Calculus

Divergence of \vec{V}, \triangledown \cdot \vec{V}=4x+5+3=4x+8
Now,
\int \int \int( \triangledown \cdot \vec{V})dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4x+8)dxdydz=\int_{0}^{1}(4x+8)dx=[2x^2+8x]_0^1=10

The function is decreasing, if f'(x) \lt 0
f'(x)=\frac{d}{dx} \int_{0}^{x}e^t(t-1)(t-2)dt
\Rightarrow \; e^x(x-1)(x-2) \lt 0
It is possible in between 1 & 2. Hence x \in (1,2)

\begin{aligned} x^{4}+2 &=4 x^{3} \\ f_{1}(x) &=x^{4}+2 \\ f_{2}(x) &=4 x^{3} \end{aligned}
It is clear that point of intersection of these graphs is solution (or) root of p(x) = 0


According to intermediate value theorem
P(0) and P(1) are having opposite signs
\therefore a root of p(x) = 0 in (0, 1)
and also from graph, there is only one point of intersection
Hence exactly one real root exists in (0, 1)

If two curves cut orthogonally then product of slopes = -1
\begin{aligned} x^{2}+y^{2} &=1 \\ 2 x+2 y \frac{d y}{d x} &=0 \\ \left(M_{1}\right)_{u, v} &=\frac{d y}{d x}=\frac{-x}{y}=\frac{-u}{v} \\ (x-1)^{2}+(y-1)^{2} &=r^{2} \\ \left(M_{2}\right)_{u, v} &=\left(\frac{d y}{d x}\right)=\left(\frac{-2(x-1)}{2(y-1)}\right)_{(u . v)}=\frac{1-u}{v-1} \\ \because \qquad \qquad M_{1} M_{2} &=-1\\ \frac{-u}{v} \times \frac{1-u}{v-1}&=-1\\ -u+u^{2} &=-v^{2}+v \\ u^{2}+v^{2} &=v+u \\ \because \qquad \qquad x^{2}+y^{2} &=1 \\ \therefore \qquad \qquad u+v &=1 \end{aligned}

x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)

\begin{aligned} \vec{F}&=y\hat{a}_{x}-x\hat{a}_{y} \\ \vec{r}&=x\hat{i}+y\hat{j} \\ \vec{F}&=y\hat{i}-x\hat{j} \\ d\vec{r}&=dx\hat{i}+dy\hat{j} \\ &=\int _{c}F.d\vec{r}\\ &=\int_{c}F_{1}dx+F_{2}dy \\ &=\int _{c}ydx-xdy \\ &\text{Where C is,}\\ y&=x^{2} \\ dy&=2x dx \\ x &\text{ varies from -1 to 2,}\\ \int _{c}\vec{F}dl&=\int_{-1}^{2}x^{2}dx-x\cdot 2xdx \\ &=\int_{-1}^{2}(x^{2}-2x)dx \\ &=\int_{-1}^{2}-x^{2}dx=\left.\begin{matrix} -\frac{x^{3}}{3} \end{matrix}\right|_{-1}^{2} \\ &=-\frac{8}{3}-\frac{1}{3}=-\frac{9}{3}\\ &=-3 \end{aligned}

\begin{aligned} \bigtriangledown \cdot (uA)&=u(\bigtriangledown \cdot A)+(\bigtriangledown A)F\\ &=(x^2+y^2+z^2)[2+3+4]\\ &+(2x\hat{i}+2y\hat{j}+2z\hat{k})(2x\hat{i}+3y\hat{j}+4z\hat{k})\\ &=9(x^2+y^2+z^2)+(4x^2+6y^2+8z^2)\\ At\; (1,1,1)&=9(3)+[4+6+8]\\ &=27+18=45 \end{aligned}

\begin{aligned} f&=2x^3+3y^2+4z\\ \Delta f&=6x^2\hat{i}+6y\hat{j}+4\hat{k}\\ \text{curl} (\Delta f)&=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 6x^2& 6y & 4\end{vmatrix}\\ \therefore \;\;\int _c\Delta f\cdot dr&=\int _c d(2x^3+3y^2+4z)=0\\ &=\int_{(-3,-3,2)}^{2,-3,2}d(2x^3+3y^2+4z)\\ &+\int_{(2,-3,2)}^{(2,6,2)}d(2x^3+3y^2+4z)\\ &+\int_{(2,6,2)}^{(2,6,-1)}d(2x^3+3y^2+4z)\\ &=70+81+(-12)=139 \end{aligned}

\begin{aligned} f(x)&=3x^3-7x^2+5x+6\\ f'(x)&=9x^2-14x+5\\ f''(x)&=18x-14\\ f'(x)&=0\\ 9x^2-14x+5&=0\\ x&=1,0.55\\ x&=1\\ f''(1)&=18-14=4 \gt 0\; \text{minima}\\ x&=0.55\\ f''(0.55)&=-4.1 \lt 0 \; \text{maxima} \end{aligned}
Maximum [f(0), f'(0.55), f(2)]
Maximum [6,7.13,12]=12

\begin{aligned} x^2+y^2&=9\\ x&=3 \cos \theta \\ y&=3 \sin \theta \\ dx&=-3 \sin \theta \; d\theta\\ dy&=3 \cos \theta \;d\theta \end{aligned}
\theta varies from 0 to \frac{\pi}{2}
\int (y^2+2xy)dx+(2xy+x^2)dy =\int_{0}^{\pi/2}(9 \sin ^2 \theta +18 \sin \theta \cos \theta )(-3 \sin \theta d\theta )+(18 \sin \theta \cos \theta +9 \cos ^2 \theta )(3 \cos \theta )d\theta
=\int_{0}^{\pi/2}(-27\sin ^3 \theta -54 \sin ^2 \theta \cos \theta +54 \sin \theta \cos ^2 \theta +27 \cos ^3 \theta )d\theta =0