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Put $h=(b-a)/n$, $g_i=g(x_i)$, and $z_0=0$. Under the endpoint-slope convention suggested by the requested connection, the recurrence is $z_{i+1}=z_i+\frac{h}{2}(g_i+g_{i+1})$. Summing telescopes, soSince $G(b)={\int }_a^{b}g(x)dx$ by the fundamental theorem of calculus, this recurrence gives exactly the composite trapezoidal rule. It replaces the average derivative over each strip by the arithmetic mean of its endpoint derivatives.

For the standard midpoint method, the literal endpoint-only request is not valid for general $g$. Both the explicit and implicit one-step midpoint formulas here give insteadthe composite midpoint quadrature rule. It requires half-grid samples. These samples cannot be recovered exactly from $g_0,\dots ,g_n$ for arbitrary smooth $g$: the zero function and $g(x)={\prod }_{j=0}^n(x-x_j{)}^2$ have identical grid values but different midpoint sums. For a smooth $g$, replacing each half-grid sample by $(g_i+g_{i+1})/2$ is an additional approximation, giving $T_n$ with an $O(h^2)$ total difference, rather than an identity. A precise useful relation, obtained by splitting the finer trapezoidal sum into old nodes and midpoints, isHere $T_{2n}$ uses the finer half-grid, so it does not eliminate the missing data.

If “Mid-Point Method” means the two-step explicit midpoint linear multistep method, the recurrence is $z_{i+1}=z_{i-1}+2h{g}_i$. It gives $z_{2r}=2h{\sum }_{j=1}^r{g}_{2j-1}$ and $z_{2r+1}=z_1+2h{\sum }_{j=1}^r{g}_{2j}$, so an odd final index additionally needs a starting value $z_1$. For even $n=2r$, the result is the midpoint quadrature on $r$ double-width strips; equivalently $z_n=2T_n-T_{n/2}$. The question specifies neither even $n$ nor a starter. Thus the trapezoidal expression above is the natural endpoint-average interpretation, while the other common midpoint meanings require the qualifications just given. The ambiguity is present in the original PDF.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.

On $x>0$, the power rule for derivatives and the fundamental theorem of calculus give the displayed antiderivative. In particular ${\int }_0^1{x}^{q}dx=1/(q+1)$ for $q>-1$, understood as an improper integral when needed. At $q=-1$, the antiderivative on $x>0$ is the natural logarithm $\log x+C$.

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.

For a twice continuously differentiable function and $h\ge 0$, the fundamental theorem of calculus givesThe nonnegative triangular weight has integral $h^2$, so its supremum norm estimate directly controls the second modulus of smoothness.

A seminorm on a real or complex vector space $E$ is a finite-valued function $p:E\to [0,\infty )$ satisfying absolute homogeneity and the triangle inequality:In particular $p(0)=0$ and $p(-x)=p(x)$. A seminorm is a norm exactly when $p(x)=0$ implies $x=0$: this is the extra definiteness of a norm axiom. The zero function is a valid seminorm. The missing definiteness is useful when a measurement deliberately ignores some directions, such as an individual coordinate or the constant term of a function whose derivative is measured.

For example, on $C^1([0,1])$ the function $p(f)={\sup }_{t\in [0,1]}|f^{\prime }(t)|$ is a seminorm that vanishes on every constant function. Adding the value measurement gives the norm $q(f)=|f(0)|+p(f)$: if $q(f)=0$, the fundamental theorem of calculus gives $f(t)=f(0)+{\int }_0^t{f}^{\prime }(s)ds=0$. A family of such partial measurements defines a seminorm topology.

For $a<b$ and a real-valued function $f$ with a finite Riemann integral or Lebesgue integral on $[a,b]$, the signed area under a graph is its definite integral: geometric area above the horizontal axis contributes positively, and geometric area below it contributes negatively. If these two nonnegative areas are $A_{+}$ and $A_{-}$, thenThe second expression is the total geometric area under a graph. Taking the absolute value of the first expression does not usually give the second: cancellation has already removed information about how much area lies on each side.^[ref]

For an exact example, take the polynomial function $f(x)=x^2-1$ on $[-2,2]$. Its polynomial roots are $-1$ and $1$: it is negative between them and positive on the two outer intervals. An antiderivative is $F(x)=x^3/3-x$, so the fundamental theorem of calculus givesThus $A_{+}=8/3$ and $A_{-}=4/3$. The signed area under a graph is $4/3$, whereas the total geometric area under a graph is $4$. To compute the latter with antiderivatives, split at each sign-changing polynomial root and reverse the sign of the negative contribution.

Cancellation is useful information when the integrand represents a signed contribution. For the polynomial function $f(x)=x$ on $[-1,1]$, the two triangular areas give ${\int }_{-1}^{1}xdx=0$, although ${\int }_{-1}^1|x|dx=1$. Zero net signed area therefore means balanced positive and negative contributions, rather than an empty region.

The short algebraic justification uses the positive part $f_{+}=\max (f,0)$ and negative part $f_{-}=\max (-f,0)$: $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. Their integrals give $A_{+}$ and $A_{-}$, and linearity of integration gives the difference and sum formulas. For Lebesgue integration, require both contributions to be finite to avoid the undefined subtraction $\infty -\infty$.^[ref]

Changing the value of a bounded Riemann-integrable function at one point preserves its Riemann integral. The difference is supported at that point: partitions isolating it in intervals of arbitrarily small total length make both Darboux sums tend to zero. Thus a nonnegative single-point spike can have zero integral without vanishing everywhere. Its indefinite integral is identically zero, so the fundamental theorem of calculus derivative identity can fail at the changed point despite differentiability of the primitive.

For a continuously differentiable function $F$ on an interval $I=[t-\delta /2,t+\delta /2]$,To prove it, average $|F(t){|}^2-|F(u){|}^2$ over $u\in I$ using the fundamental theorem of calculus. On either half of $I$, the resulting derivative kernel has absolute value at most $1/2$. Since $|(|F{|}^2)\prime |\le 2|FF\prime |$, the displayed bound follows.

A continuous function on an open interval with positive symmetric derivative everywhere is strictly increasing. One proof takes an antiderivative $F$ using the fundamental theorem of calculus. For small positive $h$, the identityfollows from $f^s(x)>0$. If $F$ exceeded a chord, its difference from the chord would attain a positive interior maximum, contradicting this symmetric difference inequality. Thus $F$ is a convex function, and its derivative $f$ is nondecreasing. Equality at two different points would make $f$ constant between them and force an interior symmetric derivative to vanish, proving strict monotonicity.

When the second partial derivatives are continuous in a neighborhood, the order of the mixed partial derivatives can be interchanged. Expressing a rectangular finite difference as either iterated integral by the fundamental theorem of calculus and shrinking the rectangle proves the equality. Mere existence of both derivatives does not imply this symmetry.

A telescoping sum cancels intermediate sequence terms, leaving boundary terms. For a finite sequence of real numbers $a_0,\dots ,a_n$, the displayed identity follows because every $a_1,\dots ,a_{n-1}$ occurs once with each sign. Applying it to an infinite series also requires checking the limit of the boundary terms; finite cancellation alone does not prove that the series is a convergent series.^[ref]

For $n\ge 1$, expanding the finite sum givesNo monotone sequence assumption is required. Consider the original sequence $4,9,2,7,5$. Its successive finite differences are $5,-7,5,-2$, soThe large intermediate rise and fall disappear from the net change. This is useful whenever one knows successive changes rather than every absolute value: adding the changes reconstructs the final value from the initial one. It also checks a cumulative-update calculation without requiring all increments to be positive.

A finite difference $\Delta a_{k-1}=a_k-a_{k-1}$ measures one discrete change. The identity ${\sum }_{k=1}^n\Delta a_{k-1}=a_n-a_0$ is the discrete counterpart of the fundamental theorem of calculus: accumulation of changes returns the endpoint difference. The related summation by parts transfers a finite difference from one factor of a product to another, with explicit boundary terms.

If finite nonnegative entropy $H(t)$ decreases, its limit $H_{\infty }$ exists. The fundamental theorem of calculus applied to $H^2$ proves the displayed absolute-integrability identity. This product conclusion does not require a zero limiting entropy.

A total derivative accounts for every dependence of a quantity on the variable of differentiation. For a differentiable function $F(t,x_1,\dots ,x_n)$ evaluated along differentiable coordinate functions $x_i(t)$, the multivariable chain rule givesAll the partial derivatives on the right are evaluated at $(t,x_1(t),\dots ,x_n(t))$. The first term captures explicit dependence on $t$; the remaining terms capture changes transmitted through the coordinates.

For example, take $F(t,x)=t{x}^2$ and $x(t)=t$. The partial derivative with respect to $t$, holding $x$ fixed, is $x^2$, whose value on the path is $t^2$. The total derivative is insteadAt $t=2$, the partial derivative is $4$ and the total derivative is $12$. The difference is the contribution from changing $x$.

If the composite function is continuously differentiable on $[a,b]$, the fundamental theorem of calculus givesThis is the sense in which an integrated total derivative contributes endpoint terms. The endpoint statement concerns the whole composite function along the path. ^[ref].

A proper Riemann integral on a compact interval requires a bounded integrand, but a derivative can be unbounded. For $f(x)=x^2\sin (1/x^2)$ away from zero and $f(0)=0$, the difference quotient gives $f^{\prime }(0)=0$, whereas $f^{\prime }(x)=2x\sin (1/x^2)-2\cos (1/x^2)/x$ for $x\ne 0$. At $x_n=(2\pi n{)}^{-1/2}$, this derivative is $-2/x_n$ and is unbounded. The fundamental theorem of calculus version that integrates a derivative therefore needs an explicit integrability hypothesis.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

For $1\le p<\infty$, boundary restriction on smooth functions extends uniquely to a bounded linear map $T:W^{1,p}({\mathbb{R}}_{+}^n)\to L^p({\mathbb{R}}^{n-1})$. The estimate follows by applying the fundamental theorem of calculus on normal line segments and then integrating over the boundary variables.