Engineering Mathematics-I

Using the beta function, $\int_0^1x^2(1-x)^3\,dx=B(3,4)=\frac{\Gamma(3)\Gamma(4)}{\Gamma(7)}=\boxed{\frac{1}{60}}$.

$\int_1^{\infty}x^{-3/2}\,dx=\left[-2x^{-1/2}\right]_1^{\infty}=2$. Hence, the improper integral converges.

A sequence converges if its terms approach a finite limit; otherwise, it diverges.

The exponential series converges for every real $x$:

$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$

The logarithmic series is

$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n},\qquad -1<x\leq1.$

For period $2\pi$, the Fourier series is

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx).$

The coefficients are

$a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,dx,\quad a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos nx\,dx,\quad b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,dx.$

The half-range sine series is

$x=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx),\qquad 0<x<\pi.$

Differentiating successively gives $\displaystyle\frac{\partial^3u}{\partial x\,\partial y\,\partial z}=e^{xyz}\left(1+3xyz+x^2y^2z^2\right)$.

At $(x_0,y_0,z_0)$, where $z_0=f(x_0,y_0)$, the tangent plane is $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$.

Reversing the order gives

$I=\int_0^1\int_0^{\sqrt y}xy\,dx\,dy+\int_1^2\int_0^{2-y}xy\,dx\,dy.$

Evaluating the two integrals gives $I=\frac16+\frac5{24}=\boxed{\frac38}$.

For a positively oriented simple closed curve $C$ bounding region $R$, $\displaystyle\oint_C(P\,dx+Q\,dy)=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy$.

Let $I=\int_0^{\infty}\frac{x^2}{(1+x^4)^3}\,dx$. Using the beta-integral identity $\int_0^{\infty}\frac{x^{a-1}}{(1+x^b)^c}dx=\frac1bB\left(\frac ab,c-\frac ab\right)$, take $a=3$, $b=4$, and $c=3$.

$I=\frac14B\left(\frac34,\frac94\right)=\frac{\Gamma(3/4)\Gamma(9/4)}{4\Gamma(3)}.$

Now $\Gamma(9/4)=\frac54\cdot\frac14\Gamma(1/4)=\frac5{16}\Gamma(1/4)$ and $\Gamma(3)=2$. Therefore,

$I=\frac5{128}\Gamma(1/4)\Gamma(3/4).$

Using $\Gamma(t)\Gamma(1-t)=\frac{\pi}{\sin(\pi t)}$, we get $\Gamma(1/4)\Gamma(3/4)=\pi\sqrt2$. Hence,

$\boxed{I=\frac{5\pi\sqrt2}{128}}.$

Let $a_n=\frac1{\sqrt n+\sqrt{n+1}}$. Rationalizing,

$a_n=\frac{\sqrt{n+1}-\sqrt n}{(n+1)-n}=\sqrt{n+1}-\sqrt n.$

The $N$th partial sum telescopes:

$S_N=\sum_{n=1}^{N}(\sqrt{n+1}-\sqrt n)=\sqrt{N+1}-1.$

As $N\to\infty$, $S_N\to\infty$. Therefore, the series diverges.

$f(x)=|x|$ is even, so all sine coefficients vanish: $b_n=0$.

$a_0=\frac1\pi\int_{-\pi}^{\pi}|x|\,dx=\pi,$

$a_n=\frac2\pi\int_0^{\pi}x\cos(nx)\,dx=\frac{2}{\pi n^2}\left((-1)^n-1\right).$

Thus $a_n=0$ for even $n$ and $a_n=-\frac4{\pi n^2}$ for odd $n$. Hence,

$\boxed{|x|=\frac\pi2-\frac4\pi\left(\cos x+\frac{\cos3x}{3^2}+\frac{\cos5x}{5^2}+\cdots\right)}.$

The gradient is $\nabla\phi=(2x,-4y,8z)$. At $(1,1,-1)$, $\nabla\phi=(2,-4,-8)$. The given direction vector is $(2,1,-1)$, whose unit vector is $\frac{(2,1,-1)}{\sqrt6}$.

The directional derivative is

$D_{\mathbf u}\phi=\nabla\phi\cdot\mathbf u=\frac{4-4+8}{\sqrt6}=\boxed{\frac8{\sqrt6}}.$

The maximum derivative equals $|\nabla\phi|=\sqrt{2^2+(-4)^2+(-8)^2}=\boxed{2\sqrt{21}}$. It occurs in the gradient direction, with unit vector

$\boxed{\frac{(1,-2,-4)}{\sqrt{21}}}.$

Along $y=x$, the function is $0$ for $x\ne0$, so the limit along this path is $0$. Along $y=-x+x^3$, for $x\ne0$,

$f(x,-x+x^3)=\frac{x^3-(-x+x^3)^3}{x^3}\longrightarrow2.$

The two paths give different limits. Therefore, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist, and $f$ is not continuous at $(0,0)$.

The curves intersect at $(0,0)$ and $(4,4)$. In the bounded first-quadrant region, $0\le y\le4$ and $y^2/4\le x\le y$. Therefore,

$I=\int_0^4\int_{y^2/4}^{y}(x^2+y^2)\,dx\,dy.$

Integrating with respect to $x$,

$I=\int_0^4\left[\frac{x^3}{3}+y^2x\right]_{y^2/4}^{y}dy$

$=\int_0^4\left(\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^6}{192}\right)dy$

$=\frac{256}{3}-\frac{256}{5}-\frac{256}{21}=\boxed{\frac{768}{35}}.$

Assume the solid lies in the first octant and is bounded by the coordinate planes, $y=z$, and $z=4-x^2$. Then $0\le x\le2$, $0\le z\le4-x^2$, and $0\le y\le z$.

$I=\int_0^2\int_0^{4-x^2}\int_0^z(2x+y)\,dy\,dz\,dx.$

The inner integral is $2xz+z^2/2$, so

$I=\int_0^2\left[x(4-x^2)^2+\frac{(4-x^2)^3}{6}\right]dx.$

Evaluating the two terms gives $32/3$ and $1024/105$, respectively. Therefore,

$\boxed{I=\frac{2144}{105}}.$

For $-1<x<1$, take the Fourier series with period $2$:

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos(n\pi x)+b_n\sin(n\pi x)\right].$

Here,

$a_0=\int_{-1}^{1}(x-x^2)\,dx=-\frac23,$

so $a_0/2=-1/3$. The cosine coefficient is

$a_n=\int_{-1}^{1}(x-x^2)\cos(n\pi x)\,dx=-2\int_0^1x^2\cos(n\pi x)\,dx=\frac{4(-1)^{n+1}}{n^2\pi^2}.$

The sine coefficient is

$b_n=\int_{-1}^{1}(x-x^2)\sin(n\pi x)\,dx=2\int_0^1x\sin(n\pi x)\,dx=\frac{2(-1)^{n+1}}{n\pi}.$

Substituting the coefficients gives the required Fourier series:

$\boxed{x-x^2=-\frac13+\sum_{n=1}^{\infty}\left[\frac{4(-1)^{n+1}}{n^2\pi^2}\cos(n\pi x)+\frac{2(-1)^{n+1}}{n\pi}\sin(n\pi x)\right]}.$

The $n$th term is

$a_n=\left(\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\right)^2.$

Since $2\cdot4\cdots2n=2^n n!$ and $1\cdot3\cdots(2n-1)=\frac{(2n)!}{2^n n!}$,

$a_n=\left[\frac{(2n)!}{4^n(n!)^2}\right]^2=\left[\frac{\binom{2n}{n}}{4^n}\right]^2.$

Using the standard central-binomial-coefficient asymptotic $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, we obtain $a_n\sim\frac1{\pi n}$. Hence, by limit comparison with the harmonic series $\sum 1/n$, the given series diverges.

Since $r=\sqrt{x^2+y^2}$,

$\frac{\partial r}{\partial x}=\frac{x}{r},\qquad \frac{\partial r}{\partial y}=\frac{y}{r}.$

Using the chain rule,

$u_x=f'(r)\frac{x}{r},\qquad u_y=f'(r)\frac{y}{r}.$

Differentiating again,

$u_{xx}=f''(r)\frac{x^2}{r^2}+f'(r)\left(\frac1r-\frac{x^2}{r^3}\right),$

$u_{yy}=f''(r)\frac{y^2}{r^2}+f'(r)\left(\frac1r-\frac{y^2}{r^3}\right).$

Adding and using $x^2+y^2=r^2$,

$u_{xx}+u_{yy}=f''(r)\frac{x^2+y^2}{r^2}+f'(r)\left(\frac2r-\frac{x^2+y^2}{r^3}\right).$

Therefore, for $r>0$,

$\boxed{\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=f''(r)+\frac1r f'(r)}.$

Let $(\pm x,\pm y,\pm z)$ be the vertices of the inscribed rectangular parallelepiped, with $x,y,z>0$. Its volume is $V=8xyz$, subject to

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$

To maximize $xyz$, maximize $\ln V=\ln8+\ln x+\ln y+\ln z$. Using a Lagrange multiplier $\lambda$,

$\mathcal L=\ln x+\ln y+\ln z-\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right).$

The stationary equations give

$\frac1x=\frac{2\lambda x}{a^2},\quad \frac1y=\frac{2\lambda y}{b^2},\quad \frac1z=\frac{2\lambda z}{c^2},$

so $x^2/a^2=y^2/b^2=z^2/c^2$. Their sum is $1$, hence each equals $1/3$. Therefore, $x=a/\sqrt3$, $y=b/\sqrt3$, and $z=c/\sqrt3$. The maximum volume is

$\boxed{V_{\max}=8\left(\frac a{\sqrt3}\right)\left(\frac b{\sqrt3}\right)\left(\frac c{\sqrt3}\right)=\frac{8abc}{3\sqrt3}}.$

Stokes’ theorem states that $\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=\oint_C\mathbf F\cdot d\mathbf r$, where $C$ is the positively oriented boundary circle.

First, calculate the curl. With $P=x^2+y-4$, $Q=3xy$, and $R=2xz+z^2$,

$\nabla\times\mathbf F=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)=(0,-2z,3y-1).$

For the sphere of radius $4$, the outward unit normal is $(x,y,z)/4$. Thus

$(\nabla\times\mathbf F)\cdot\mathbf n=\frac{yz-z}{4}.$

Use spherical coordinates $x=4\sin\phi\cos\theta$, $y=4\sin\phi\sin\theta$, $z=4\cos\phi$, with $0\le\phi\le\pi/2$, $0\le\theta\le2\pi$, and $dS=16\sin\phi\,d\phi\,d\theta$. Then

$\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=-16\pi,$

because the term containing $yz$ integrates to zero by symmetry and $\iint_S z\,dS=64\pi$.

Now evaluate the line integral along the boundary circle $C$: $x=4\cos t$, $y=4\sin t$, $z=0$, for $0\le t\le2\pi$. Then $dx=-4\sin t\,dt$ and $dy=4\cos t\,dt$. Since $dz=0$,

$\oint_C\mathbf F\cdot d\mathbf r=\int_0^{2\pi}\left(128\cos^2t\sin t-16\sin^2t+16\sin t\right)dt=-16\pi.$

Both integrals are equal: $\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=\oint_C\mathbf F\cdot d\mathbf r=-16\pi$. Therefore, Stokes’ theorem is verified.