Direct differentiation gives $\boxed{\Phi =x^3{y}^{2}z+C}$ with $\nabla \Phi =\mathbf{G}$, so $\mathbf{G}$ is a conservative vector field and its integral depends only on the endpoints. In contrast,is nonzero, for example at $(1,1,1)$. Thus $\mathbf{F}$ is not conservative. Both earlier paths lie in the surface $y=z$, where the two fields happen to coincide. This illustrates why coinciding path integrals need not imply a conservative field.

For $\mathbf{r}(t)=(t,t^2,t^2)$, the velocity is $(1,2t,2t)$. Because $y=z$ throughout this path, both fields again agree pointwise, now as $(3t^8,2t^7,t^7)$. ThusThese equal values alone do not establish path independence for either field on all of space.

On the straight path $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, both fields restrict to $(3t^5,2t^5,t^5)$. Dotting with ${\mathbf{r}}^{\prime }=(1,1,1)$ gives

The trace is unchanged by transposition and is cyclic under a product. With $S^T=S$ and $A^T=-A$,Thus $\boxed{\mathrm{tr}(AS)=0}$. In components this is the vanishing contraction of a symmetric tensor with an antisymmetric tensor.

At $k=6$, the second row is twice the first. The independent equations are $x+3y=0$ and $4x+z=0$. Therefore the null space isIt has dimension one, consistent with matrix rank two.

For $k=1$, the required right-hand matrix has twice the first row of $A$, the same second row, and twice the third row. Left multiplication scales rows, soEquivalently, right multiply the given equation $BA=C$ by $A^{-1}$ to obtain the same answer.

The determinant is $k-6$, so for $k\ne 6$ the inverse matrix isThe specified equations use $k=4$, with right-hand side $(2,3,1{)}^T$. Multiplying by $A^{-1}{|}_{k=4}$ givesIn particular the parameter in the inverse formula must be set to four, not left arbitrary for this numerical system.

The cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

For $1<x<e$, the natural logarithm satisfies $0<\ln x<1$. Squaring therefore decreases its value, while $\ln (x^2)=2\ln x$. Pointwise comparison gives $(\ln x{)}^2<\ln x<\ln (x^2)$, soAs a check, integration by parts gives $I_1=1$, $I_2=2$ and $I_3=e-2$, consistent with this order.

Use ${\tan }^{2}x={\sec }^{2}x-1$ and the derivative of the tangent:The antiderivative applies on each interval avoiding the poles of $\tan x$.

The partial fraction decomposition is $(x-1)/[x(x+1)]=-1/x+2/(x+1)$. Since both endpoints are positive, integration gives

Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-{\mathbf{e}}_x$ or $-{\mathbf{e}}_y$, so their flux of the curl is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)dxdy$, with $x,y\ge 0$, $x+y\le 1$. Thus the direct surface integral isThis surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf{A}=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral ${\int }_0^{1}2tdt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving ${\int }_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. ThereforeBoth sides have been calculated directly, verifying the Stokes theorem with consistent orientations.