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Here $I_0=x$ up to a constant. Substitution of $n=0$ in the integration by parts recurrence givesIndeed, its derivative is $\ln x+1-1=\ln x$.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

For integration by parts, choose $u=\ln x$ and $dv=x^{-4}dx$. Then $du=dx/x$ and $v=-1/(3x^3)$, givingHere $C$ is an arbitrary constant of integration.

Evenness makes every sine coefficient zero. For the constant coefficient,An integration by parts gives, for $n\ge 1$,The even coefficients vanish and the odd coefficients are $-4L/(n^2{\pi }^2)$. ThereforeThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem identifies the Fourier series with $f$ at every point, including the endpoints. The absolute summability of the coefficients also gives uniform convergence.

The expected value is zero, so the variance is $\mathbb{E}[X^2]$. Symmetry givesThe polynomial term contributes $8/3$. Two applications of integration by parts giveso its definite value on $[0,2]$ is $2-10e^{-2}$. Therefore

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

Let $I_n=\int e^{nx}\sin xdx$ and $J_n=\int e^{nx}\cos xdx$. Integrating the trigonometric functions and differentiating $e^{nx}$ in integration by parts gives, modulo constants,Substitution yields $(1+n^2)I_n=e^{nx}(n\sin x-\cos x)$ and $(1+n^2)J_n=e^{nx}(n\cos x+\sin x)$. ThereforeNo division by $n$ was used, so this formula also applies at $n=0$.

For $n\ge 2$, write $x^n{e}^{-x^2}=x^{n-1}(x{e}^{-x^2})$ and apply integration by parts, using the antiderivative $-e^{-x^2}/2$ for the second factor:The boundary term is zero: at zero, $n-1>0$, and at infinity Gaussian decay dominates every power. Consequently the Gaussian moment recurrence is

Separate the already finite sine terms from the even part $g(x)=x^2-\pi |x|$. Since $g$ is an even function, it has no sine coefficients. Its constant coefficient isFor $n\ge 1$, integration by parts givesTherefore the cosine Fourier coefficients areOnly even frequencies survive: $a_{2m}=1/m^2$. The sine Fourier coefficients are $b_1=1$, $b_2=-2$, with all others zero. HenceThe periodic function is continuous at zero and at its endpoint joins, and its coefficients are absolutely summable. Thus this Fourier series represents it everywhere, including those joins.

For a separated nonzero solution $T=f(x)g(y)$, substitution and division by $fg$ givesThe homogeneous Dirichlet boundary conditions impose $g(0)=g(1)=0$. For a nonzero spatial factor, integration by parts gives ${\lambda }^2{\int }_0^1{g}^{2}dy={\int }_0^1(g^{\prime }{)}^{2}dy$, so the separation value is nonnegative. The zero value gives only a trivial linear $g$ under these boundary conditions. The nontrivial separated factors, for $a\ne 0$, are thereforeFor the intended forward diffusion equation, assume $a>0$. The superposition principle gives a Fourier sine series, whose coefficients must expand the initial ramp in the interior:Consequently the standard heat-flow solution isIt satisfies both lateral boundary conditions for $x>0$, and tends to $y$ as $x\downarrow 0$ for $0<y<1$, also converging to the ramp in the L2 norm. The exponential coefficient decay justifies differentiating the series for positive $x$.

There is a genuine incompatibility in the printed data: at $(0,1)$ the lateral condition demands zero while the initial condition demands one. Thus no function satisfies every stated value at that corner. The displayed series solves the usual consistent interpretation with initial data on $0<y<1$ and lateral data for $x>0$; it cannot be continuous at the upper initial corner. This discrepancy is present in the original PDF.

The sign of $a$ is also unstated. If $a=0$, the spatial equation with two zero endpoints forces $T=0$, incompatible with the ramp. If $a<0$, the separated Fourier coefficients grow like $e^{n^2{\pi }^{2}x/|a|}/n$ and fail to be square summable for every $x>0$; these data therefore have no forward solution in the standard L2 space class. The positive-$a$ assumption is essential to the stated heat-flow interpretation.

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 function graphs intersect when $2x=4x{e}^{-x}$, so $x=0$ or $x=\ln 2$. For $0<x<\ln 2$, the exponential function satisfies $e^{-x}>1/2$, making $4x{e}^{-x}$ the upper graph. The enclosed area is thereforeThe antiderivative of $x{e}^{-x}$ follows by integration by parts.

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

For $L=1$, the even function $h(x)=x^2$ has $b_n=0$, because $x^2\sin (n\pi x)$ is an odd function. Its constant coefficient is $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. The remaining Fourier coefficients areTwo applications of integration by parts give the antiderivative, with $k=n\pi$,Its endpoint difference is $2(-1{)}^n/(n\pi {)}^2$, so $a_n=4(-1{)}^n/(n^2{\pi }^2)$. ThereforeThe periodic extension is continuous and piecewise smooth, so pointwise convergence of a piecewise smooth Fourier series gives equality, including at the endpoints.

Write $I_n=\int {\sin }^{n}xdx$. Apply integration by parts with ${\sin }^{n-1}x$ and $\sin xdx$:Using ${\cos }^{2}x=1-{\sin }^{2}x$ on the right gives $I_n=-\cos x{\sin }^{n-1}x+(n-1)(I_{n-2}-I_n)$. Rearranging provesFor $J_n=\int {\cos }^{n}xdx$, the analogous integration by parts uses ${\cos }^{n-1}x$ and $\cos xdx$, giving $J_n=\sin x{\cos }^{n-1}x+(n-1)(J_{n-2}-J_n)$. HenceAn arbitrary constant of integration absorbs the choices in the lower-order antiderivatives.

Starting from $I_0=J_0=x$, successive use of $n=2,4,6$ givesThese expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.

For integration by parts, take $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$, giving the antiderivative

The function is an odd function, so $a_0=a_n=0$. Here $L=1$, and the Fourier coefficients areWrite $k=n\pi$. One integration by parts has zero endpoint terms because $x(1-x)$ vanishes at both endpoints:A second integration by parts givesTherefore $b_n=4(1-(-1{)}^n)/(n^3{\pi }^3)$, zero for even $n$ and $8/(n^3{\pi }^3)$ for odd $n$. The Fourier series isThe coefficients are absolutely summable, and the periodic function is continuous even at the joins $x=\pm 1$, so the displayed Fourier series represents it everywhere.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

Apply integration by parts twice. Taking an antiderivative of $e^{x/2}$ to be $2e^{x/2}$ givesand the second application gives $\int x{e}^{x/2}dx=2x{e}^{x/2}-4e^{x/2}+C$. Therefore the required definite integral is

Here $a=0$, so the probability density function is the even function $\cos x$ on $[-h,h]$, where $h=\pi /6$. Hence $\mathbb{E}[X]=0$ by oddness, and the variance is just $\mathbb{E}[X^2]$. Two applications of integration by parts giveThus