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The printed arguments $x-t$ and $x+t$ assume unit wave speed. For the stated wave equation with general $c>0$, the correct arguments are $x-ct$ and $x+ct$, as in the paper's subsequent formula. Equivalently, use rescaled time $\tilde{t}=ct$ to obtain the printed unit-speed form.

Introduce coordinates $\xi =x-ct$, $\eta =x+ct$ along the characteristic lines for the one-dimensional wave equation, and write $u(x,t)=U(\xi ,\eta )$. The chain rule givesso the wave equation becomesThus $U_{\xi }$ is independent of $\eta$, and integration gives the countertravelling decomposition of the one-dimensional wave equationConversely, differentiating this expression twice verifies the wave equation for any sufficiently differentiable $F,G$.

The two initial conditions giveDifferentiating the first and solving these two equations for the derivatives givesFor a fixed reference point $x_0$, their antiderivatives can be writtenSubstituting the two travelling arguments cancels the constants and combines the integrals. This proves D'Alembert's formula with initial velocity:At $t=0$ the integral vanishes, giving $u=f$; differentiating by the fundamental theorem of calculus gives $u_t=g$ there. These checks also fix the sign and the factor $1/(2c)$.

We prove the symmetric derivative criterion for monotonicity using an antiderivative. Fix $a<u<v<b$ and defineThe fundamental theorem of calculus gives $F^{\prime }=f$ because $f$ is continuous. For each interior $x$,Since $f^s(x)>0$, the definition of symmetric derivative gives $\delta >0$ such that $f(x+t)-f(x-t)>t{f}^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,

We next show that $F$ is a convex function. On any closed interval $[r,s]\subset (a,b)$ let $\ell$ be the affine function joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the extreme value theorem would give a positive maximum at an interior point $x$. For every sufficiently small $h>0$, that maximum would implyThe affine function contributes zero to this symmetric difference, contradicting the strict positive inequality above. Thus $F\le \ell$ on every such interval, exactly convexity.

For a differentiable convex function, secant slopes bound endpoint derivatives. Explicitly, the chord inequality at $u+h$ and at $v-h$, followed by $h\downarrow 0$, givesSo $f$ is nondecreasing. If $f(u)=f(v)$ for some $u<v$, monotonicity would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is strictly increasing. This proof requires no continuity of $f^s$ and never applies the ordinary mean value theorem to a merely symmetrically differentiable function.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

For the cardioid, $r^{\prime }=-a\sin \theta$. The expression under the square root simplifies toSince $0\le \theta \le {\theta }_2\le \pi$, the cosine is nonnegative. ThereforeThe upper half of the cardioid starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when ${\theta }_2=\pi$. Its full upper-half arc length is $4a$; reflecting gives $8a$ for the full cardioid.

The fundamental theorem of calculus givesThis decreases from $2a$ to zero, so the arc length increases fastest at ${\theta }_2=0$, interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as ${\theta }_2\downarrow 0$, rather than an attained interior maximum.

For an oriented piecewise smooth parametric curve $\mathbf{r}:[a,b]\to {\mathbb{R}}^3$ and a continuous vector field $\mathbf{F}$ along it, the vector-field line integral isFor a piecewise smooth curve, sum these integrals over its smooth pieces. The dot product selects the component of the vector field in the direction of motion.

If $\mathbf{F}=\nabla \varphi$ for a smooth function $\varphi$, the multivariable chain rule givesThe fundamental theorem of calculus, with cancellation at intermediate endpoints, therefore proves the fundamental theorem for line integrals:On a closed curve $\mathbf{r}(a)=\mathbf{r}(b)$, so $\boxed{{\oint }_C\nabla \varphi \cdot d\mathbf{r}=0}$.

The fundamental theorem of calculus differentiates the upper endpoint, while differentiation under the integral sign differentiates the parameter:These formulas hold, for example, when $g$ and its displayed parameter partial derivative are continuous on a neighborhood of the integration interval and the parameter value.

The fundamental theorem of calculus and part (a) giveThe difference is constant in $\theta$. Since $I(0,k)=E(0,k)=0$, the constant is zero, provingThis is simply the chain rule integrated, with $E$ itself as the antiderivative of the square-root factor.

The numerator and denominator both vanish at zero. The fundamental theorem of calculus differentiates the numerator, while the denominator derivative is nonzero near zero. L'Hôpital's rule therefore gives

The chain rule gives $f{f}^{\prime }=\frac{1}{2}(f^2{)}^{\prime }$. Both $f(0)$ and ${\lim }_{x\to \infty }f(x)$ vanish, so the fundamental theorem of calculus givesThe positive contribution while $f$ rises is cancelled by the negative contribution while it falls.

For $x>0$ write $f(x)=x^{-a}I(x)$, where $I(x)={\int }_0^x{e}^{-t}{t}^{a-1}dt$. Because $a>0$, the lower endpoint is integrable. At each positive upper endpoint the fundamental theorem of calculus gives $I^{\prime }(x)=e^{-x}{x}^{a-1}$. The product rule therefore yieldsDifferentiating once more and substituting this expression for $f^{\prime }$ givesSubstitution now cancels the $e^{-x}$ terms and the terms proportional to $f$, proving $x{f}^{\prime \prime }+(1+a+x)f^{\prime }+af=0$.

The change of variables $t=xu$ identifies $f$ as the scaled lower incomplete gamma integral ${\int }_0^1{e}^{-xu}{u}^{a-1}du$. It has a smooth extension at zero, with $f(0)=1/a$, $f^{\prime }(0)=-1/(a+1)$ and $f^{\prime \prime }(0)=1/(a+2)$. Thus the apparent singularities in the derivative formulas cancel there; those formulas with denominators $x$ were derived for $x>0$.

Differentiate the rewritten expression. Its new kernel vanishes at the moving upper endpoint, and its partial derivative is an integrable inverse square root. The derivative of an Abel integral isFor a rigorous route, rewrite $2{\int }_0^x{h}^{\prime }(t)\sqrt{x-t}dt$ as ${\int }_0^x{\int }_0^s{h}^{\prime }(t)/\sqrt{s-t}dtds$ using Fubini's theorem, and apply the fundamental theorem of calculus to the continuous inner integral. The assumption $h(0)\ne 0$ implies $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$, so the extension $g(0)=0$ has no finite right derivative at zero.

The derivative of $\tan x$ is $1/{\cos }^{2}x$. The fundamental theorem of calculus gives

An antiderivative of $\sin x$ is $-\cos x$. The fundamental theorem of calculus gives

The integrand is a continuous function, so the fundamental theorem of calculus differentiates the upper limit directly:No explicit antiderivative is needed.

Apply the fundamental theorem of calculus separately to the upper and lower limits. No evaluation of the integral is needed:The first result has no $y$ dependence and the second no $x$ dependence, so the mixed partial derivative is

The chain rule identifies an antiderivative directly:The upper limit in the PDF is $\sqrt{\pi /2}$. By the fundamental theorem of calculus, the required integral is

The fundamental theorem of calculus states that integrating a continuous function from a fixed lower endpoint produces an antiderivative. More precisely, if $f$ is continuous at the interior point $x$, thenIndeed, the difference quotient is the average of $f$ over the short interval from $x$ to $x+h$, which tends to $f(x)$ by continuity. The accompanying evaluation rule is ${\int }_a^{b}f=F(b)-F(a)$ whenever $F^{\prime }=f$ on the interval.

The sum is finite, so differentiation passes through it without a convergence issue. Each lower limit $n$ is constant with respect to $x$, and the fundamental theorem of calculus gives the same upper-end integrand. Hence the derivative isThe binomial theorem at $1+1$ gives the sum $2^N$. Therefore

Put $L=5\pi /2$. The curve tangent is ${\mathbf{r}}^{\prime }=(-b\sin \theta ,b\cos \theta ,c)$. For the first line integral, the scalar triple product identity gives $(\mathbf{a}\times \mathbf{r})\cdot {\mathbf{r}}^{\prime }=\mathbf{a}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })$. Its cross product isThe needed definite integrals areand ${\int }_0^L(-\theta \sin \theta -\cos \theta )d\theta =[\theta \cos \theta -2\sin \theta {]}_0^L=-2$. HenceFor the second line integral, substitute the curve into ${\mathbf{v}}_2\cdot {\mathbf{r}}^{\prime }$. The cubic terms combine to $b^3\cos \theta$, leavingThe fundamental theorem of calculus therefore givesBoth calculations retain the full path from $\theta =0$ to $5\pi /2$, rather than inadvertently closing the helix after one turn.

On the negative half-line, an antiderivative of $1/x$ is $\ln |x|$. The fundamental theorem of calculus therefore givesThe negative sign agrees with the negative integrand throughout the interval.