Calculus

\begin{aligned} I & =\iint_{R} x y d x d y \\ & =\int_{y=0}^{1} \int_{x=-y}^{y} x y d x d y+\int_{y=1}^{2} \int_{x=y-2}^{2-y} x y d x d y \\ & =\int_{0}^{1} y\left(\frac{x^{2}}{2}\right)_{-y}^{y} d y+\int_{1}^{2} y\left(\frac{x^{2}}{2}\right)_{y-2}^{2-y} d y \\ & =0+0=0 \end{aligned}

\int_{P}^{Q} z^{2} d x+3 y^{2} d y+2 x z d z along the line joining the points P(1,1,2) and Q(2,3,1) is
\begin{aligned} & =\int_{P(1,2)}^{P(2,1)} z^{2} d x+2 x y d z+\int_{y=1}^{3} 3 y^{2} d y \\ & =\left(x z^{2}\right)_{(1,2)}^{(2,1)}+\left(y^{3}\right)_{1}^{3} \\ & =\left(2 \times 1^{2}-1 \times 2^{2}\right)+\left(3^{3}-1^{3}\right) \\ & =-2+26=24 \end{aligned}

I=\oint_{c} \frac{z+2}{z^{2}+2 z+2} d z ; \quad c=\left|z+1-\frac{3}{2} i\right|=1

Poles are given (z+1)^{2}+1=0
\begin{aligned} & z+1= \pm \sqrt{-1} \\ & z=-1+j,-1-j \end{aligned}
where -1-i lies outside ' c '
z=(-1,1) \text { lies inside } 'c'.

by \mathrm{CRT}
\begin{aligned} \oint_{c} f(z) d z & =2 \pi i \operatorname{Res}(f(z), z=-1+j) \\ & =2 \pi i\left(\frac{z+2}{2(z+1)}\right)_{z=-1+i}\\ & =2 \pi i\left(\frac{-1+j+2}{2(-1+j+1)}\right) \\ & =\pi(1+j) \end{aligned}

\begin{aligned} I&=\int_{0}^{4}\int_{-x}^{x}(3x^2+3y^2)dydx\\ &=\int_{0}^{4}\left [ 3x^y+\frac{3y^3}{3} \right ]_{-x}^xdx\\ &=\int_{0}^{4}[3x^2(2x)+2x^3]dx\\ &=\int_{0}^{4}8x^3 dx\\ &=2 \times 4^4\\ &=512 \end{aligned}

f(x)=8 \log _e x-x^2+3 when x \in [1,e]
Differentiating both side,
\begin{aligned} f(x)&=\frac{8}{x}-2x=0 \; where\; x \gt 0\\ f'(x)&=0\\ \frac{8}{x}-2x&=0\\ 8-2x^2&=0\\ x^2&=4\\ x&=\pm 2 \end{aligned}
\begin{aligned} f''(x)&=\frac{-8}{x^2}-2 \\ f''(2)&=-6 \lt 0\\ \end{aligned}
f(x) is maximum for x = 2
Minimum of f(x) will be in [1, e] = min [f(1), f(e)]
f(e)=8 \ln e -e^2+3=3.61
Hence, minimum value of f(x) occurs at x=1