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The first field has vanishing curl and is the gradient of a monomial:For the second field, direct component differentiation givesIt is not identically zero. Thus ${\mathbf{F}}_1$ is conservative and ${\mathbf{F}}_2$ is not on ${\mathbb{R}}^3$. The conditional path-integral requests therefore apply to ${\mathbf{F}}_1$.

Factor out $x^2$ before expanding the logarithm:for large positive $x$. Using $\log (1+u)=u-u^2/2+O(u^3)$ then givesThese are the first three nonzero terms, including the leading logarithm. Replacing $\log x$ by $\log |x|$ yields the corresponding two-sided large-$|x|$ expansion.

Put $q(x)=\sqrt{{\pi }^2/16+x}$. The original PDF confirms that the $+x$ is inside the radical. For the composite cosine,At $x=0$, $q=\pi /4$ and $\sin q=\cos q=\sqrt{2}/2$. Substituting into $f(0)+f^{\prime }(0)x+f^{\prime \prime }(0)x^2/2$ givesAll three displayed coefficients are nonzero.

The generalized binomial series givesIn particular,The recurrence $u_{n+1}=u_n(1/3-n)/(n+1)$ follows directly from the coefficient product. The power series has radius of convergence one and represents the real cube root for $|x|<1$.

With sufficiently many derivatives, Taylor's theorem gives a finite approximation and a controlled remainder. For example, if $f$ is $N+1$ times continuously differentiable between $a$ and $a+\delta$, thenThus derivatives at the nearby base point determine the polynomial approximation; a bound on the next derivative bounds its error. The smallness of $\delta$ makes this useful, provided the derivatives remain controlled.

The statement's bare assumption of differentiability alone guarantees only $f(a+\delta )=f(a)+f^{\prime }(a)\delta +o(\delta )$. Higher-order formulas need higher regularity. Even smoothness does not by itself guarantee equality with an infinite Taylor series; that stronger conclusion requires the remainder to tend to zero.

Here $M_y-N_x=(6xy+2)-(4xy+1)=2xy+1$, whereas $N=x(2xy+1)$. Thus the ratio in (b) is $1/x$, and one may choose the nonzero integrating factor $\boxed{\mu (x)=x}$ on either half-line. A negative constant multiple on a half-line is harmless.

The multiplied form isHence $x^3{y}^2+x^{2}y=C$, a quadratic equation for $y$. Solving it givesChoose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels $2xy+1$, the displayed exact differential verifies the multiplying factor directly wherever $x\ne 0$. At $C=0$ the branches are $y=0$ and $y=-1/x$. Only $y=0$ has a finite smooth extension across $x=0$, where the original equation requires $y=0$.

Exactness after multiplication requires ${\partial }_y(\mu M)={\partial }_x(\mu N)$. Since $\mu$ depends only on $x$,For $\mu =\exp (\int f(x)dx)$, the ratio ${\mu }^{\prime }/\mu$ is $f(x)$, soThe assumed existence means this ratio is a function only of $x$ wherever the division is valid. For a factor depending only on $y$, the corresponding equation is ${\psi }^{\prime }M+\psi M_y=\psi N_x$, givingThese are the integrating factor for a differential one-form formulas. Nonzero constant multiples are equivalent. Exactness follows locally from closedness, and globally on a suitable simply connected domain.

Writing the two coefficients as $M,N$, their mixed derivatives satisfy $M_y=2xy=N_x$, so the smooth form is exact on ${\mathbb{R}}^2$. Integrating the first coefficient with respect to $x$ gives $f=2x^2+x^2{y}^2/2+h(y)$. Matching $f_y=y+x^{2}y$ gives $h^{\prime }(y)=y$. One potential is thereforeThe integral curves of $df=0$ are its level curves. Absorbing a factor of two into the constant giveson real graph branches. The initial condition fixes $C=12$ and selects the positive branch:This is the maximal smooth graph interval containing $x=1$; at either endpoint the level curve has a vertical tangent, so $y^{\prime }$ cannot remain finite.

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

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 argument $3x^3+2x$ is odd, so its squared sine is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because $|x|\le 1/\pi <1$. Therefore