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The fundamental theorem of calculus, using continuity of $f$ near $x$, givesThe lower endpoint $a$ is constant, so it contributes no additional term.

The fundamental theorem of calculus gives $y^{\prime }(x)=e^{-x^2}>0$, so the graph is strictly increasing. Its second derivative is $y^{\prime \prime }(x)=-2x{e}^{-x^2}$: it curves upward for $x<0$ and downward for $x>0$, with an inflection point at zero.

By the Gaussian integral and the evenness of its integrand,Thus the only axis intersection is $(0,\sqrt{\pi }/2)$; there is no finite $x$-axis intersection because the integral is strictly positive for every finite $x$. The error function gives $y(x)=\frac{\sqrt{\pi }}{2}(1+\mathrm{erf}x)$.

For $x>-1$, interpret the endpoint $-1$ by a one-sided limit. Part (iii) shows that $g$ extends continuously there with value zero, so both nested integrals converge. Let $F(x)$ denote the nested expression inside the derivative. The fundamental theorem of calculus, applied twice with moving lower endpoints, givesBy part (iii), $g^{\prime }(x)=1$, henceA direct check is ${\int }_s^{-1}(t+1)dt=-(s+1{)}^2/2$, so $F(x)=(x+1{)}^3/6$, with the same third derivative.

The fundamental theorem of calculus applies because the integrand is continuous. ThusThe requested derivative is $dI/dx$, as shown in the PDF; the local OCR corrupts its notation.

Use the product rule and the fundamental theorem of calculus:Since the integral is $1-\cos x$, this simplifies to

Use the two pieces of the surface integral directly; this avoids introducing an unnecessary second derivative of $f$. On the unit sphere, $\mathbf{n}=\mathbf{r}$ and $\mathbf{F}\cdot \mathbf{n}=f^{\prime }(z)$. A spherical band between heights $z$ and $z+dz$ has area $2\pi dz$, so the curved flux isby the fundamental theorem of calculus. On the horizontal top disk, $\mathbf{F}\cdot \mathbf{n}=z_0{f}^{\prime }(z_0)$ and the area is $\pi (1-z_0^2)$. ThusAs usual for these surface integrals, the first derivative is assumed sufficiently integrable for the displayed fundamental theorem of calculus evaluation; a continuous first derivative suffices. Mere pointwise differentiability without such an integrability assumption need not make an ordinary flux integral exist.

At $t=0$, the initial conditions implyDifferentiate the first relation and solve for the two derivatives:Fix a reference point $x_{\ast }$ and set $V(x)={\int }_{x_{\ast }}^{x}v(s)ds$. Taking antiderivatives givesThe constants must be opposite to preserve the first initial condition; the arbitrary constant of integration $K$ does not change $\psi$. Substitution yields D'Alembert's formula:The fundamental theorem of calculus verifies that its time derivative at zero is $v(x)$, and its value at zero is $u(x)$. The oriented integral also makes the formula valid for negative $t$.

Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

Integrate the ordinary differential equation across one step, using the fundamental theorem of calculus:The midpoint quadrature rule approximates the integral by $hf(x_n+h/2,y(x_n+h/2))$. Approximate the unknown midpoint state by the midpoint value of linear interpolation between the endpoint states:Combining these approximations gives the implicit midpoint ruleIt is an implicit time-stepping method because its new state occurs inside the evaluation of $f$. An equivalent derivation uses a straight-line approximation on the step and imposes the ordinary differential equation at its midpoint, a one-stage collocation method.

Choose the bounded, nonnegative nondecreasing function $\tau$ from part (d), and defineThe function $h$ is convex and nondecreasing. For $0\le a\le 1$, the triangle inequality, monotonicity, and convexity giveHence $U$ is a convex function. By the gradient of a radial function formula and the fundamental theorem of calculus,for $x\ne 0$ whenever $h^{\prime }(|x|)=\tau (|x|)$. A monotone function has at most countably many discontinuities, so part (d) makes the exceptional radii $\tilde{\mu }$-null. The origin is also $\mu$-null. Define $\mathcal{T}(0)=0$; thus $\mathcal{T}=\nabla U$ $\mu$-almost everywhere.

For every $Q$ in the orthogonal group, $\mathcal{T}(Qx)=Q\mathcal{T}(x)$, including at zero. Consequently ${\mathcal{T}}_{\#}\mu$ is invariant under the orthogonal group. Moreover $|\mathcal{T}(x)|=\tau (|x|)$ away from the null origin, and composition of pushforward measures givesPart (c) therefore gives ${\mathcal{T}}_{\#}\mu =\nu$. The Knott–Smith optimality criterion makes a transport map given by a convex function's gradient optimal for the quadratic transport cost. Uniqueness from Brenier's theorem yieldsSince directions agree, this also identifies the optimal value:

Fix $x$ and put $g={\nabla }_{\mathcal{B}}f(x)$, the block coordinate gradient. For $t\ge 0$, define the one-variable function $q(t)=f(x-tg)$. By the chain rule,The last equality holds because the coordinates of $g$ outside $\mathcal{B}$ are zero. The fundamental theorem of calculus and the Cauchy-Schwarz inequality therefore giveFor $\alpha \ge 0$, the supplied condition (6), applied at the fixed starting point $x$ with parameter $t$, bounds the integrand by $L_{\mathcal{B}}t\Vert g{\Vert }^2$. Integration yieldsThis argument needs only the given directional Lipschitz continuity and continuous differentiability; convexity is not needed for this part. If the block coordinate gradient is zero, the inequality is equality.

Use the nonnegative increasing truncations $f_n=x^{-2}{1}_{[1/n,1]}$. They increase almost everywhere to the specified function. On each compact interval the continuous function has equal Riemann and Lebesgue integrals, so the fundamental theorem of calculus givesMonotone convergence yields $\int f=+\infty$. Since a nonnegative function is Lebesgue integrable exactly when its integral is finite, this function is not Lebesgue integrable.

A dominated differentiation theorem is the following. Suppose $\varphi (x,t)$ is measurable in $x$ and differentiable in $t$ on an open interval, is integrable for at least one parameter value, and its parameter derivative is dominated in absolute value by an integrable function independent of $t$. Then its integral is differentiable andLocal versions require the domination only on a neighbourhood of each parameter. The proof bounds the difference quotients by the dominating function, using the fundamental theorem of calculus or the mean-value theorem, then applies dominated convergence. Apply the argument to real and imaginary parts for a complex integrand.

For fixed $t$, $f(x)e^{itx}$ is measurable and has absolute value $|f(x)|$, so is integrable. Its derivative with respect to $t$ is $ixf(x)e^{itx}$, whose absolute value is the assumed integrable function $|xf(x)|$. HenceDominated convergence also shows this derivative is continuous, though only differentiability is required.

Use $u=x-y$, $v=x+y$. It sends the diamond-shaped open square onto $(-1,1{)}^2$, and the inverse Jacobian has absolute determinant $1/2$. On this rectangle $|u|<1$ and $v+2>1$, so the original function is measurable, bounded by one and supported on a finite-area set, proving integrability. The transformed integral separates:The two one-dimensional integrals follow from the fundamental theorem of calculus; Fubini's theorem applies by the established integrability.

A naive absolute bound on the original derivative would be $1/x$, which is not integrable on $[1,\infty )$. Instead make the change of variable $s=tx$, valid for $t>0$:The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the fundamental theorem of calculus givesThe product rule now yieldsThis proves differentiability and the equation without an unjustified differentiation of a conditionally convergent derivative integral.

Similarly set $g_n(x)=x^{1/2}{1}_{[1/n,1]}(x)$. These nonnegative measurable functions increase to $g$ almost everywhere. The fundamental theorem of calculus givesBy monotone convergence,The integral is finite and $g\ge 0$, so $g$ is Lebesgue integrable. Identifying Lebesgue and ordinary integrals of continuous functions on compact intervals justifies each truncated calculation.

Use the increasing nonnegative measurable functions $f_n(x)=x^{-2}{1}_{[1/n,1]}(x)$. Their pointwise limit is the given function except at irrelevant endpoint choices, and the fundamental theorem of calculus on $[1/n,1]$ givesmonotone convergence therefore gives $\int f=\infty$. Since $f\ge 0$, this is also $\int |f|$, and $f$ is not Lebesgue integrable. Measurability alone does not make a finite integral.

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

By the fundamental theorem of calculus,