Engineering Mathematics-I

Divide numerator and denominator by $n$: $\displaystyle\lim_{n\to\infty}\frac{5-3/n}{7+8/n}=\boxed{\frac57}$.

For $|x|<1$, $\displaystyle\boxed{\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots}$. It also converges at $x=1$.

$\displaystyle a_n=\frac1\pi\int_{-\pi}^{\pi}(x+x^2)\cos nx\,dx$. The $x\cos nx$ term is odd, so its integral is zero; the $x^2$ term gives $\boxed{a_n=\frac{4(-1)^n}{n^2}}$.

$f$ is continuous at $a$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$.

If $u(x,y,z)$ is homogeneous of degree $n$, then $\boxed{x\,u_x+y\,u_y+z\,u_z=nu}$.

Using $\displaystyle\int_0^\infty x^{n-1}e^{-ax}\,dx=\Gamma(n)/a^n$, we get $\displaystyle I=\Gamma(7)/2^7=6!/128=\boxed{45/8}$.

The region is $0\le y\le1$, $0\le x\le1-y$. Hence

$I=\int_0^1\int_0^{1-y}xy\,dx\,dy=\frac12\int_0^1y(1-y)^2\,dy=\boxed{\frac1{24}}.$

The region is $0\le x\le4$, $x\le y\le2\sqrt{x}$. Reversing the inequalities gives $0\le y\le4$ and $y^2/4\le x\le y$. Therefore

$\boxed{\int_0^4\int_{y^2/4}^{y}f(x,y)\,dx\,dy}.$

$\nabla\cdot\mathbf F=2xy-4xyz+3x^2$. At $(3,-1,-2)$ this is $-6-24+27=\boxed{-3}$.

For an oriented surface $S$ with boundary $C$, $\displaystyle\oint_C\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS$, with consistent orientations.

By definition, $B(m,n)=\int_0^1t^{m-1}(1-t)^{n-1}\,dt$. Also

$\Gamma(m)\Gamma(n)=\int_0^\infty\int_0^\infty e^{-(x+y)}x^{m-1}y^{n-1}\,dx\,dy.$

Put $x=rt$, $y=r(1-t)$; the Jacobian is $r$, with $r>0$ and $0<t<1$. Then

$\Gamma(m)\Gamma(n)=\left(\int_0^\infty e^{-r}r^{m+n-1}\,dr\right)\left(\int_0^1t^{m-1}(1-t)^{n-1}\,dt\right)$

$=\Gamma(m+n)B(m,n).\quad\therefore\quad\boxed{B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}}.$

The $n$th term is $\displaystyle a_n=\frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots2n}=\frac{(2n)!}{4^n(n!)^2}$. By the central-binomial-coefficient asymptotic, $a_n\sim1/\sqrt{\pi n}$. Since $\sum n^{-1/2}$ diverges, the given series diverges by limit comparison.

Since $x^2$ is even, $b_n=0$. Its Fourier series is

$\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n\cos nx}{n^2}}.$

Put $x=0$:

$0=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}.$

Thus $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\boxed{\frac{\pi^2}{12}}$.

Let $F=x^3+y^3+3xyz-3$. Then $\nabla F=(3x^2+3yz,3y^2+3xz,3xy)$. At $(1,2,-1)$, $\nabla F=(-3,9,6)$.

Tangent plane:

$-3(x-1)+9(y-2)+6(z+1)=0\quad\Rightarrow\quad\boxed{x-3y-2z+3=0}.$

Normal line:

$\boxed{\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z+1}{2}}.$

For an interior stationary point,

$f_x=x^2y^2(3-4x-3y)=0,\qquad f_y=x^3y(2-2x-3y)=0.$

Solving $4x+3y=3$ and $2x+3y=2$ gives $x=1/2$, $y=1/3$. Then

$f\left(\frac12,\frac13\right)=\frac18\cdot\frac19\cdot\frac16=\boxed{\frac1{432}}.$

The function vanishes on the boundary of the triangular domain $x\ge0$, $y\ge0$, $x+y\le1$, so this is the maximum there.

The region $0\le y\le x\le1$ gives $0\le\theta\le\pi/4$, $0\le r\le\sec\theta$. Since $x=r\cos\theta$ and $dx\,dy=r\,dr\,d\theta$,

$I=\int_0^{\pi/4}\int_0^{\sec\theta}r^3\cos^3\theta\,dr\,d\theta=\frac14\int_0^{\pi/4}\sec\theta\,d\theta$

$=\boxed{\frac14\ln(1+\sqrt2)}.$

Write $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$. Expanding the determinant for $\mathbf a\times\mathbf b$ and differentiating its components gives

$\nabla\cdot(\mathbf a\times\mathbf b)=b_1(\partial_y a_3-\partial_z a_2)+b_2(\partial_z a_1-\partial_x a_3)+b_3(\partial_x a_2-\partial_y a_1)$

$\quad-a_1(\partial_y b_3-\partial_z b_2)-a_2(\partial_z b_1-\partial_x b_3)-a_3(\partial_x b_2-\partial_y b_1).$

The first three terms are $\mathbf b\cdot(\nabla\times\mathbf a)$ and the last three are $-\mathbf a\cdot(\nabla\times\mathbf b)$. Hence proved.

In the first quadrant, $y^2=(a^{2/3}-x^{2/3})^3$. Revolving the upper half about the $x$-axis gives

$V=2\pi\int_0^a(a^{2/3}-x^{2/3})^3\,dx.$

Put $x=at^3$, so $dx=3at^2dt$:

$V=6\pi a^3\int_0^1t^2(1-t^2)^3\,dt=6\pi a^3\left(\frac{16}{315}\right)=\boxed{\frac{32\pi a^3}{105}}.$

Factor

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$

Let $D=\partial_x+\partial_y+\partial_z$. The second factor has $D$-derivative zero, while $D(x+y+z)=3$. Hence

$Du=D\ln(x+y+z)=\frac3{x+y+z}.$

Applying $D$ again gives

$\boxed{D^2u=-\frac9{(x+y+z)^2}}.$

For $0<x<1$, the half-range cosine coefficients are $a_0=0$ and

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

Thus only odd cosine terms remain:

$\boxed{2x-1=-\frac8{\pi^2}\left(\cos\pi x+\frac{\cos3\pi x}{3^2}+\frac{\cos5\pi x}{5^2}+\cdots\right)}.$

At $x=0$, the series equals $-1$, so $-1=-\frac8{\pi^2}\sum_{k=0}^\infty(2k+1)^{-2}$, giving the result.

The intercepts are $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$. The tetrahedron's volume is one-sixth the product of its three intercepts:

$\boxed{V=\frac{abc}{6}}.$

Gauss's theorem states $\iint_S\mathbf F\cdot\mathbf n\,dS=\iiint_V\nabla\cdot\mathbf F\,dV$ for a closed surface with outward normal.

Here $\nabla\cdot\mathbf F=y+0+2y=3y$. Over the simplex $x,y,z\ge0$, $x+y+z\le1$,

$\iiint_V3y\,dV=3\left(\frac1{24}\right)=\frac18,$

using the simplex volume $1/6$ and centroid coordinate $\bar y=1/4$.

Surface flux: On $x=0$ and $z=0$, the flux is zero. On $y=0$, the outward flux is $-\int_0^1z^2(1-z)\,dz=-1/12$. On $x+y+z=1$, the outward flux is

$\int_0^1\int_0^{1-x}[xy+z^2+2yz]_{z=1-x-y}\,dy\,dx=\frac5{24}.$

Total outward flux is $5/24-1/12=\boxed{1/8}$, equal to the volume integral. Gauss's theorem is verified.