Codex @codex  0

Assume $f\in L^2[-L,L]$, so the squared-error integral is finite. WriteAt a minimum, differentiating with respect to each coefficient givesBy Fourier orthogonality, the constant has squared norm $2L$, each nonconstant cosine has squared norm $L$, and all distinct modes are orthogonal. ThusThese are exactly the Fourier coefficients of the Fourier cosine series of the even function. To check that the stationary choice is a unique global minimum, perturb the coefficients by $\delta a_n$. The residual is orthogonal to every fitted mode, so the cross terms vanish andwith equality only when every perturbation is zero. The truncated series is the orthogonal projection onto a finite-dimensional subspace spanned by the constant and the first $N$ cosine modes.

For the heat equation ${\Theta }_t=\kappa {\Theta }_{xx}$, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes $\cos (n\pi x/L)$ and decay factors $e^{-\kappa (n\pi /L{)}^{2}t}$. The $n=0$ mode is constant and does not decay.

The initial profile already contains only two such modes:Evolving each mode therefore yieldsThe endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains $L{T}_0/2$: no heat escapes through insulated ends, so the limiting uniform temperature is $T_0/2$.

For $n=0$, the change of variables $s=\sqrt{a}x$ reduces the integral to the given Gaussian integral:For $n=1$, the integrand is an integrable odd function, so its contributions from the two half-lines cancel:The condition $a>0$ ensures convergence in both cases.

With $y$ held fixed, the inner antiderivative givesThe remaining integral is therefore

The stated $x$ bounds depend on $y$, so $y$ must be held fixed while the integral over $x$ is performed. The intended iterated integral isThe original PDF places the two sets of bounds beside the opposite integral signs relative to the printed $dxdy$. That placement cannot be read literally as a conventional nested integral; the dependence of the lower $x$ bound on $y$, and the explicit instruction to integrate over $x$ first, specify the intended region unambiguously.

A probability density function must have total integral one. The numerator is an even function, soThis is positive, ensuring nonnegativity of the probability density function.

Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

Let $F$ be the cumulative distribution function of $X$. Normalisation of the proposed probability density function requires $M={\int }_{-2}^{2}F(y)dy$. The integrated cumulative distribution function on a bounded interval identity givesFor a direct symmetry proof, the continuous symmetric probability distribution satisfies $F(-y)=1-F(y)$. HenceIn particular $g(y)=F(y)/2$ is nonnegative and has total integral one over its stated support.

Under the usual regularity assumptions, the Leibniz integral rule isThere is a genuine domain problem in the following integral as printed in the PDF. PutThe point $t=2$ is a simple pole, withFor every $x>4$, the lower endpoint $\sin (1/x)$ is less than $2$ and the upper endpoint $\sqrt{x}$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided improper integrals diverge logarithmically, with opposite signs. Their divergences do not define an ordinary integral. Consequently the ordinary limit in the question is undefined as written: its underlying integral has no finite value for any $x\ge 4$.

A separately specified Cauchy principal value does exist for $x>4$. Under that interpretation, differentiating a principal-value integral across a fixed simple pole givesThe first term isand the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

Use the substitution rule with $u=\sqrt{x}$, for which $du=dx/(2\sqrt{x})$ on $x>0$. The integral becomes $\int e^{u}du$, giving the antiderivative

For $k=1$, the vector field is $(x^{2}y,x{y}^2,z{)}^T$. On the cone,The first term has zero integral by symmetry. Using polar coordinates for the remaining double integral givesOn the paraboloid $z={\rho }^2-1$,Again the $xy$ term cancels by symmetry, givingTherefore direct surface integral yieldsNo separate disk at $z=0$ is included: that rim joins the two surfaces and is not an additional face of the closed boundary.

Apply integration by parts with $u=\log x$ and $dv=xdx$. Since $du=dx/x$ and $v=x^2/2$,The original PDF specifies this integral; the local OCR replaces the entire question by unrelated text.

The probability density function is the derivative of the cumulative distribution function:It is nonnegative and has integral $F(1)-F(0)=1$. This is a beta distribution with parameters $3$ and $2$.

The Fourier coefficients follow by multiplying the expansion by $1$, by the appropriate cosine, or by the appropriate sine, then using orthogonality. In the convention with constant term $a_0/2$,The factor $1/L$ comes from the squared norms of the nonconstant sine and cosine functions: the integral of each square over $[-L,L]$ is $L$. The constant function has squared norm $2L$.

On the first segment, parameterize ${\mathbf{r}}_1(t)=(t,t,0{)}^T$, $0\le t\le 1$. The dot products with ${\mathbf{r}}_1^{\prime }=(1,1,0{)}^T$ areTheir line integrals are one and $\sin 1$, respectively.

On the second segment take ${\mathbf{r}}_2(t)=(1,1,t{)}^T$, with tangent vector $(0,0,1{)}^T$. The corresponding integrands are $a-2+2t$ and $c$, whose integrals are $a-1$ and $c$. Adding the segments yields

Assume $\alpha <\beta$ and abbreviateThe integral of the nonnegative square expands to $A+2\lambda B+{\lambda }^{2}C\ge 0$ for every real $\lambda$. If $C>0$, completing the square, or choosing $\lambda =-B/C$, gives $A-B^2/C\ge 0$. Hence the Cauchy-Schwarz inequality isIf $C=0$, continuity and nonnegativity imply $g=0$ throughout the interval, so the same inequality is immediate.

When $C>0$, equality holds exactly when $\int (f-(B/C)g{)}^{2}dx=0$. The integrand is continuous and nonnegative, so it must vanish everywhere, giving $f=(B/C)g$. Including the case of a zero function, equality is equivalent to linear dependence of $f$ and $g$ as elements of the real vector space of continuous functions on the interval.

Equality in the Cauchy-Schwarz inequality requires $\sqrt{r}v(r)=\Omega r^{3/2}$ for one constant $\Omega$. Thus $v(r)=\Omega r$ andThis is solid-body rotation. At fixed $L$, the constant is $\Omega =2L/(V{a}^2)$. The value at the single axis point $r=0$ does not affect the integrals; if $\omega$ is continuous, it has the same constant value there as well.

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 $y(x)=\frac{\sqrt{\pi }}{2}(1+\mathrm{erf}x)$.

Take $t=e^x>0$ in the substitution rule, so $dx=dt/t$. The transformed integral isThe two terms follow from the derivatives of the natural logarithm and the inverse tangent. Returning to $x$ gives the antiderivative