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Evaluate the given integral$鈭�xydx+xdy$.

A line integral is integral in which the function to be integrated is determined along a curve in thecoordinate system.In mathematics and physics, scalar field or scalar-valued function referred to a scalarvalueto everypointin aspace鈥� possiblyphysical space. The scalar may either be a (dimensionless) mathematical number or aphysical quantity.

$$\begin{gathered} a)Thepathisfrom\left(0,0\right)to\left(1,2\right). \\ Findtheequationoflinefrompoint\left(0,0\right)to\left(1,2\right). \\ \frac{y-0}{x-0}=\frac{2-0}{1-0} \\ y=2x \\ Substituteintheintegralwithy=2x. \\ Itsderivativeisdy=2dx. \\ {I}_1=鈭�xydx+xdy \\ ={鈭�}_0^{1}x\left(2x\right)dx+2xdx \\ ={\left[\frac{2x^3}{3}\right]}_0^1 \\ =\frac{5}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{b})\mathrm{The}\mathrm{path}\mathrm{is}\mathrm{from}\left(0,0\right)\mathrm{to}\left(0,2\right). \\ \mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{from}\mathrm{point}\left(0,0\right)\mathrm{to}\left(0,2\right). \\ \mathrm{x}=0 \\ \mathrm{Substitute}\mathrm{in}\mathrm{the}\mathrm{integral}\mathrm{with}\mathrm{x}=0. \\ \mathrm{Its}\mathrm{derivative}\mathrm{is}\mathrm{dx}=0. \\ {\mathrm{I}}_{\mathrm{i}}=鈭�\mathrm{xydx}+\mathrm{xdy} \\ ={鈭�}_0^{2}0\mathrm{dx} \\ =0 \\ \mathrm{The}\mathrm{path}\mathrm{is}\mathrm{from}\left(0,2\right)\mathrm{to}\left(1,2\right). \\ \mathrm{Substitute}\mathrm{in}\mathrm{the}\mathrm{integral}\mathrm{with}\mathrm{y}=2. \\ \mathrm{Its}\mathrm{derivative}\mathrm{is}\mathrm{dy}=0. \\ {\mathrm{I}}_{\mathrm{ii}}=鈭�\mathrm{xydx}+\mathrm{xdy} \\ ={鈭�}_0^{1}2\mathrm{xdx} \\ \mathrm{Simplify}\mathrm{further}. \\ {\mathrm{I}}_2={\mathrm{I}}_{\mathrm{i}}+{\mathrm{I}}_{\mathrm{ii}} \\ =0+1 \\ =1 \\ \mathrm{The}\mathrm{path}\mathrm{is}\mathrm{from}\left(0,0\right)\mathrm{to}\left(1,2\right). \\ \mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{from}\mathrm{point}(0,0)\mathrm{to}(1,2). \\ \frac{\mathrm{y}-0}{\mathrm{x}-0}=\frac{2-0}{1-0} \\ \mathrm{y}=2\mathrm{x} \\ \mathrm{Substitute}\mathrm{in}\mathrm{the}\mathrm{integral}\mathrm{with}\mathrm{y}=2\mathrm{x}. \\ \mathrm{Its}\mathrm{derivative}\mathrm{is}\mathrm{dy}=2\mathrm{dx}. \\ {I}_1=鈭�xydx+xdy \\ ={鈭�}_0^{1}x\left(2x\right)dx+2xdx \\ ={\left[\frac{2x^3}{3}\right]}_0^1 \\ =\frac{5}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{c})\mathrm{The}\mathrm{path}\mathrm{is}\mathrm{from}(0,0)\mathrm{to}(3,0). \\ \mathrm{Substitute}\mathrm{in}\mathrm{the}\mathrm{integral}\mathrm{with}\mathrm{y}=0. \\ \mathrm{Its}\mathrm{derivative}\mathrm{is}\mathrm{dy}=0. \\ {\mathrm{I}}_{\mathrm{i}}=鈭�\mathrm{xydx}+\mathrm{xdy} \\ ={鈭�}_0^{3}0\mathrm{dx} \\ =0 \\ \mathrm{The}\mathrm{path}\mathrm{is}\mathrm{from}\left(3,0\right)\mathrm{to}\left(1,2\right) \\ \mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{from}\mathrm{point}\left(3,0\right)\mathrm{to}\left(1,2\right). \\ \frac{\mathrm{y}-0}{\mathrm{x}-3}=\frac{2-0}{1-3} \\ \mathrm{x}=-\mathrm{y}+3 \\ \mathrm{Substitute}\mathrm{in}\mathrm{the}\mathrm{integral}\mathrm{with}\mathrm{x}=-\mathrm{y}+3. \\ \mathrm{Its}\mathrm{derivative}\mathrm{is}\mathrm{dx}=-\mathrm{dy}. \\ {\mathrm{I}}_{\mathrm{ii}}=鈭�\mathrm{xydx}+\mathrm{xdy} \\ ={鈭�}_3^1\mathrm{x}\left(3-\mathrm{x}\right)\mathrm{dx}-\mathrm{xdx} \\ =\frac{2{\mathrm{x}}^2}{2}-{\overline{)\frac{{\mathrm{x}}^3}{3}}}_3^1 \\ =\frac{2}{3} \\ \mathrm{Simplify}\mathrm{further}. \\ {\mathrm{I}}_3={\mathrm{I}}_{\mathrm{i}}+{\mathrm{I}}_{\mathrm{ii}} \\ =0+\frac{2}{3} \\ =\frac{2}{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{this}\mathrm{problem}\mathrm{is}\mathrm{mentioned}\mathrm{below}. \\ \mathrm{a}){\mathrm{I}}_1=\frac{5}{3}, \\ \mathrm{b}){\mathrm{I}}_2=1, \\ \mathrm{c}){\mathrm{I}}_3=\frac{2}{3}. \end{gathered}$$