eta =2/L$ it guarantees only nonincrease; the quadratic example has a nonshrinking alternating component at this endpoint. At $\eta =1/L$ the guaranteed decrease is $\Vert \nabla f(u){\Vert }_2^2/(2L)$. A bound valid only on $D$ cannot justify a step leaving $D$.

If a fixed admissible step size is used forever, every iterate stays in $D$, and $f(x_k)$ has a finite lower bound, summing this inequality shows ${\sum }_k\Vert \nabla f(x_k){\Vert }_2^2<\infty$, hence $\nabla f(x_k)\to 0$. That conclusion alone gives neither convergence of the iterates nor a global minimum. For example, the smooth polynomial $f(x)=x^4-x^2$ has $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=-2$: gradient descent started at zero never moves, although $f(1/\sqrt{2})=-1/4<0$. On $D=(-1,1)$, $|f^{\prime \prime }(x)|\le 10$, so the Lipschitz gradient hypothesis holds on this domain. Convex functions and strongly convex functions provide additional structure; the exact quadratic convergence result above uses a positive-definite matrix.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. ThenApply the chain rule and the fundamental theorem of calculus to getThe same boundary expression as in the Leibniz integral rule thus applies to this explicitly specified Cauchy principal value. If $A\ne 0$, the two ordinary one-sided improper integrals diverge, so their sum is not an ordinary integral. A finite boundary derivative alone does not make that divergent integral meaningful.

The differentiation operator sends a differentiable function to its derivative; its domain and the norms must be specified. On $C^1([-1,1])$ with values in $C([-1,1])$, using the supremum norm on both, it is an unbounded linear operator: $f_n(x)=\sin (nx)$ satisfies $\Vert f_n{\Vert }_{\infty }\le 1$ and $\Vert D{f}_n{\Vert }_{\infty }=n$. Its graph norm is $\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, which is complete. Indeed, the kernel consists of constants and is complete, while $Sg(x)={\int }_0^{x}g(t)dt$ is a bounded right inverse by the fundamental theorem of calculus; apply completeness of a graph norm from a bounded right inverse.

If $g(t,x)$ and its partial derivative $D_{i}g(t,x)$ are continuous for $t$ in a compact interval and $x$ in an open set, thenApply the fundamental theorem of calculus to the difference quotient in the $i$th coordinate and use uniform continuity on a compact neighbourhood.

For the one-dimensional Sobolev representative with $v(c)=0$, the fundamental theorem of calculus gives $v(x)=-{\int }_x^c{v}^{\prime }(t)dt$. The Cauchy-Schwarz inequality then gives $|v(x){|}^2\le (c-x)\Vert v^{\prime }{\Vert }_2^2$. Integrating proves the displayed, generally nonoptimal Poincaré inequality. In particular $\Vert v{\Vert }_{H^1}^2\le [1+(c-b{)}^2/2]\Vert v^{\prime }{\Vert }_2^2$. Only one endpoint must be fixed, which is useful for mixed boundary conditions and for controlling a negative potential.

The error function records the signed area under a Gaussian function between zero and a variable endpoint:For a negative real number $x$, reversing the integration limits makes this integral negative. The normalization uses the Gaussian integral: the full area of $e^{-t^2}$ is $\sqrt{\pi }$, so its positive half contributes one after multiplication by $2/\sqrt{\pi }$. The same definition extends to a complex number $z$ along any path from zero to $z$, giving an entire function.^[ref]

For example, $\mathrm{erf}(0)=0$, $\mathrm{erf}(1)\approx 0.842701$, and $\mathrm{erf}(-1)\approx -0.842701$. A finite positive endpoint captures only part of the positive Gaussian integral; $\mathrm{erf}(x)$ approaches one as $x\to +\infty$ but never reaches it at a finite real $x$.

The useful idea is to name a recurring integral once instead of carrying it through every calculation. The error function and the normal distribution express probabilities and quantiles through this area, while an error-function diffusion front describes how diffusion smooths a sharp jump. The complementary error function retains the small area left in the tail.

The fundamental theorem of calculus givesThus the real-axis error function is strictly increasing, with values in $(-1,1)$. Its integrand is an even function, so the error function is an odd function, $\mathrm{erf}(-x)=-\mathrm{erf}(x)$. The derivative also gives ${\mathrm{erf}}^{\prime \prime }(x)=-4x{e}^{-x^2}/\sqrt{\pi }$: the curve is concave for $x>0$ and convex for $x<0$. These real-axis properties do not imply boundedness throughout the complex plane; the existing quadrant asymptotics of the error function describe a different behavior in other directions.

For a continuously differentiable function $\Phi$ and a piecewise smooth parametric curve $\gamma$ from $a$ to $b$,The chain rule converts the integrand into $d(\Phi (\gamma (t)))/dt$, then the fundamental theorem of calculus supplies the endpoint difference. This is why a conservative vector field has a path independent line integral. For $\Phi (x,y)=x^2+y^2$, every path from $(0,0)$ to $(1,2)$ gives $5$, provided it lies in the domain where the potential is defined.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

On a star-shaped neighbourhood of $a$, a smooth function satisfiesThe fundamental theorem of calculus along the segment proves the identity, and differentiation under the integral sign proves smoothness of the coefficients.